Zadatak 061 (Matea, gimnazija) - · PDF file6 Broj x 0 je nulto čka kvadratne funkcije f ako...

1

Zadatak 061 (Matea gimnazija) Luk mosta ima oblik parabole Visina mosta je 2 metra a duljina raspona 24 metra Napiši

jednadžbu luka mosta

Rješenje 061 Ponovimo

Polinom drugog stupnja (kvadratna funkcija) ima oblik

( )2

f x a x b x c= sdot + sdot +

gdje su a ne 0 b i c realni brojevi Broj a naziva se vodeći koeficijent polinoma drugog stupnja b

linearni koeficijent a c slobodni koeficijent polinoma

Nultočka grafa je točka u kojoj graf siječe os apscisa (y = 0) Vrijednost x za koju je f(x) = 0 zove se

nulište funkcije Najčešće se za oba pojma rabi izraz nultočka

Graf kvadratne funkcije

( ) ( )2

0 0f x a x x y= sdot minus +

parabola je

( )2

0 0y a x x y= sdot minus +

s tjemenom u točki T(x0 y0) dobivena translacijom parabole y = a middot x2 U točki x0 funkcija f poprima

najmanju vrijednost y0 ako je a gt 0 a najveću vrijednost y0 ako je a lt 0

1inačica

O

y

x

N2(12 0)N1(- 12 0)

T(0 2)

Postavimo most u pravokutni koordinatni sustav kao na slici Visina luka mosta je ordinata tjemena

T parabole Budući da tjeme parabole ima koordinate T(0 2) jednadžba luka mosta (parabole)

je

( ) ( )

( )( )

0 20 0 2 2

0 2 22

0 0

T x y T

y a x y a x

y a x x y

=

rArr = sdot minus + rArr = sdot +

= sdot minus +

Da bismo izračunali vodeći koeficijent a parabole uvrstit ćemo koordinate jedne od nultočaka (N1 ili

N2) u jednadžbu parabole

( ) ( ) 12 02 2 2

0 12 2 0 144 2 144 22

2

N x y Na a a

y a x

=

rArr = sdot + rArr = sdot + rArr minus sdot = rArr= sdot +

( ) 1442 1

144 2 144 72

a a arArr minus sdot = rArr = minus rArr = minusminus

Tada jednadžba luka mosta (parabole) glasi

1 22

72y x= minus sdot +

2inačica

2

P(12 0)

T(12 2)

N(24 0)O(0 0)

y

x

Postavimo most u pravokutni koordinatni sustav kao na slici Tjeme parabole ima koordinate T(12 2)

pa jednadžba luka mosta glasi

( ) ( )

( )( )

12 20 0 2

12 22

0 0

T x y T

y a x

y a x x y

=

rArr = sdot minus +

= sdot minus +

Budući da graf sadrži točku O(0 0) (isto tako možemo uzeti točku N(24 0)) uvrštavanjem dobivamo

vrijednost vodećeg koeficijenta a

( ) ( )

( )( ) ( )

0 02 2

0 0 12 2 0 12 2 0 144 22

12 2

O x y O

a a a

y a x

=

rArr = sdot minus + rArr = sdot minus + rArr = sdot + rArr= sdot minus +

( )2 1

144 2 144 2 14

44 72

14a a a aminusrArr minus sdot = rArr minus sdot = rArr = minus rArr = minus

Jednadžba luka mosta glasi

( )1 2

12 272

y x= minus sdot minus +

Uočimo da se parabola

( )1 2

12 272


dobije translacijom parabole

1 22

72y x= minus sdot +

za broj x0 = 12 u pozitivnom smjeru x ndash osi

Vježba 061 Luk mosta ima oblik parabole Visina mosta je 5 m a duljina raspona 40 m Napiši


Rezultat 1 2

580

y x= minus sdot +

Zadatak 062 (Matea gimnazija) Ako su ndash 1 i 2 nultočke polinoma drugog stupnja a najveća vrijednost polinoma iznosi 3

odredi taj polinom



( )2


3



Kvadratna funkcija (polinom drugog stupnja)

( )2


ima ekstrem u točki s apscisom

0 2

bx

a= minus

sdot

Vrijednost ekstrema iznosi

2

0

4

4

a c by

a

sdot sdot minus=

sdot

Broj x0 je nultočka kvadratne funkcije f ako vrijedi

( ) 20 0

0 0

0f x a x b x c= rArr sdot + sdot + =

Kvadratna funkcija

( )2


ima dvije nultočke x1 i x2 Tada za apscisu x0 točke u kojoj funkcija ima ekstrem vrijedi

0 21 2

x xx

+=

Ordinata tjemena iznosi

( ) 20 0 0 0 0

y f x y a x b x c= rArr = sdot + sdot +

Da bi umnožak bio jednak nuli dovoljno je da jedan faktor bude jednak nuli

0 0 ili 0 il i 0a b a b a bsdot = hArr = = = =

1inačica

Budući da su zadane obje nultočke x1 = ndash 1 i x2 = 2 kvadratne funkcije lako se izračuna apscisa x0

tjemena

1 21 2

1 2 1

0 01 2 2 20 2

x x

x xx xx

= minus =minus +

rArr = rArr =+=

Točke x1 i x2 su nultočke funkcije pa vrijedi

( ) ( )0 01 2

f x f x= =

Za točku x0 vrijednost funkcije je

( ) 30

f x =

Postavimo sustav od tri jednadžbe sa tri nepoznanice

( ) ( )2

1 1 1 01

x a b c= minus rArr sdot minus + sdot minus + =

22 2 2 0

2x a b c= rArr sdot + sdot + =

21 1 1

30 2 2 2

x a b c= rArr sdot + sdot + =

Riješimo sustav jednadžbi

4

( ) ( )2

1 1 00 0

22 2 0 4 2 0 4 2 0

2 1 1 1 11 1 3 3

3 4 2 4 22 2

4

a b ca b c a b c

a b c a b c a b c

a b c a b ca b c

sdot minus + sdot minus + =minus + = minus + =

sdot + sdot + = rArr sdot + sdot + = rArr sdot + sdot + = rArr

sdot + sdot + = sdot + sdot + =sdot + sdot + =

sdot

0

4 2 0 4 2 0

2 4

metoda

supstituci12

j1

e2 4 2

a b c c b a

a b c a b c

a b c a b c

minus + = = minus

rArr sdot + sdot + = rArr sdot + sdot + = rArr rArr

+ sdot + sdot = + sdot + sdot =

( )

4 2 0 4 2 0 3 3 0

2 4 12 2 4 4 12 3 6 12

a b b a a b b a a b

a b b a a b b a a b

sdot + sdot + minus = sdot + sdot + minus = sdot + sdot =rArr rArr rArr rArr

+ sdot + sdot minus = + sdot + sdot minus sdot = minus sdot + sdot =

metoda suprotnih 9

koeficij

12 49

ena12 9 12

3a 9tb b b brArr rArr sdot = rArr sdot = rArr = rArr =

Računamo koeficijent a

3 3 04 4

3 3 0 3 0 333

4 0 3 443

3

a b

a a a ab

sdot + sdot =

rArr sdot + sdot = rArr sdot + sdot = rArr sdot + = rArr sdot = minus rArr=

34

3 4 3

a arArr sdot = minus rArr = minus

Računamo koeficijent c

4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus

rArr = minus minus rArr = + rArr == minus =

Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +

rArr = minus sdot + sdot +

= minus = =

2inačica

Budući da su zadane obje nultočke x1 = ndash 1 i x2 = 2 kvadratne funkcije vrijedi

( ) ( )2

1 1 1 0 01

x a b c a b c= minus rArr sdot minus + sdot minus + = rArr minus + =

22 2 2 0 4 2 0

2x a b c a b c= rArr sdot + sdot + = rArr sdot + sdot + =

Vrijednost ekstrema kvadratne funkcije iznosi 3 pa slijedi

2 24 4 2

3 3 4 12 4 4

4a c b a c b

a c b aa a

asdot sdot minus sdot sdot minus

= rArr = rArr sdot sdot minus = sdotsdot sdot

sdot sdot


( )

metoda

supstitucije

04 2 0

4 2 0 4 2 0 24 12

2 24 12 4 12

a b c c b aa b b a

a b c a b ca b a b a

a c b a a c b a

minus + = = minussdot + sdot + minus =

sdot + sdot + = rArr sdot + sdot + = rArr rArr rArrsdot sdot minus minus = sdot

sdot sdot minus = sdot sdot sdot minus = sdot

5

3 3 0 3 3 0

2 2 2 24 4 12 4 4

3

12 0

a b a b

a b a b a a b a b a

sdot + sdot = sdot + sdot =rArr rArr rArr

sdot sdot minus sdot minus = sdot sdot sdot minus sdot minus minus sdot =

metoda

supstituci

0

2 2 2 24 4 12 0 4 je4 12 0

a b b a

a b a b a a b a b a

+ = = minusrArr rArr rArr rArr

sdot sdot minus sdot minus minus sdot = sdot sdot minus sdot minus minus sdot =

( ) ( )22 2 2 2

4 4 12 0 4 4 12 0a a a a a a a a arArr sdot sdot minus minus sdot minus minus minus sdot = rArr minus sdot minus sdot minus minus sdot = rArr

( )2 2 2

9 12 0 9 12 0 3 4 0 3a a a a a arArr minus sdot minus sdot = rArr minus sdot minus sdot = rArr sdot + sdot =minus rArr

( )nema smisla00 413 4 0 3 4

3 4 0 33 3

4

aaa a a a

a a

==rArr sdot sdot + = rArr rArr rArr sdot = minus rArr = minus

sdot + = sdot = minus

Računamo koeficijent b

4 44

3 33

b a

b ba

= minus

rArr = minus minus rArr == minus


4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus


Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +


= minus = =

Vježba 062 Ako su ndash 2 i 2 nultočke polinoma drugog stupnja a najmanja vrijednost polinoma iznosi ndash 4 odredi taj polinom

Rezultat ( )2

4f x x= minus

Zadatak 063 (Matea gimnazija)

Odredi jednadžbu parabole koja dira os apscisa u toči s apscisom 3 a prolazi točkom (5 12)



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

6


( ) 20 0

0 0




( )2 2 2

2 a b a a b bminus = minus sdot sdot +


( ) ( )2

0f x a x x= sdot minus

je parabola

( )2

0y a x x= sdot minus

čije je tjeme (točka u kojoj funkcija poprima najmanju ili najveću vrijednost) točka

( )0 0

T x


( )2


je parabola

2y a x b x c= sdot + sdot +

Parabola


simetrična je s obzirom na os koja je paralelna s osi y i prolazi njezinim tjemenom T Os koja prolazi

tjemenom parabole paralelno s y osi zove se os simetrije parabole Jednadžba osi (pravca) glasi

0

20 2

x xb

xbax

a

=

rArr = minussdot= minus

sdot

1inačica

Budući da parabola dira os apscisa u točki s apscisom 3 ta točka je njezino tjeme i vrijedi

( ) ( ) 3 0 0 0

T x y T=

Parabola prolazi i točkom A(5 12) pa je

( ) ( )

