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Graphing Parabolas
Using the VertexAxis of Symmetry
& y-Intercept
By: Jeffrey Bivin
Lake Zurich High School
Last Updated: October 15, 2007
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Graphing Parabolas
y = x2 + 4x - 7
With your graphing calculator, grapheach of the following quadraticequations and identify the vertex andaxis of symmetry.
y = 2x2 + 10x + 4
y = -3x2 + 5x + 9
Jeff Bivin -- LZHS
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Graph the following parabola
y = x2 + 4x - 7
axis of symmetry:
vertex:
2!x
)11,2(
)7,0( y-intercept:
Jeff Bivin -- LZHS
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Graph the following parabola
y = 2x2 + 10x + 4
axis of symmetry:
vertex:
25!x
),( 21725
)4,0(y-intercept:
25
!x
),(217
25
)4,0(
Jeff Bivin -- LZHS
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Graph the following parabola
y = -3x2 + 5x + 9
axis of symmetry:
vertex:
6
5!x
),( 1213365
)9,0(y-intercept:
6
5!x
),( 1213365
)9,0(
Jeff Bivin -- LZHS
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Graphing Parabolas
y = x2 + 4x - 7
Now look at the coefficients of theequation and the value of the axis ofsymmetry especially a and b
y = ax2 + bx + c
y = 2x2 + 10x + 4
y = -3x2 + 5x + 9
2!x
25
!x
6
5!x
Jeff Bivin -- LZHS
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Graphing Parabolas
y = ax2 + bx + c
Vertex:
Axis of symmetry: ab
x 2
!
)(,22 ab
ab f
Jeff Bivin -- LZHS
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Graph the following parabola
y = x2 + 4x - 7
axis of symmetry:
vertex:
224
)1(24
2!!!!
abx
)11,2(
)7,0( y-intercept:
117)2(4)2()2( 2 !!f
Jeff Bivin -- LZHS
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Graph the following parabola
y = 2x2 + 10x + 4
axis of symmetry:
vertex:
25
410
)2(210
2 !!!!
abx
),(217
25
)4,0(y-intercept:
217
252
25
25 4)(10)(2)( !!f
25!x
Jeff Bivin -- LZHS
),(217
25
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Graph the following parabola
y = -3x2 + 5x + 9
axis of symmetry:
vertex:
6
56
5)3(2
52
!!!!
abx
),(12
1336
5
)9,0(y-intercept:
12133
6
52
6
56
5 9)(5)(3)( !!f
6
5!x
),( 1213365
)9,0(
Why did this parabola opendownward instead of upward as
did the previous two?
Jeff Bivin -- LZHS
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Graph the following parabolay = x2 + 6x - 8
Axis of symmetry:
Vertex:
3)1(2 62 !!! abx
178)3(6)3()3(2
!!f
)17,3(
)8,0( y-intercept:Jeff Bivin -- LZHS
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Graph the following parabolay = -2x2 + 7x + 12
Axis of symmetry:
Vertex:
47
47
)2(27
2 !!!!
abx
8145472
4747 12)(7)(2)( !!f
),(8
14547
)12,0(y-intercept:
),(8
14547
47
!x
Jeff Bivin -- LZHS
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Graphing Parabolas
In Vertex Form
Jeff Bivin -- LZHS
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Graphing Parabolas
y = x2
With your graphing calculator, grapheach of the following quadraticequations and identify the vertex andaxis of symmetry.
y = (x - 5)2 - 4
y = -3(x + 2)2 + 5
y = (x - 3)2 + 1
Jeff Bivin -- LZHS
0!x)0,0(
5!x)4,5(
2!x)5,2(
3!x)1,3(
vertex axis of sym.
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Graph the following parabola
y = (x - 5)2 - 4
axis of symmetry:
vertex:
5!x
)4,5(
)21,0(y-intercept:
Jeff Bivin -- LZHS
05 !x
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Graph the following parabola
y = -3(x + 2)2 + 5
axis of symmetry:
vertex:
2!x
)5,2(
)7,0( y-intercept:
Jeff Bivin -- LZHS
02 !x
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Graph the following parabola
y = (x - 3)2 - 1
axis of symmetry:
vertex:
3!x
)1,3(
),0(8
19y-intercept:
Jeff Bivin -- LZHS
03 !x
),0(8
19
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Graphing Parabolas
In Intercept Form
Jeff Bivin -- LZHS
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Graph the following parabola
y = (x 4)(x + 2)
x-intercepts:
vertex:
)8,0( y-intercept:Jeff Bivin -- LZHS
04 !x 02!x
)0,4( )0,2(
axis of symmetry:
2
24 !x
1!x)9,1(
9)3)(3()21)(41( !!!y
8)2)(4()20)(40( !!!y
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Graph the following parabola
y = (x - 1)(x - 9)
x-intercepts:
vertex:
)9,0(y-intercept:Jeff Bivin -- LZHS
01!x 09!x
)0,1( )0,9(
axis of symmetry:
2
91 !x
5!x)16,5(
16)4)(4()95)(15( !!!y
9)9)(1()90)(10( !!!y
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Graph the following parabola
y = -2(x + 1)(x - 5)
x-intercepts:
vertex:
)10,0(y-intercept:Jeff Bivin -- LZHS
01!x 05!x
)0,1( )0,5(
axis of symmetry:
2
51 !x
2!x)18,2(
18)3)(3(2)52)(12(2 !!!y
10)5)(1(2)50)(10(2 !!!y
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Convert to standard form
y = -2(x + 1)(x - 5)
Jeff Bivin -- LZHS
y = -2(x2 5x + 1x 5)
y = -2(x2 4x 5)
y = -2x2 + 8x + 10
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Now graph from standard form.
y = -2x2 + 8x + 10
Axis of symmetry:
Vertex:
248)2(2 82 !!!! abx
1810)2(8)2(2)2(2
!!f
)18,2(
)10,0(y-intercept:Jeff Bivin -- LZHS
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A taxi service operates between two airports transporting200 passengers a day. The charge is $15.00. The ownerestimates that 10 passengers will be lost for each $2increase in the fare. What charge would be most profitablefor the service? What is the maximum income?
Jeff Bivin -- LZHS
Income = Price Quantity
f(x) = ( 15 + 2x ) ( 200 10x )
Define the variable
x