( ) ( )

( )

( )

0 3 00

2 2 5 12 12 5 3 12 2 12 4 4 12

2

0

T x T

A x y A a a a a

y a x x

=

= rArr = sdot minus rArr = sdot rArr = sdot rArr sdot = rArr

= sdot minus

4 312 3a arArr sdot = rArr =

Jednadžba tražene parabole glasi

( ) ( )

( )( ) ( )

3 0 3 00 2 2

3 3 3 6 92

0

a T x T

y x y x x

y a x x

= =

rArr = sdot minus rArr = sdot minus sdot + rArr

= sdot minus

23 18 27y x xrArr = sdot minus sdot +

2inačica


7

( ) ( )( ) 2

4

3 00 0

3 3 32 2 2

2 2 20 2 4 4 40 0 02

4 4 44

0 4

T x y T b b b

a a abx

a a c b a c b a c b

a a aa c by

a

a

a

== minus minus = minus =

sdot sdot sdot= minus rArr rArr rArr rArr

sdot sdot sdot minus

sdot minus sdot

sdot sdotsdot sdot minus sdot sdot minus

= = =sdot sdot sdotsdot sdot minus

=sdot

6

24 0

b a

a c b

= minus sdotrArr

sdot sdot minus =


( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2

A x y Aa b c a b c a b c

y a x b x c

=rArr = sdot + sdot + rArr = sdot + sdot + rArr sdot + sdot + =

= sdot + sdot +

Riješimo sustav od tri jednadžbe sa tri nepoznanice

( )

( )

metoda

supstitucije

62

4 6 024 0

25 5 6 1225 5 12

b a

a c aa c b

a a ca b c

= minus sdot

sdot sdot minus minus sdot =sdot sdot minus = rArr rArr rArr

sdot + sdot minus sdot + =sdot + sdot + =

2 2 24 36 0 4 36 0 4 36 0

25 30 12 5 12

4

12

5

a c a a c a a c a

a a c a c c a

sdot sdot minus sdot = sdot sdot minus sdot = sdot sdot minus sdot =rArr rArr rArr rArr

sdot minus sdot + = minus sdot + = = + sdot

( )metoda

supstitucij

22 2 29 0

12 5 9 0 12 5 9 01 e2 5

a c aa a a a a a

c a

sdot minus sdot =rArr rArr rArr sdot + sdot minus sdot = rArr sdot + sdot minus sdot = rArr

= + sdot

( ) 42 2 2

12 4 0 12 4 0 3 0 3 0a a a a a a a arArr sdot minus sdot = rArr sdot minus sdot = rArr sdot minus = rArr sdot minus = rArr

( )nema smisla00 1 3 3

3 0 3 1

aaa a

a a

==rArr rArr rArr minus = minus rArr =

minus =sdot minus

minus = minus

Računamo koeficijente b i c

66 3 18

12 5 12 5 3 27

3

b ab b

c ac c

a

= minus sdot= minus sdot = minus

= + sdot rArr rArr= + sdot =

=


3 18 27 23 18 27

2

a b cy x x

y a x b x c

= = minus =rArr = sdot minus sdot +

= sdot + sdot +

3inačica

Budući da parabola dira os apscisa u točki s apscisom 3 ta točka je njezino tjeme pa je

( ) ( )( )

3 00 0

3 3 3 6 2 2 2

0

2

2

T x y Tb b b

b ab a a a

a

xa

=

rArr = minus rArr minus = rArr minus = rArr = minus sdotsdot sdot sdot= minus

sdot

sdot minus sdot

Prema uvjetu zadatka na paraboli leže dvije točke T(3 0) i A(5 12) pa vrijede jednadžbe

8

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2

T x y Ta b c a b c a b c

y a x b x c


= sdot + sdot +

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucij

69 3 6 0

9 3 025 5 6 12

25 5 1e

2

b aa a c

a b ca a c

a b c

= minus sdotsdot + sdot minus sdot + =

sdot + sdot + = rArr rArr rArrsdot + sdot minus sdot + =

sdot + sdot + =

9 18 0 9 0 9

25 30 12 5 12

metoda

supstituc je5 12 i

a a c a c c a

a a c a c a c

sdot minus sdot + = minus sdot + = = sdotrArr rArr rArr rArr rArr

sdot minus sdot + = minus sdot + = minus sdot + =

metoda 4

supstitucij

95 9 12 4 12 4 12 3

5 12 e

c aa a a a a

a c

= sdotrArr rArr rArr minus sdot + sdot = rArr sdot = rArr sdot = rArr =

minus sdot + =


66 3 18

9 9 3 27

3

b ab b

c ac c

a


= sdot rArr rArr= sdot =

=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

4inačica

Budući da parabola dira os apscisa u točki s apscisom 3 ta točka je njezino tjeme T(3 0) Pravac x = 3

je os simetrije parabole pa je točki A(5 12) parabole simetrična točka B(1 12)

12

11

10

9

8

7

6

5

4

3

2

1

-1

1 2 3 4 5

x = 3

y

x

51 3 Budući da tri točke B(1 12) T(3 0) i A(5 12) pripadaju paraboli vrijede jednadžbe

( ) ( ) 1 12 212 1 1 12 12

2

B x y Ba b c a b c a b c

y a x b x c

=rArr = sdot + sdot + rArr = + + rArr + + =

= sdot + sdot +

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c

+ + = = minus minussdot + sdot + minus minus =

sdot + sdot + = rArr sdot + sdot + = rArr rArr rArrsdot + sdot + minus minus =

sdot + sdot + = sdot + sdot + =

8 2 12 8 metoda suprotnih

koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b

sdot + sdot = minus sdot + sdot = minusrArr rArr rArr rArr

sdot + sdot = minus sdot + sdot =

( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b

sdot + sdot = minus minus sdot minus =rArr rArr rArr sdot = rArr sdot = rArr =

sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a

sdot + sdot = minusrArr sdot + sdot = minus rArr + sdot = minus rArr sdot = minus minus rArr sdot = minus rArr

=

22 36 18b brArr sdot = minus rArr = minus


( )12

12 3 18 12 3 18 273 18

c a bc c c

a b

= minus minusrArr = minus minus minus rArr = minus + rArr =

= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2

3 18 27y x x= sdot minus sdot +

Zadatak 064 (Boris gimnazija)

Za svaki realni broj m m ne 0 jednadžbom y = m middot x2 + x + m određena je neka parabola

a) Za koji m se dobije parabola s tjemenom na osi apscisa

b) Za koji m se dobije parabola kojoj je pravac 2 middot x ndash 1 = 0 os simetrije



( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot

Diskriminanta kvadratne jednadžbe 2

0a x b x csdot + sdot + = je broj 2

4 D b a c= minus sdot sdot

bull Ako je D gt 0 jednadžba ima dva realna rješenja (korijena)

bull Ako je D = 0 jednadžba ima jedno dvostruko realno rješenje (korijen)

bull Ako je D lt 0 jednadžba ima kompleksno-konjugirana rješenja (korijene)

1

n m n m

a a a a a+

= sdot =

Parametar

Vladimir Anić Ivo Goldstein Rječnik stranih riječi Novi Liber Zagreb 2002

Veličina obično realna varijabla čije vrijednosti služe za razlikovanje elemenata nekog skupa točaka

funkcija jednadžbi ili drugih matematičkih objekata

Bratoljub Klaić Rječnik stranih riječi Nakladni zavod MH Zagreb 1983

Veličina o kojoj ovisi funkcija ili oblik krivulje

a) Računamo za koji m se dobije parabola s tjemenom na osi apscisa

1inačica

Budući da tjeme parabole leži na osi apscisa (osi x) njegova je ordinata jednaka nuli

( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus

rArr =sdot sdot minus sdot

=sdot

Računamo vrijednost parametra m

22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a

= sdot + +sdot sdot minus sdot minus sdot minus

= = = rArr sdot sdot= rArr = rArr = rArrsdot sdot sdot

sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1

4 4 4 m m m m mrArr sdot minus = rArr sdot = rArr sdot = rArr = rArr = rArr

11

1 1

12 124 2m mrArr = plusmn rArr = plusmn

2inačica

Budući da tjeme parabole leži na osi apscisa (osi x) pripadna kvadratna jednadžba ima jedno

dvostruko realno rješenje (korijen) pa je njezina diskriminanta jednaka nuli

2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +

= = = rArr minus sdot sdot = rArr minus sdot = rArr minus sdot = minus rArr

minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2

4m m m m mrArr minus sdot minus rArr = rArr = rArr = plusmn rArr = plusmnminus=

b) Računamo za koji m se dobije parabola kojoj je pravac 2 middot x ndash 1 = 0 os simetrije

Parabola

2y m x x m= sdot + +


tjemenom parabole paralelno s y osi zove se os simetrije parabole i njezina jednadžba je

2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot

= sdot + + rArr = minussdot

= = =

Jednadžba pravca koji je os simetrije zadane parabole glasi

12 1 0 2 1 2 1

2 2x x x xsdot minus = rArr sdot = rArr sdot = rArr =

Vrijednost parametra m iznosi

( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot

rArr rArr minus sdot minus sdot= rArr = minus rArr = minussdot

=

Vježba 064 Za svaki realni broj m m ne 0 jednadžbom y = x2 + x + m određena je neka parabola Za koji

m se dobije parabola s tjemenom na osi apscisa

Rezultat 1

4

Zadatak 065 (Tea gimnazija)

Funkcija f(x) = ndash x2 + 4 middot x ima tjeme u točki

A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Koordinate tjemena T(x0 y0) mogu se izračunati i na sljedeći način

bull apscisa x0 tjemena je

0 21 2

x xx

+=

gdje su x1 i x2 nultočke kvadratne funkcije

( )2


bull ordinata y0 tjemena je

( ) 20 0 0 0 0


1inačica

Apscisa tjemena parabole iznosi

( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot

= minus = = rArr = minus rArr =sdot minus

= minussdot

U toj točki funkcija ima vrijednost

( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=

rArr = minus + sdot rArr = minus + rArr == minus + sdot

Koordinate su tjemena (2 4)

Odgovor je pod B

2inačica

Najprije odredimo nultočke funkcije

( )0 02

4 0 4 04 0 4

x xx x x x

x x

= =minus + sdot = rArr sdot minus + = rArr rArr rArr

minus + = minus = minus

13

( ) 1

00 1

4 42

xx

x x

==rArr rArr

minus = minus =sdot minus


0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C

Zadatak 066 (Mimi medicinska škola)

Luk na slici ima jednadžbu 2

03 18 y x x= minus sdot + sdot gdje je y udaljenost točke na luku od

x ndash osi izražena u metrima Kolika je maksimalna visina luka

y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +

sdot minus sdot minus= minus sdot + sdot rArr rArr = rArr

sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica

Najprije odredimo nultočke kvadratne funkcije

( )2 2 2

03 18 0 03 1 8 0 3 1810 0x x x x x xsdotminus sdot + sdot = rArr minus sdot + sdot = rArr sdot minus sdot =minus rArr

( )2 2

3 18 0 6 0 03 6 x x x x x xrArr sdot minus sdot = rArr minus sdot = rArr sdot minus = rArr

( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x

==rArr sdot minus sdot = rArr minus sdot = rArr sdot minus = rArr rArr

minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=

Maksimalna visina luka y0 ima vrijednost

30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=

rArr = minus sdot + sdot rArr = minus sdot + sdot rArr= minus sdot + sdot

15

27 54 270 0

y yrArr = minus + rArr =

Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C

Zadatak 067 (Vesna gimnazija)

Za x = 4 funkcija f(x) = x2 + b x + c postiže najmanju vrijednost jednaku ndash 9 Koliki je c


( ) ( ) ( )2 2

22 2

2

n na a

a b a a b b a b a b nb b

+ = + sdot sdot + minus minus = + =

( )22 2

2 a a b b a bminus sdot sdot + = minus



Zakon distribucije množenja prema zbrajanju

( ) ( ) a b c a b a c a b a c a b csdot + = sdot + sdot sdot + sdot = sdot +


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Za x = 4 funkcija f(x) = x2 + b x + c ima vrijednost jednaku ndash 9 pa vrijedi

( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f

= + sdot +rArr rArr + sdot + = minus rArr + sdot + = minus rArr

= minus

( )4 9 16 4 25 1b c b crArr sdot + = minus minus rArr sdot + = minus

16

Odredimo koeficijente kvadratne funkcije

( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=

Budući da za x0 = 4 funkcija f(x) = x2 + b x + c postiže najmanju vrijednost jednaku y0 = ndash 9 slijedi

4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a

= minus = minus = minus =sdot sdot

rArr rArr rArrsdot sdot minus sdot sdot minus sdot minus

= = minus = minus = minussdot sdot

( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus

rArr rArrsdot minus sdot minus = minus

=

sdot

sdotminus

minus

1inačica

Iz sustava jednadžbi (1) i (2) dobije se c

( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b

sdot + = minusrArr rArr sdot minus + = minus rArr minus + = minus rArr = minus + rArr =

= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b

minus minussdot + = minus sdot = minus minus sdot = minus minus =

rArr rArr rArr rArrsdot minus = minus sdot minus = minus sdot minus = minus

sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c

minus minusminus minusrArr rArr sdot minus = minus rArr sdot minus = minus rArr

( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c

+ + sdot + + sdot +rArr sdot minus = minus rArr sdot minus = minus rArr = minus sdotsdot minus rArr

( )2 264 625 50 576 64 625 50 576c c c c c crArr sdot minus + sdot + = minus rArr sdot minus minus sdot minus = minus rArr

( )2 2 2

64 625 50 576 0 14 49 0 14 4 19 0c c c c c c crArr sdot minus minus sdot minus + = rArr minus + sdot minus = rArr minus + minus = sdot minussdot rArr

( ) ( )2 22

14 49 0 7 0 7 7 0 70c c c c c crArr minus sdot + = rArr minus = rArr minus = rArr minus = rArr =

3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b

= minusrArr rArr sdot minus minus = minus rArr sdot minus = minus rArr

sdot minus = minus

4 36 64 4 28 4 28 74c c c crArr sdot = minus + rArr sdot = rArr sdot = rArr =

Vježba 067

Za x = 4 funkcija f(x) = x2 + b x + c postiže najmanju vrijednost jednaku ndash 9 Koliki je b

Rezultat ndash 8

17

Zadatak 068 (Franjo srednja škola)

Temperatura T (u degC) u stakleniku t sati nakon početka sumraka dana je formulom

( )1 2

5 30 0 124

T t t t t= sdot minus sdot + le le Uzima se da sumrak počinje u 19 h

a) Kolika je temperatura bila u 21 h

b) U koliko je sati temperatura bila minimalna

c) Koliko je iznosila minimalna temperatura u stakleniku


1

a

n a c a c a dbncb d b d b c

d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema

bull minimum ako je a gt 0

bull maksimum ako je a lt 0

a) Računamo temperaturu u 21 h Budući da sumrak počinje u 19 h od početka sumraka do 21 h

prošla su 2 h

21 19 2 t h h t h= minus rArr =

( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t

= sdot minus sdot +rArr rArr = sdot minus sdot + rArr

=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30

4 4 1 1T T T TrArr = sdot minus + rArr = sdot minus + rArr = sdot minus + rArr = minus + rArr

( ) ( ) 02 21 2 21 T T CrArr = rArr =

b) Računamo u koliko je sati temperatura bila minimalna

Budući da je

( )1 2

5 304

T t t t= sdot minus sdot +

kvadratna funkcija (varijabla je t) a vodeći je koeficijent pozitivan

10

4a a= rArr gt

funkcija ima minimum Temperatura je bila minimalna u 10 h

18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca

= sdot minus sdot +minus

rArr rArr = minus rArr = rArr = rArr

sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =

c) Računamo minimalnu temperaturu u stakleniku

1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot

= sdot minus sdot + sdot sdot minus minus sdot sdot minus

rArr rArr = rArr = rArr

sdot sdot= = minus =

130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C

sdot sdot minussdot minus minus

rArr = rArr = rArr = rArr = rArr = rArr =

sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t

= sdot minus sdot +rArr = sdot minus sdot + rArr = sdot minus + rArr

= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4

T T TrArr = sdot minus + rArr = minus + rArr = minus + rArr

( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124


Kolika je temperatura bila u 20 h

Rezultat 2525 degC

Zadatak 069 (Mario srednja škola)

Za koji realni parametar b polinom ( )1 2

52

f x x b x= minus + sdot + poprima najveću vrijednost 13

1 2 3 4A B C Dplusmn plusmn plusmn plusmn


1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2

vrijednost maksimuma

2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +

sdot minus sdot minus = minus sdot + sdot + rArr rArr = rArr sdot minus= minus = = =

= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb

sdot minus sdot minus sdot minus sdot minus sdot minus sdot minus

minus sdot minus rArr = rArr = rArr = rArr = rArrminus

sdot minus sdot minus sdot minus

( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot

minus minus minus minusrArr = rArr = rArr minus = minus minus rArr = minus +

minus minusminus rArr

2 2

16 16 16 4b b b brArr = rArr = rArr = plusmn rArr = plusmn

Odgovor je pod D

Vježba 069


52

f x x b x= + sdot + poprima najmanju vrijednost ndash 13


Rezultat C

Zadatak 070 (Ivan srednja škola)

Tjeme parabole y = 2 x ndash x2 je točka

( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20

Graf kvadratne funkcije ( )2

f x a x b x c= sdot + sdot + je parabola koju dobivamo translacijom parabole

2y a x= sdot tako da joj tjeme bude u točki T(x0 y0) pri čemu je

24

0

0 2 4

b a c bx y

a a

sdot sdot minus= minus =

sdot sdot

Odredimo koeficijente jednadžbe parabole

12 2

2 2 2

0

2

22

a

y x x y x xy a x b x c

x cx

b

y

= sdot + sdot +

=

=

minus

minus

= sdot minus rArr = minus + sdot rArr

+ sdot

rArr =

=

Koordinate tjemena T(x0 y0) iznose

bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =

rArr = minus rArr = minus rArr =sdot minus minus= minus

sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a

= minus = =sdot minus sdot minus minus minus

rArr = rArr = rArr = rArr =sdot sdot minussdot minus minus minus=

sdot

Tjeme parabole je točka

( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C

Vježba 070 Tjeme parabole y = 2 x + x

2 je točka

( ) ( ) ( ) ( ) 2 0 2 1 1 1 1 1A B C Dminus minus minus minus minus

Rezultat D

Zadatak 071 (Antonio srednja škola)

Ovisnost brzine kočenja automobila i zaustavnog puta dana je formulom

( )2

0004 03f x x x= sdot + sdot pri čemu je x brzina kočenja automobila u ms a f(x) zaustavni put u

metrima Ako je automobil išao 72 kmh koliko mu je trebalo da se zaustavi


1 1000 1 3 600km m h s= =


( )2




Mjernu jedinicu brzine kmh moramo pretvoriti u ms

100072 72 20

3600

km m mx x x

h s s= rArr = rArr =

Sada računamo zaustavni put automobila

( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x

= sdot + sdotrArr = sdot + sdot rArr =

=

Automobil je trebao 76 m da se zaustavi

21

Vježba 071 Ovisnost brzine kočenja automobila i zaustavnog puta dana je formulom

( )2



Rezultat 34 m

Zadatak 072 (Ivona srednja škola)

Zadana je funkcija ( )2

2 3f x x x= + sdot minus Izračunajte koordinate tjemena grafa zadane

funkcije

Rješenje 072

Ponovimo

( )2 2 2

2 x y x x y y+ = + sdot sdot +


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2

3 5 3 5 3 5f x a x f x a x T= sdot minus + rArr = sdot minus + rArr

bull ( ) ( ) ( ) ( ) ( ) ( )2 2

3 3 55 3 5f x a x f x a x T= sdot minus minus minusrArr = sdot + rArr minusminus

bull ( ) ( ) ( ) ( )( ) ( )22

3 3 55 3 5f x a x f x a x T= sdot + + rArr sdot + rArr minus= minusminus

bull ( ) ( ) ( ) ( )( ) ( ) ( )22

3 3 5 35 5f x a x f x a x T= sdot + minus rArr = sdot rArrminus minus+minus minusminus

22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=

= minus== + sdot minus sdot

rArr = rArr rArr rArrsdot sdot minus minus= sdot + sdot + = minus =

sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y

= minus = minus = minusrArr rArr rArr rArr = minus minus

minus minus = minus= = minus

2inačica

Nadopunjavanjem trinoma na potpuni kvadrat zadanu funkciju transformiramo u sljedeći oblik

( ) ( ) ( ) ( )12 2 2

2 3 2 3 2 1 11 3f x x x f x x x f x x x= + sdot minus rArr = + sdot minus rArr = + sdot + minusminus minus+ rArr

( ) ( ) ( ) ( ) ( ) ( )( ) ( )222

2 1 4 1 4 1 4 f x x x f x x f x xrArr = + sdot + minus rArr = + minus rArr = minus minus + minus

Koordinate tjemena su

( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y

= minus minus + minus = minusrArr rArr = minus minus

= minus= sdot minus +

3inačica

Odredimo nultočke zadane funkcije

( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x

= + sdot minus + sdot minus =rArr + sdot minus = rArr rArr

= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA

f(x) = a sdotsdotsdotsdot (x - x0)2 + y0

23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa

= = = minusminus plusmn minus sdot sdot minus minus plusmn +

rArr rArr = rArr = rArrminus plusmn minus sdot sdot sdot=

sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x

minus += = =minus plusmn minus plusmn

rArr = rArr = rArr rArr rArrminus minus = minus

= = minus

Računamo apscisu x0 tjemena zadane funkcije

( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx

= = minus+ minus minus

rArr = rArr = rArr = minus rArr = minus+=

Računamo ordinatu y0 tjemena te funkcije

( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x

= + sdot minus= + sdot minus

= minus rArr rArr = minus + sdot minus minus rArr= minus

=

1 2 3 40 0

y yrArr = minus minus rArr = minus

Koordinate tjemena glase

( ) ( ) 1 4 0 0

T x y T= minus minus

Vježba 072



funkcije

Rezultat ( )1 2 T minus minus

Zadatak 073 (Anita srednja škola)

Visina na kojoj se nalazi projektil t sekundi nakon ispaljivanja dana je formulom

( ) ( )2

2 11 310h t t= minus sdot minus + (h je izraženo u metrima) Koliko će sekundi projektil biti na visini iznad

182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24

Graf kvadratne funkcije f(x) = a middot (x ndash x0)2 + y0 parabola je s tjemenom u točki T(x0 y0) dobivena

translacijom parabole y = a middot x2 U točki x0 funkcija f(x) = a middot (x ndash x0)2 + y0 poprima najveću vrijednost

y0 ako je a lt 0 a najmanju vrijednost y0 ako je a gt 0

Računamo vrijeme t za koje je visina projektila jednaka 182 m

( ) ( ) ( ) ( )2 2 2

182 2 11 310 182 2 11 182 310 2 11 128h t t t t= rArr minus sdot minus + = rArr minus sdot minus = minus rArr minus sdot minus = minus rArr

( ) ( ) ( ) ( )2 2

2

2 11 128 11 64 11 64 112 64t t t trArr minus sdot minus = minus rArr minus = rArr minus = rArr minus = plusmnminus rArr

311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s

=minus = minus = minus +rArr minus = plusmn rArr rArr rArr

minus = = + =

Prvi put projektil dosegne visinu 182 m nakon 3 s poslije ispaljivanja Penje se do maksimalne visine

a onda pri padu ponovno dođe na visinu 182 m nakon 19 s od ispaljivanja Ukupno vrijeme ∆t za koje

će projektil biti na visini iznad 182 m iznosi

19 3 16 2 1

t t t t s s t s∆ = minus rArr ∆ = minus rArr ∆ =

Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C

Zadatak 074 (Anđelka Katarina HTT)

Koliki su koeficijenti funkcije prikazane na slici

a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2

a b a b a bminus = minus sdot +

Funkcija zadana formulom

( )2

0f x a x b x c a b c R a= sdot + sdot + isin ne

zove se kvadratna funkcija ili kvadratni polinom Za broj a kažemo da je vodeći koeficijent b je

linearni a c slobodni koeficijent kvadratne funkcije Graf kvadratne funkcije ( )2


je parabola 2

y a x b x c= sdot + sdot +

Broj x0 je nultočka funkcije f ako vrijedi

( ) 00

f x =

Realna nultočka funkcije apscisa je točke u kojoj graf funkcije siječe (ili dira) x ndash os Grafički nultočke

određujemo tako da nacrtamo parabolu


i odredimo točke na x ndash osi u kojima parabola siječe (ili dira) x ndash os Ako su poznate nultočke x1 i x2

kvadratne funkcije tada se ona može faktorizirati

( )( ) ( ) ( )

2

1 2 nultočke funkcije

1

2

f x a x b x cf x a x x x x

x x

= sdot + sdot +rArr = sdot minus sdot minus

minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica

Sa slike vidi se da su nultočke funkcije x1 = ndash 1 i x2 = 1 pa funkciju možemo zapisati u obliku

( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2

f x a x x x xf x a x x f x a x x

x x

= sdot minus sdot minusrArr = sdot minus minus sdot minus rArr = sdot + sdot minus

= minus =

Budući da točka T(0 ndash 1) pripada grafu funkcije (paraboli) njezine koordinate uvrstit ćemo u

jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x

= minusrArr minus = sdot + sdot minus rArr minus = sdot sdot minus rArr minus = minus rArr =

= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1

af x x x f x x x f x x

f x a x x

=rArr = sdot + sdot minus rArr = + sdot minus rArr = minus

= sdot + sdot minus

Koeficijenti a b i c su

( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica

Sa slike vidi se da točke A B i T pripadaju paraboli Njihove koordinate glase

( ) ( ) ( ) ( ) ( ) ( ) 1 0 1 0 0 1 A x y A B x y B T x y T= minus = = minus

Ako uvrstimo koordinate točaka u jednadžbu parabole dobit ćemo sustav od tri jednadžbe sa tri

nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c

= minusrArr sdot minus + sdot minus + = rArr minus + =

= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c

=rArr sdot + sdot + = rArr + + =

= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c

= minusrArr sdot + sdot + = minus rArr + + = minus rArr = minus

= sdot + sdot +

Riješimo sustav

metoda metoda suprotnih

supstitucije koeficije

01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c

minus + =minus minus = minus =

+ + = rArr rArr rArr rArr rArr+ minus = + =

= minus

2 2 2 2 1 2a a arArr sdot = rArr sdot = rArr =

Računamo b

11 1 1 1 0

1

ab b b

a b

=rArr + = rArr = minus rArr =

+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus

Zadatak 075 (Terminator tehnička škola)

Odredi nultočke funkcije ( )2

2 3f x x x= minus + sdot +

Rješenje 075

Ponovimo

Broj x0 nazivamo nultočka funkcije f ako je

( ) 00

f x =


( )2



linearni a c slobodni koeficijent kvadratne funkcije

Jednadžba oblika

20a x b x csdot + sdot + =

(a ne 0 b i c realni brojevi) naziva se kvadratna jednadžba Svaki broj x (realni ili kompleksni) koji

zadovoljava tu jednadžbu naziva se rješenje kvadratne jednadžbe Rješenja kvadratne jednadžbe


su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a

minus + minus sdot sdot minus minus minus sdot sdot= =

sdot sdot

ili kraće

24

12 2

b b a cx

a

minus plusmn minus sdot sdot=

sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x

= minus + sdot +rArr minus + sdot + = rArr minus + sdot + sdot= rArr

=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus

minus sdot minus =rArr minus sdot minus = rArr rArr rArr

minus plusmn minus sdot sdot= = minus = minus =

sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x

minus minus plusmn minus minus sdot sdot minus plusmn + plusmnrArr = rArr = rArr = rArr

sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn

rArr = rArr rArr rArrminus = minus

= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =


Odredi polinom drugog stupnja ( )2

f x a x b x c= sdot + sdot + ako je zadano f(0) = ndash 4 f(3) = ndash 4

f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =

Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i

jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus

= minus rArr rArr sdot + sdot + = minus rArr

minus = sdot minus + sdot minus + =

0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c

sdot + sdot + = minus = minussdot + sdot minus = minus

rArr sdot + sdot + = minus rArr sdot + sdot + = minus rArr rArr rArrsdot minus sdot minus =

sdot minus sdot + = sdot minus sdot + =

metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih

koeficijenata 36 6 18

2a b a b a b

a b a b a b

sdot + sdot = minus + sdot + sdot = sdot + sdot =rArr rArr rArr rArr rArr

sdot

sdot minus sdot = + sdot minus sdot = sdot minus sdot =

29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b

sdot + sdot =rArr rArr sdot = rArr sdot = rArr = rArr = rArr =

sdot minus sdot =sdot

Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =

rArr sdot + sdot = rArr sdot + sdot = rArr + sdot = rArr sdot = minus rArr=


Polinom drugog stupnja glasi

( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c

= = minus = minusrArr = sdot minus minus

= sdot + sdot +

Vježba 076


f x a x b x c= sdot + sdot + ako je zadano f(0) = 7 f(1) = 10

f(ndash 1) = 6

Rezultat ( )2

2 7f x x x= + sdot +

Zadatak 077 (Real gimnazija)

Točka T(2 3) je točka maksimuma funkcije ( )2

f x a x b x= sdot + sdot Odredite vrijednost

koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot





Odredimo koeficijente zadane funkcije

30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =

Budući da je točka T(2 3) točka maksimuma navedene funkcije vrijedi

( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c

= = minus = minus = minus sdot sdot sdot

= minus rArr rArr rArr rArr rArr sdot sdot sdot minus sdot sdot minus minus = = = sdot sdot sdot sdot sdot minus =sdot

=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a

= minus = minus = minus = minus sdot sdot sdot sdot

rArr rArr rArr rArr rArr rArr = minus sdot = minus = minus = minus sdot sdot sdot

sdot sdot

sdot sdot

( )2 2 2 2

4 12 16 12 16 12 0 16 1 0 42a a a a a a a arArr minus sdot = minus sdot rArr sdot = minus sdot rArr sdot + sdot = rArr sdot + sdot = rArr

( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a

= rArr sdot + sdot = rArr sdot sdot + = rArr rArr sdot + = rArr

sdot + =

43

4 3 4 3 4

a a arArr sdot = minus rArr sdot = minus rArr = minus

Vježba 077



koeficijenta b

Rezultat 3

Zadatak 078 (4A 4B TUPŠ)

Koja slika prikazuje kvadratnu funkciju ( )2

f x a x b x c= sdot + sdot + kojoj je diskriminanta

negativna i koeficijent c pozitivan

31

Rješenje 078

Ponovimo


( )2





( )2


je parabola





bull Ako je D gt 0 jednadžba ima dva realna rješenja (korijena) tj parabola siječe os x u dvije

točke

bull Ako je D = 0 jednadžba ima jedno dvostruko realno rješenje (korijen) tj parabola dira os x u

jednoj točki

bull Ako je D lt 0 jednadžba ima kompleksno ndash konjugirana rješenja (korijene) tj parabola ne

siječe os x

Uočimo da parabola siječe os y za x = 0 odakle slijedi

22

0 0 0

y a x b x cy a b c y c

x

= sdot + sdot +rArr = sdot + sdot + rArr =

=

Dakle sjecište je u točki (0 c) tj c je odsječak parabole na osi y Analizirat ćemo svaku sliku posebno

c gt 0c gt 0c gt 0c gt 0

D lt 0D lt 0D lt 0D lt 0


D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0

c lt 0c lt 0c lt 0c lt 0



Diskriminanta je negativna i koeficijent c pozitivan na slici A

32

Odgovor je pod A

Vježba 078



negativna i koeficijent c negativan

Rezultat D

Zadatak 079 (Martina hotelijerska škola)

Odredite najmanju vrijednost funkcije ( )12

32

f x a x x= sdot minus sdot + ako se ta vrijednost postiže za

x = 2

5 5 3 5

2 2A B C Dminus minus

Rješenje 079

Ponovimo

1

a

n a c a d b c a c a c a dbncb d b d b d b d b c

d

sdot minus sdot sdot sdot= minus = sdot = =

sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc

== sdot minus sdot +

rArr = minus

= sdot + sdot +=

Budući da zadana funkcija ima ekstremnu vrijednost za x = 2 vrijedi

23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus

= minus minusrArr minus = rArr rArr minus = rArr = rArr = rArrsdot

sdot sdot sdot sdot=

33 4 4 3 4 3

4 4a a a arArr = sdot rArr sdot = rArr sdot = rArr =

Sada su poznata sva tri koeficijenta funkcije pa njezina najmanja vrijednost iznosi

( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya

= = minus = sdot sdot minus minus sdot sdot minus minus

rArr = rArr = rArrsdot sdot minus sdot sdot=

sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y

minusminus minus minus

rArr = rArr = rArr = rArr = rArr = minus rArr = minus rArr = minus

Odgovor je pod B

Vježba 079

Odredite ekstremnu vrijednost funkcije ( )12

32

f x a x x= sdot + sdot + ako se ta vrijednost postiže

za x = 2

7 7 3 7


Rezultat C

Zadatak 080 (Nina gimnazija)

Polinom f(x) = a middot x2 + b middot x + c prima negativne vrijednosti ako i samo ako je 11 5 x isin minus

Za funkciju f vrijedi

( ) ( ) ( ) ( ) 0 55 0 6 0 25 0 5A f B f C f D f= minus = = = minus

Rješenje 080

Ponovimo

Broj x0 je nultočka funkcije f ako vrijedi f(x0) = 0

Prima li polinom ( )2

f x a x b x c= sdot + sdot + negativne vrijednosti ako i samo ako je 1 2

x x xisin

onda su njegove nultočke x1 i x2 Na intervalima i 1 2

x xminus infin + infin polinom prima pozitivne

vrijednosti

34

Rješenja x1 i x2 kvadratne jednadžbe

bull 2

0a x b x csdot + sdot + = zadovoljavaju Viegraveteove formule 2

1 2 1

b cx x x x

a a+ = minus sdot =

bull 2

0x b x c+ sdot + = zadovoljavaju Viegraveteove formule 1

2 1 2

x x b x x c+ = minus sdot =

1inačica

Budući da polinom f(x) = x2 + b middot x + c prima negativne vrijednosti ako i samo ako je 11 5 x isin minus

točke x1 = ndash 11 i x2 = 5 su nultočke pa vrijedi

( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c

minus = minus + sdot minus + = minus sdot + =rArr rArr rArr rArr

+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot

minus sdot + = minus minus sdot + = minusrArr rArr rArr rArr

sdot + = minus sdot + = minus

55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c

minus sdot + sdot = minusrArr rArr sdot = minus rArr sdot = minus rArr = minus

sdot + sdot = minus

Sada imamo

( )( )

22

5555

f x x b x cf x x b x

c

= + sdot +rArr = + sdot minus

= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x

= + sdot minusrArr = + sdot minus rArr = minus

=

Odgovor je pod A

2inačica


točke x1 = ndash 11 i x2 = 5 su nultočke pa po drugoj Viegraveteovoj formuli vrijedi

11 51 2

11 5 55

1 2

x xc c

x x c

= minus =rArr minus sdot = rArr = minus

sdot =

Odgovor je pod A

35

Vježba 080



( ) ( ) ( ) ( ) 0 28 0 20 0 9 0 5A f B f C f D f= minus = minus = minus =

Rezultat A

2

P(12 0)

T(12 2)

N(24 0)O(0 0)

y

x

Postavimo most u pravokutni koordinatni sustav kao na slici Tjeme parabole ima koordinate T(12 2)

pa jednadžba luka mosta glasi

( ) ( )

( )( )

12 20 0 2

12 22

0 0

T x y T

y a x

y a x x y

=

rArr = sdot minus +

= sdot minus +

Budući da graf sadrži točku O(0 0) (isto tako možemo uzeti točku N(24 0)) uvrštavanjem dobivamo

vrijednost vodećeg koeficijenta a

( ) ( )

( )( ) ( )

0 02 2

0 0 12 2 0 12 2 0 144 22

12 2

O x y O

a a a

y a x

=

rArr = sdot minus + rArr = sdot minus + rArr = sdot + rArr= sdot minus +

( )2 1

144 2 144 2 14

44 72

14a a a aminusrArr minus sdot = rArr minus sdot = rArr = minus rArr = minus

Jednadžba luka mosta glasi

( )1 2

12 272


Uočimo da se parabola

( )1 2

12 272


dobije translacijom parabole

1 22

72y x= minus sdot +

za broj x0 = 12 u pozitivnom smjeru x ndash osi

Vježba 061 Luk mosta ima oblik parabole Visina mosta je 5 m a duljina raspona 40 m Napiši


Rezultat 1 2

580

y x= minus sdot +

Zadatak 062 (Matea gimnazija) Ako su ndash 1 i 2 nultočke polinoma drugog stupnja a najveća vrijednost polinoma iznosi 3

odredi taj polinom



( )2


3




( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0




1inačica


tjemena

1 21 2

1 2 1

0 01 2 2 20 2

x x

x xx xx

= minus =minus +

rArr = rArr =+=


( ) ( )0 01 2

f x f x= =


( ) 30

f x =


( ) ( )2

1 1 1 01


22 2 2 0


21 1 1

30 2 2 2



4

( ) ( )2

1 1 00 0

22 2 0 4 2 0 4 2 0

2 1 1 1 11 1 3 3

3 4 2 4 22 2

4

a b ca b c a b c

a b c a b c a b c

a b c a b ca b c




sdot

0

4 2 0 4 2 0

2 4

metoda

supstituci12

j1

e2 4 2

a b c c b a

a b c a b c

a b c a b c

minus + = = minus



( )

4 2 0 4 2 0 3 3 0

2 4 12 2 4 4 12 3 6 12

a b b a a b b a a b

a b b a a b b a a b



metoda suprotnih 9

koeficij

12 49

ena12 9 12



3 3 04 4

3 3 0 3 0 333

4 0 3 443

3

a b

a a a ab

sdot + sdot =


34

3 4 3



4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus


Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +


= minus = =

2inačica


( ) ( )2

1 1 1 0 01


22 2 2 0 4 2 0



2 24 4 2

3 3 4 12 4 4

4a c b a c b

a c b aa a



sdot sdot


( )

metoda

supstitucije

04 2 0

4 2 0 4 2 0 24 12

2 24 12 4 12

a b c c b aa b b a


a c b a a c b a




5

3 3 0 3 3 0

2 2 2 24 4 12 4 4

3

12 0

a b a b

a b a b a a b a b a



metoda

supstituci

0

2 2 2 24 4 12 0 4 je4 12 0

a b b a

a b a b a a b a b a



( ) ( )22 2 2 2


( )2 2 2


( )nema smisla00 413 4 0 3 4

3 4 0 33 3

4

aaa a a a

a a




4 44

3 33

b a

b ba

= minus



4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus


Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +


= minus = =


Rezultat ( )2

4f x x= minus





( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

6


( ) 20 0

0 0




( )2 2 2



( ) ( )2


je parabola

( )2



( )0 0

T x


( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot

1inačica


( ) ( ) 3 0 0 0

T x y T=


( ) ( )

( ) ( )

( )

( )

0 3 00

2 2 5 12 12 5 3 12 2 12 4 4 12

2

0

T x T

A x y A a a a a

y a x x

=


= sdot minus



( ) ( )

( )( ) ( )

3 0 3 00 2 2

3 3 3 6 92

0

a T x T

y x y x x

y a x x

= =


= sdot minus


2inačica


7

( ) ( )( ) 2

4

3 00 0

3 3 32 2 2

2 2 20 2 4 4 40 0 02

4 4 44

0 4

T x y T b b b

a a abx

a a c b a c b a c b

a a aa c by

a

a

a




sdot minus sdot



=sdot

6

24 0

b a

a c b

= minus sdotrArr

sdot sdot minus =


( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucije

62

4 6 024 0

25 5 6 1225 5 12

b a

a c aa c b

a a ca b c

= minus sdot



2 2 24 36 0 4 36 0 4 36 0

25 30 12 5 12

4

12

5

a c a a c a a c a

a a c a c c a



( )metoda

supstitucij

22 2 29 0

12 5 9 0 12 5 9 01 e2 5

a c aa a a a a a

c a


= + sdot

( ) 42 2 2



3 0 3 1

aaa a

a a


minus =sdot minus

minus = minus


66 3 18

12 5 12 5 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

3inačica


( ) ( )( )

3 00 0

3 3 3 6 2 2 2

0

2

2

T x y Tb b b

b ab a a a

a

xa

=


sdot

sdot minus sdot


8

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucij

69 3 6 0

9 3 025 5 6 12

25 5 1e

2

b aa a c

a b ca a c

a b c



sdot + sdot + =

9 18 0 9 0 9

25 30 12 5 12

metoda

supstituc je5 12 i

a a c a c c a

a a c a c a c



metoda 4

supstitucij

95 9 12 4 12 4 12 3

5 12 e

c aa a a a a

a c


minus sdot + =


66 3 18

9 9 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

4inačica



12

11

10

9

8

7

6

5

4

3

2

1

-1

1 2 3 4 5

x = 3

y

x


( ) ( ) 1 12 212 1 1 12 12

2


y a x b x c


= sdot + sdot +

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c





koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b



( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b


sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a


=



( )12

12 3 18 12 3 18 273 18

c a bc c c

a b


= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2








( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

3




( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0




1inačica


tjemena

1 21 2

1 2 1

0 01 2 2 20 2

x x

x xx xx

= minus =minus +

rArr = rArr =+=


( ) ( )0 01 2

f x f x= =


( ) 30

f x =


( ) ( )2

1 1 1 01


22 2 2 0


21 1 1

30 2 2 2



4

( ) ( )2

1 1 00 0

22 2 0 4 2 0 4 2 0

2 1 1 1 11 1 3 3

3 4 2 4 22 2

4

a b ca b c a b c

a b c a b c a b c

a b c a b ca b c




sdot

0

4 2 0 4 2 0

2 4

metoda

supstituci12

j1

e2 4 2

a b c c b a

a b c a b c

a b c a b c

minus + = = minus



( )

4 2 0 4 2 0 3 3 0

2 4 12 2 4 4 12 3 6 12

a b b a a b b a a b

a b b a a b b a a b



metoda suprotnih 9

koeficij

12 49

ena12 9 12



3 3 04 4

3 3 0 3 0 333

4 0 3 443

3

a b

a a a ab

sdot + sdot =


34

3 4 3



4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus


Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +


= minus = =

2inačica


( ) ( )2

1 1 1 0 01


22 2 2 0 4 2 0



2 24 4 2

3 3 4 12 4 4

4a c b a c b

a c b aa a



sdot sdot


( )

metoda

supstitucije

04 2 0

4 2 0 4 2 0 24 12

2 24 12 4 12

a b c c b aa b b a


a c b a a c b a




5

3 3 0 3 3 0

2 2 2 24 4 12 4 4

3

12 0

a b a b

a b a b a a b a b a



metoda

supstituci

0

2 2 2 24 4 12 0 4 je4 12 0

a b b a

a b a b a a b a b a



( ) ( )22 2 2 2


( )2 2 2


( )nema smisla00 413 4 0 3 4

3 4 0 33 3

4

aaa a a a

a a




4 44

3 33

b a

b ba

= minus



4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus


Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +


= minus = =


Rezultat ( )2

4f x x= minus





( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

6


( ) 20 0

0 0




( )2 2 2



( ) ( )2


je parabola

( )2



( )0 0

T x


( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot

1inačica


( ) ( ) 3 0 0 0

T x y T=


( ) ( )

( ) ( )

( )

( )

0 3 00

2 2 5 12 12 5 3 12 2 12 4 4 12

2

0

T x T

A x y A a a a a

y a x x

=


= sdot minus



( ) ( )

( )( ) ( )

3 0 3 00 2 2

3 3 3 6 92

0

a T x T

y x y x x

y a x x

= =


= sdot minus


2inačica


7

( ) ( )( ) 2

4

3 00 0

3 3 32 2 2

2 2 20 2 4 4 40 0 02

4 4 44

0 4

T x y T b b b

a a abx

a a c b a c b a c b

a a aa c by

a

a

a




sdot minus sdot



=sdot

6

24 0

b a

a c b

= minus sdotrArr

sdot sdot minus =


( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucije

62

4 6 024 0

25 5 6 1225 5 12

b a

a c aa c b

a a ca b c

= minus sdot



2 2 24 36 0 4 36 0 4 36 0

25 30 12 5 12

4

12

5

a c a a c a a c a

a a c a c c a



( )metoda

supstitucij

22 2 29 0

12 5 9 0 12 5 9 01 e2 5

a c aa a a a a a

c a


= + sdot

( ) 42 2 2



3 0 3 1

aaa a

a a


minus =sdot minus

minus = minus


66 3 18

12 5 12 5 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

3inačica


( ) ( )( )

3 00 0

3 3 3 6 2 2 2

0

2

2

T x y Tb b b

b ab a a a

a

xa

=


sdot

sdot minus sdot


8

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucij

69 3 6 0

9 3 025 5 6 12

25 5 1e

2

b aa a c

a b ca a c

a b c



sdot + sdot + =

9 18 0 9 0 9

25 30 12 5 12

metoda

supstituc je5 12 i

a a c a c c a

a a c a c a c



metoda 4

supstitucij

95 9 12 4 12 4 12 3

5 12 e

c aa a a a a

a c


minus sdot + =


66 3 18

9 9 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

4inačica



12

11

10

9

8

7

6

5

4

3

2

1

-1

1 2 3 4 5

x = 3

y

x


( ) ( ) 1 12 212 1 1 12 12

2


y a x b x c


= sdot + sdot +

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c





koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b



( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b


sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a


=



( )12

12 3 18 12 3 18 273 18

c a bc c c

a b


= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2








( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

4

( ) ( )2

1 1 00 0

22 2 0 4 2 0 4 2 0

2 1 1 1 11 1 3 3

3 4 2 4 22 2

4

a b ca b c a b c

a b c a b c a b c

a b c a b ca b c




sdot

0

4 2 0 4 2 0

2 4

metoda

supstituci12

j1

e2 4 2

a b c c b a

a b c a b c

a b c a b c

minus + = = minus



( )

4 2 0 4 2 0 3 3 0

2 4 12 2 4 4 12 3 6 12

a b b a a b b a a b

a b b a a b b a a b



metoda suprotnih 9

koeficij

12 49

ena12 9 12



3 3 04 4

3 3 0 3 0 333

4 0 3 443

3

a b

a a a ab

sdot + sdot =


34

3 4 3



4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus


Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +


= minus = =

2inačica


( ) ( )2

1 1 1 0 01


22 2 2 0 4 2 0



2 24 4 2

3 3 4 12 4 4

4a c b a c b

a c b aa a



sdot sdot


( )

metoda

supstitucije

04 2 0

4 2 0 4 2 0 24 12

2 24 12 4 12

a b c c b aa b b a


a c b a a c b a




5

3 3 0 3 3 0

2 2 2 24 4 12 4 4

3

12 0

a b a b

a b a b a a b a b a



metoda

supstituci

0

2 2 2 24 4 12 0 4 je4 12 0

a b b a

a b a b a a b a b a



( ) ( )22 2 2 2


( )2 2 2


( )nema smisla00 413 4 0 3 4

3 4 0 33 3

4

aaa a a a

a a




4 44

3 33

b a

b ba

= minus



4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus


Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +


= minus = =


Rezultat ( )2

4f x x= minus





( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

6


( ) 20 0

0 0




( )2 2 2



( ) ( )2


je parabola

( )2



( )0 0

T x


( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot

1inačica


( ) ( ) 3 0 0 0

T x y T=


( ) ( )

( ) ( )

( )

( )

0 3 00

2 2 5 12 12 5 3 12 2 12 4 4 12

2

0

T x T

A x y A a a a a

y a x x

=


= sdot minus



( ) ( )

( )( ) ( )

3 0 3 00 2 2

3 3 3 6 92

0

a T x T

y x y x x

y a x x

= =


= sdot minus


2inačica


7

( ) ( )( ) 2

4

3 00 0

3 3 32 2 2

2 2 20 2 4 4 40 0 02

4 4 44

0 4

T x y T b b b

a a abx

a a c b a c b a c b

a a aa c by

a

a

a




sdot minus sdot



=sdot

6

24 0

b a

a c b

= minus sdotrArr

sdot sdot minus =


( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucije

62

4 6 024 0

25 5 6 1225 5 12

b a

a c aa c b

a a ca b c

= minus sdot



2 2 24 36 0 4 36 0 4 36 0

25 30 12 5 12

4

12

5

a c a a c a a c a

a a c a c c a



( )metoda

supstitucij

22 2 29 0

12 5 9 0 12 5 9 01 e2 5

a c aa a a a a a

c a


= + sdot

( ) 42 2 2



3 0 3 1

aaa a

a a


minus =sdot minus

minus = minus


66 3 18

12 5 12 5 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

3inačica


( ) ( )( )

3 00 0

3 3 3 6 2 2 2

0

2

2

T x y Tb b b

b ab a a a

a

xa

=


sdot

sdot minus sdot


8

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucij

69 3 6 0

9 3 025 5 6 12

25 5 1e

2

b aa a c

a b ca a c

a b c



sdot + sdot + =

9 18 0 9 0 9

25 30 12 5 12

metoda

supstituc je5 12 i

a a c a c c a

a a c a c a c



metoda 4

supstitucij

95 9 12 4 12 4 12 3

5 12 e

c aa a a a a

a c


minus sdot + =


66 3 18

9 9 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

4inačica



12

11

10

9

8

7

6

5

4

3

2

1

-1

1 2 3 4 5

x = 3

y

x


( ) ( ) 1 12 212 1 1 12 12

2


y a x b x c


= sdot + sdot +

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c





koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b



( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b


sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a


=



( )12

12 3 18 12 3 18 273 18

c a bc c c

a b


= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2








( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

5

3 3 0 3 3 0

2 2 2 24 4 12 4 4

3

12 0

a b a b

a b a b a a b a b a



metoda

supstituci

0

2 2 2 24 4 12 0 4 je4 12 0

a b b a

a b a b a a b a b a



( ) ( )22 2 2 2


( )2 2 2


( )nema smisla00 413 4 0 3 4

3 4 0 33 3

4

aaa a a a

a a




4 44

3 33

b a

b ba

= minus



4 4 4 4 84 4

3 3 3 3 33 3

c b a

c c ca b

= minus


Polinom glasi

( )( )

2

4 4 82

4 4 8 3 3 3 3 3 3

f x a x b x c

f x x x

a b c

= sdot + sdot +


= minus = =


Rezultat ( )2

4f x x= minus





( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

6


( ) 20 0

0 0




( )2 2 2



( ) ( )2


je parabola

( )2



( )0 0

T x


( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot

1inačica


( ) ( ) 3 0 0 0

T x y T=


( ) ( )

( ) ( )

( )

( )

0 3 00

2 2 5 12 12 5 3 12 2 12 4 4 12

2

0

T x T

A x y A a a a a

y a x x

=


= sdot minus



( ) ( )

( )( ) ( )

3 0 3 00 2 2

3 3 3 6 92

0

a T x T

y x y x x

y a x x

= =


= sdot minus


2inačica


7

( ) ( )( ) 2

4

3 00 0

3 3 32 2 2

2 2 20 2 4 4 40 0 02

4 4 44

0 4

T x y T b b b

a a abx

a a c b a c b a c b

a a aa c by

a

a

a




sdot minus sdot



=sdot

6

24 0

b a

a c b

= minus sdotrArr

sdot sdot minus =


( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucije

62

4 6 024 0

25 5 6 1225 5 12

b a

a c aa c b

a a ca b c

= minus sdot



2 2 24 36 0 4 36 0 4 36 0

25 30 12 5 12

4

12

5

a c a a c a a c a

a a c a c c a



( )metoda

supstitucij

22 2 29 0

12 5 9 0 12 5 9 01 e2 5

a c aa a a a a a

c a


= + sdot

( ) 42 2 2



3 0 3 1

aaa a

a a


minus =sdot minus

minus = minus


66 3 18

12 5 12 5 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

3inačica


( ) ( )( )

3 00 0

3 3 3 6 2 2 2

0

2

2

T x y Tb b b

b ab a a a

a

xa

=


sdot

sdot minus sdot


8

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucij

69 3 6 0

9 3 025 5 6 12

25 5 1e

2

b aa a c

a b ca a c

a b c



sdot + sdot + =

9 18 0 9 0 9

25 30 12 5 12

metoda

supstituc je5 12 i

a a c a c c a

a a c a c a c



metoda 4

supstitucij

95 9 12 4 12 4 12 3

5 12 e

c aa a a a a

a c


minus sdot + =


66 3 18

9 9 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

4inačica



12

11

10

9

8

7

6

5

4

3

2

1

-1

1 2 3 4 5

x = 3

y

x


( ) ( ) 1 12 212 1 1 12 12

2


y a x b x c


= sdot + sdot +

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c





koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b



( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b


sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a


=



( )12

12 3 18 12 3 18 273 18

c a bc c c

a b


= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2








( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

6


( ) 20 0

0 0




( )2 2 2



( ) ( )2


je parabola

( )2



( )0 0

T x


( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot

1inačica


( ) ( ) 3 0 0 0

T x y T=


( ) ( )

( ) ( )

( )

( )

0 3 00

2 2 5 12 12 5 3 12 2 12 4 4 12

2

0

T x T

A x y A a a a a

y a x x

=


= sdot minus



( ) ( )

( )( ) ( )

3 0 3 00 2 2

3 3 3 6 92

0

a T x T

y x y x x

y a x x

= =


= sdot minus


2inačica


7

( ) ( )( ) 2

4

3 00 0

3 3 32 2 2

2 2 20 2 4 4 40 0 02

4 4 44

0 4

T x y T b b b

a a abx

a a c b a c b a c b

a a aa c by

a

a

a




sdot minus sdot



=sdot

6

24 0

b a

a c b

= minus sdotrArr

sdot sdot minus =


( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucije

62

4 6 024 0

25 5 6 1225 5 12

b a

a c aa c b

a a ca b c

= minus sdot



2 2 24 36 0 4 36 0 4 36 0

25 30 12 5 12

4

12

5

a c a a c a a c a

a a c a c c a



( )metoda

supstitucij

22 2 29 0

12 5 9 0 12 5 9 01 e2 5

a c aa a a a a a

c a


= + sdot

( ) 42 2 2



3 0 3 1

aaa a

a a


minus =sdot minus

minus = minus


66 3 18

12 5 12 5 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

3inačica


( ) ( )( )

3 00 0

3 3 3 6 2 2 2

0

2

2

T x y Tb b b

b ab a a a

a

xa

=


sdot

sdot minus sdot


8

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucij

69 3 6 0

9 3 025 5 6 12

25 5 1e

2

b aa a c

a b ca a c

a b c



sdot + sdot + =

9 18 0 9 0 9

25 30 12 5 12

metoda

supstituc je5 12 i

a a c a c c a

a a c a c a c



metoda 4

supstitucij

95 9 12 4 12 4 12 3

5 12 e

c aa a a a a

a c


minus sdot + =


66 3 18

9 9 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

4inačica



12

11

10

9

8

7

6

5

4

3

2

1

-1

1 2 3 4 5

x = 3

y

x


( ) ( ) 1 12 212 1 1 12 12

2


y a x b x c


= sdot + sdot +

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c





koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b



( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b


sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a


=



( )12

12 3 18 12 3 18 273 18

c a bc c c

a b


= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2








( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

7

( ) ( )( ) 2

4

3 00 0

3 3 32 2 2

2 2 20 2 4 4 40 0 02

4 4 44

0 4

T x y T b b b

a a abx

a a c b a c b a c b

a a aa c by

a

a

a




sdot minus sdot



=sdot

6

24 0

b a

a c b

= minus sdotrArr

sdot sdot minus =


( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucije

62

4 6 024 0

25 5 6 1225 5 12

b a

a c aa c b

a a ca b c

= minus sdot



2 2 24 36 0 4 36 0 4 36 0

25 30 12 5 12

4

12

5

a c a a c a a c a

a a c a c c a



( )metoda

supstitucij

22 2 29 0

12 5 9 0 12 5 9 01 e2 5

a c aa a a a a a

c a


= + sdot

( ) 42 2 2



3 0 3 1

aaa a

a a


minus =sdot minus

minus = minus


66 3 18

12 5 12 5 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

3inačica


( ) ( )( )

3 00 0

3 3 3 6 2 2 2

0

2

2

T x y Tb b b

b ab a a a

a

xa

=


sdot

sdot minus sdot


8

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucij

69 3 6 0

9 3 025 5 6 12

25 5 1e

2

b aa a c

a b ca a c

a b c



sdot + sdot + =

9 18 0 9 0 9

25 30 12 5 12

metoda

supstituc je5 12 i

a a c a c c a

a a c a c a c



metoda 4

supstitucij

95 9 12 4 12 4 12 3

5 12 e

c aa a a a a

a c


minus sdot + =


66 3 18

9 9 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

4inačica



12

11

10

9

8

7

6

5

4

3

2

1

-1

1 2 3 4 5

x = 3

y

x


( ) ( ) 1 12 212 1 1 12 12

2


y a x b x c


= sdot + sdot +

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c





koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b



( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b


sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a


=



( )12

12 3 18 12 3 18 273 18

c a bc c c

a b


= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2








( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

8

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


( )

( )

metoda

supstitucij

69 3 6 0

9 3 025 5 6 12

25 5 1e

2

b aa a c

a b ca a c

a b c



sdot + sdot + =

9 18 0 9 0 9

25 30 12 5 12

metoda

supstituc je5 12 i

a a c a c c a

a a c a c a c



metoda 4

supstitucij

95 9 12 4 12 4 12 3

5 12 e

c aa a a a a

a c


minus sdot + =


66 3 18

9 9 3 27

3

b ab b

c ac c

a



=


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

4inačica



12

11

10

9

8

7

6

5

4

3

2

1

-1

1 2 3 4 5

x = 3

y

x


( ) ( ) 1 12 212 1 1 12 12

2


y a x b x c


= sdot + sdot +

( ) ( ) 3 0 20 3 3 0 9 3 9 3 0

2


y a x b x c


= sdot + sdot +

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c





koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b



( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b


sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a


=



( )12

12 3 18 12 3 18 273 18

c a bc c c

a b


= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2








( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

9

( ) ( ) 5 12 212 5 5 12 25 5 25 5 12

2


y a x b x c


= sdot + sdot +


12 129 3 12 0

9 3 0 9 3 025 5 12 12

25 5 12 25 5 12

metoda

supstitucije

a b c c a ba b a b

a b c a b ca b a b

a b c a b c





koeficijenata

2 12

24 4 12 12 24 4 0

a b a b

a b a b



( ) 2 2

8 2 12 4 62 6 2 6 3

6 024 4 0 4

a b a ba a a

a ba b


sdot + =sdot +

minus

sdot =


8 2 128 3 2 12 24 2 12 2 12 24 2 36

3

a bb b b b

a


=



( )12

12 3 18 12 3 18 273 18

c a bc c c

a b


= = minus


3 18 27 23 18 27

2

a b cy x x

y a x b x c


= sdot + sdot +

Vježba 063


Rezultat 2








( )2





( )2



0 2

bx

a= minus

sdot


10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

10

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola


Parabola




0

20 2

x xb

xbax

a

=


sdot







1

n m n m

a a a a a+

= sdot =

Parametar







1inačica


( ) ( ) 0 20 0 04

024 4

0 4

T x y T xa c b

a c b ay

a

=sdot sdot minus


=sdot


22 2 2

4 1 4 1 4 1 1 0 0 0 4

4 4 42

40

4

y m x x mm m m m

a m b c mm m m

a c b

m

a



sdot sdot minus=

sdot

1 12 2 2 2 24 1 0 4 1 4 1


11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

11

1 1


2inačica



2

2 2 2 1 1 4 0 1 4 0 4 1

24 0

y m x x m

a m b c m m m m m

b a c

= sdot + +


minus sdot sdot =

( )1 1 1 12 2 2

4 1 12 124 4 4

2



Parabola




2

12

2 1

bx

a

y m x x m xm

a m b c m

= minussdot


= = =


12 1 0 2 1 2 1



( )metoda

2komparacij

1

1 121 1

1 2 2

2

e

xm

m m mm

x

= minussdot


=



Rezultat 1

4



A (2 2) B (2 4) C (ndash 2 2) D (ndash 2 4)





12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

12

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica


( )

( )

24

41 4 0 2

0 02 1

0 2

f x x x

a b c x x

bx

a

= minus + sdot


= minussdot


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

2inačica


( )0 02

4 0 4 04 0 4

x xx x x x

x x



13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

13

( ) 1

00 1

4 42

xx

x x

==rArr rArr



0 41 2

0 42

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


( )( ) ( ) ( )

20 2

2 2 4 2 2 4 8 2 424

0 0 0

x

f f ff x x x

=



Odgovor je pod B

Vježba 065


A (4 8) B (4 32) C (4 16) D (4 4)

Rezultat C





y

x

1

10

17 23 27 33A m B m C m D m





( )2





( )2



0 2

bx

a= minus

sdot


14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

14

2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) 20 0

0 0



( )2


je parabola




0 21 2

x xx

+=


( )2



( ) 20 0 0 0 0


1inačica

( )

( )( )

( )

2

24 03 0 182

0

24

03 18

0 4 0303 18 0

4

f x a x b x c

f x x x y

a b

a c by

ac

sdot sdot minus

= sdot + sdot +


sdot minus=

=sdot

minus = =

0 32427

0 012y y

minusrArr = rArr =

minus

Odgovor je pod C

2inačica


( )2 2 2


( )2 2


( )002 2 1

3 18 0 6 0 6 0 6 0 6

2

3xx

x x x x x xx x


minus = =


0 61 2

0 63

0 01 2 20 2

x x

x xx xx

= =+

rArr = rArr =+=


30 2

03 3 18 3 03 9 18 32 0 0

03 180 0 0

x

y yy x x

=


15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

15

27 54 270 0


Odgovor je pod C

Vježba 066




y

x

1

10

34 46 54 66A m B m C m D m

Rezultat C




( ) ( ) ( )2 2

22 2

2

n na a



( )22 2







( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( )

( )

metoda

komparacije

24 4 4 2

4 4 9 16 4 94 9

f b cb c b c

f


= minus


16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

16


( )

12

4

a

f x x b c b b

c c

=

= + sdot + rArr =

=


4 4 40 02 2 1 2

2 2 24 4 1 4

9 9 90 04 4 1 4

b b bx x

a

a c b c b c by y

a




( ) ( )

( )

2

4

48 22

2 24 4 36 3

94

b

b

c b c b

minus == minus


=

sdot

sdotminus

minus

1inačica


( )metoda

supstitucije

4 254 8 25 32 25 25 32 7

8

b cc c c c

b


= minus

2inačica


254 25 4 25 4 25

42 2 2

4 36 4 36 4 36 24 3

6

4c

b c b c b c b

c b c b c bc b



sdot minus = minus

( )22

25254 36 4 36

24

metoda

supstitucije 4

ccc c


( )2 2 2

25 625 50 625 504 36 4 36 4 36

16 1 16

6 16

c c c c cc c c



( )2 2 2


( ) ( )2 22


3inačica


( )metoda

supstitucije

8 24 8 36 4 64 36

24 36

bc c

c b


sdot minus = minus


Vježba 067


Rezultat ndash 8

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

17



( )1 2

5 30 0 124






1

a


d

sdot sdot= sdot = =

sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema




prošla su 2 h


( )( )

metoda

supstitucije

1 25 30 1 2

2 2 5 2 3044

2

T t t tT

t


=

( ) ( ) ( ) ( )1 1 4 1

2 4 10 30 2 10 30 2 10 30 2 14

410 30


( ) ( ) 02 21 2 21 T T CrArr = rArr =


Budući da je

( )1 2

5 304



10

4a a= rArr gt


18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

18

( )1 2 5 5

5 3054 1 1

0 0 01 2 1 212 5 30

4 44

0

1 4

2

bT t t t

t t t

a b

t

ca



sdot sdot= = minus

= minussdot

=

10 10 0 0

t t hrArr = rArr =


1inačica

( ) ( )1 2 1 4 12

5 30 4 30 5 30 254 4 1 4

0 0

24

0 4 1 4 114 5 30

4 1 44

T t t t

Ta c b

Ta

T

a b c

sdot sdot minus=

sdot




130 25

1 30 25 30 25 5 01 5 5 0 0 0 0 0 01 1

4

1

44 1

1 4

T T T T T T C



sdot

2inačica

( )( ) ( )

1 25 30 1 12

4 10 10 5 10 30 10 100 50 304 4

100

T t t tT T

t t


= =

( ) ( ) ( )1 100 100

10 50 30 10 50 30 10 25 50 304 1 4


( ) ( ) 010 5 10 5 T T CrArr = rArr =

Vježba 068


( )1 2

5 30 0 124



Rezultat 2525 degC



52




1

n a c a cn

b d b d

sdot= sdot =

sdot


19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

19

( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

Vrste ekstrema



( )

( )

10

2


2

1 21 2 4 5522 13

11 4 522

10

43

24

0

a

a c by

f x a x b x c

bf x x b x

a

a b b c

y

= sdot + sdot +


= minus lt

sdot sdot minus

sdot

=

4 1 1 2 12 2 25 5 5 2

2 51 2 1 1 113 13 13 13

4 1 1 2 1 2

1 2 1 1 1

4

2

4

2

b b bb




( )2 2

10 10 2 213 13 26 10 10 2 6

22

2

b bb bsdot



2 2


Odgovor je pod D

Vježba 069


52



Rezultat C



( ) ( ) ( ) ( ) 0 2 2 1 1 1 1 2A B C Dminus minus



( )2




20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

20




24

0

0 2 4

b a c bx y

a a


sdot sdot


12 2

2 2 2

0

2

22

a


x cx

b

y

= sdot + sdot +

=

=

minus

minus


+ sdot

rArr =

=


bull ( )

1 2 02 2

10 0 02 1 2

0 2

a b c

x x xbx

a

= minus = =


sdot

bull ( )

( )

1 2 0 24 1 0 2 0 4 4

2 14 0 0 0 04 1 4 40 4

a b c

y y y ya c by

a



sdot


( ) ( ) 1 1 0 0

T x y T=

Odgovor je pod C


2 je točka


Rezultat D



( )2




1 1000 1 3 600km m h s= =


( )2





100072 72 20

3600

km m mx x x



( )( ) ( )

20004 03 2

20 0004 20 03 20 20 7620

f x x xf f

x


=


21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

21


( )2



Rezultat 34 m




funkcije

Rješenje 072

Ponovimo

( )2 2 2



( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot


( ) ( )2


parabola je

( )2




Primjeri

bull ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( ) ( ) ( )2 2


bull ( ) ( ) ( ) ( )( ) ( )22


bull ( ) ( ) ( ) ( )( ) ( ) ( )22


22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

22




( ) 20 0

0 0


Kvadratna funkcija

( )2



0 21 2

x xx

+=


( ) 20 0 0 0 0


1inačica

( )

( ) ( )

212 02 3 2 12 22 4 1 3 2

30

0 2

24

0 4 4 1

bx

a

a c by

a

xaf x x x

b

f x a x b x c c y

= minussdot

sdot sdot minus=



sdotsdot

( ) ( )

2 210 0 02 2

1 4 0 012 4 16 4

00 04 4

x x xT x y T

yy y



2inačica


( ) ( ) ( ) ( )12 2 2


( ) ( ) ( ) ( ) ( ) ( )( ) ( )222



( ) ( )( ) ( )

( ) ( )( ) ( )

21 4 1

0 1 4

0 02 40

0 0

f x x xT x y T

yf x a x x y



3inačica


( )

( )

2 22 3 2 2 3 02 3 0

1 2 30

f x x x x xx x

a b cf x


= = = minus=

T(x0 y0)

ORDINATATJEMENA

APSCISATJEMENA


23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

23

( )1 2 3 2

2 2 4 1 3 2 4 122

4 12 122 1 212 2

a b c

x xb b a c

xa



sdot

2 4 211 12 16 2 4 2 2 1

12 12 2 4 6 32 2

22 22 2

x x xx x

xx x



= = minus


( )1 3

1 2 1 3 1 3 21

0 0 0 01 2 2 2 20 2

x x

x x x xx xx




( )

( )

( )

( )( ) ( )

22 3 2

2 3 21 1 2 1 3

0 01

0

0 0

f x x xf x x x

x yy f

y f x



=

1 2 3 40 0



( ) ( ) 1 4 0 0


Vježba 072



funkcije




( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rješenje 073

Ponovimo

2 0a a a= ge


( )2





( )2


parabola je


24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

24





( ) ( ) ( ) ( )2 2 2


( ) ( ) ( ) ( )2 2

2


311 8 8 11 111 8

11 8 8 11 192

t st tt

t t t s


minus = = + =




19 3 16 2 1


Odgovor je pod C

Vježba 073


( ) ( )2


182 m

4 10 16 22A s B s C s D s

Rezultat C



a = ___

b = ___

c = ___

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

25

yyyy

xxxx11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

Rješenje 074

Ponovimo

( ) ( )2 2



( )2





je parabola 2



( ) 00

f x =






( )( ) ( ) ( )

2


1

2


x x


minus

TTTT

BBBBAAAA

xxxx

yyyy

11110000

- 1- 1- 1- 1

- 1- 1- 1- 1

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

26

1inačica


( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )1 2 1 1 1 1

1 11 2


x x


= minus =


jednadžbu parabole

( ) ( )

( ) ( )( ) ( ) ( )

0 11 0 1 0 1 1 1 1 1 1

1 1

T x y Ta a a a

y a x x


= sdot + sdot minus

Funkcija glasi

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

1 21 1 1 1 1 1

1 1


f x a x x


= sdot + sdot minus


( )

( )

12

0 2

1 1

af x a x b x c

b

f x x c

== sdot + sdot +

rArr =

= minus = minus

2inačica




nepoznanice a b i c

bull

( ) ( )( ) ( )

1 0 21 1 0 0

2

A x y Aa b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 1 0 21 1 0 0

2

B x y Ba b c a b c

y a x b x c


= sdot + sdot +

bull

( ) ( ) 0 1 20 0 1 0 0 1 1

2

T x y Ta b c c c

y a x b x c


= sdot + sdot +

Riješimo sustav



01 0 1

01 nat1 a0

1

a b ca b a b

a b ca b a b

c



= minus


Računamo b

11 1 1 1 0

1

ab b b

a b


+ =


1 0 1a b c= = = minus

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

27

Vježba 074


a = ___

b = ___

c = ___

yyyy

xxxx22220000

- 2- 2- 2- 2

- 2- 2- 2- 2

Rezultat 1

0 22

a b c= = = minus




Rješenje 075

Ponovimo


( ) 00

f x =


( )2




Jednadžba oblika





su brojevi

2 24 4

1 22 2

b b a c b b a cx x

a a


sdot sdot

ili kraće

24

12 2

b b a cx

a


sdot

Tražimo nultočke

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

28

( )

( )( )

22 3 2 2

2 3 0 2 3 00

1f x x x

x x x xf x


=

minus

1 2 32

2 2 3 022 3 0

41 2 3

12 2

a b c

x xx x

b b a ca b c x

a

= = minus = minus



sdot

( ) ( ) ( )2

2 2 4 1 3 2 4 12 2 16

12 12 122 1 2 2x x x


sdot

2 4 631 12 4 2 2 1

12 2 4 2 12

22 22 2

x x xx

xx x

+= = =plusmn


= = minus

Vježba 075



Rezultat 3 51 2

x x= minus =




f(ndash 6) = 14

Rješenje 076

Ponovimo

b a b

ac c

sdotsdot =


jedinice

0 1a n a

n nb n b

sdot= ne ne

sdot


( )2




( )

( )

( )

( )

( ) ( )

2 0 0 40 4

23 4 3 3 4

26 14

0

23

6 46 6 1

x

f x a x b x c x

x

a b cf

f a b c

f a b c

sdot + sdot=

= sdot + sdot + =

= minus

+ = minus= minus



0 0 4 49 3 4 4

9 3 4 9 3 4metoda

za 36 6 4 1436 6 1

mjen4 36 4

e6 1

a b c ca b

a b c a b ca b

a b c a b c




metoda sup9 3 4 4 9 3 0 9 3 0

36 6 14 4 36 6 18

rotnih


2a b a b a b

a b a b a b


sdot


29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

29

1 1818 6 0 18 15

54 544 18 54 18

36 6 18 54 3

a ba a a a a

a b



Računamo b

9 3 01 1

9 3 0 3 0 3 393

0 3 313

3

a b

b b b ba

sdot + sdot =




( )

( )

1 1 4 1 23 4

32

a b cf x x x

f x a x b x c


= sdot + sdot +

Vježba 076



f(ndash 1) = 6

Rezultat ( )2





koeficijenta a

Rješenje 077

Ponovimo

( ) n n n

a b a bsdot = sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot






30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

30

( )

( )

2

2

0

a af x a x b x c

b b

f x a x b x c

== sdot + sdot +

rArr =

= sdot + sdot =


( ) ( ) 2 30 0

2 2 22 2 2

2 2 20 20

4 4 0 03 3 32 4 4 44

0 4

aT x y T b b b

a a abx

a

ba a c b a b b

a a aa c by

a

b

c



=

=

=

2met

2 2 242 2 2

2 2 2 212

oda

supstitucije3 3 3

4 44

4

ab b b

b aa a a

b b b b a

a aa

a



sdot sdot

sdot sdot

( )2 2 2 2


( )nema s02 mi

4 3 0 4 3 0 4 3 04

a

3

l

0

saa a a a a

a


sdot + =

43

4 3 4 3 4


Vježba 077



koeficijenta b

Rezultat 3





31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

31

Rješenje 078

Ponovimo


( )2





( )2


je parabola






točke


jednoj točki


siječe os x


22

0 0 0


x


=





D = 0D = 0D = 0D = 0

D = 0D = 0D = 0D = 0





32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

32

Odgovor je pod A

Vježba 078




Rezultat D



32


x = 2

5 5 3 5


Rješenje 079

Ponovimo

1

a


d


sdot sdot sdot


( )2





( )2



0 2

bx

a= minus

sdot


2

0

4

4

a c by

a

sdot sdot minus=

sdot

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

33

( )

( )

123

2 3 2

1

2

a af x a x x

b

f x a x b x cc


rArr = minus

= sdot + sdot +=


23

3 3 32 2 2 22

2 2 2 22

bx b

aa a

aba a a a

x

=sdot sdot

= minus



33 4 4 3 4 3



( ) ( )3 1 3 1 3 12 2 3 4 3 34 2 4 2 2

2 3 34 44

4

44

4 4

a b c

y ya c b

ya



sdot

3 3 9 3 18 159

15 52 2 1 2 2 3 3 33 6 2

1 1

15

6

1

y y y y y y y



Odgovor je pod B

Vježba 079


32


za x = 2

7 7 3 7


Rezultat C





Rješenje 080

Ponovimo




x x xisin



vrijednosti

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

34


bull 2


1 2 1

b cx x x x

a a+ = minus sdot =

bull 2


2 1 2


1inačica



( )

( )( )

( ) ( )2

11 0 11 11 0 121 11 0

2 25 5 05 0 5 5

2

0

f b cf x x b x c

b c

b cfb c


+ sdot + ==+ sdot +

= + sdot +

=

metoda suprotnih 5

koeficijenata

11 121 11 121

5 25 15 5 12

b c b c

b c b c

sdot

sdot



55 5 60516 880 16 880 55

55 11 16

275

b cc c c

b c


sdot + sdot = minus

Sada imamo

( )( )

22

5555


c


= minus

Za funkciju vrijedi

( )( ) ( )

255 2

0 0 0 55 0 550

f x x b xf b f

x


=

Odgovor je pod A

2inačica



11 51 2

11 5 55

1 2

x xc c

x x c


sdot =

Odgovor je pod A

35

Vježba 080




Rezultat A

35

Vježba 080




Rezultat A

Zadatak 061 (Matea, gimnazija) - · PDF file6 Broj x 0 je nulto čka kvadratne funkcije f ako...

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Transcript of Zadatak 061 (Matea, gimnazija) - · PDF file6 Broj x 0 je nulto čka kvadratne funkcije f ako...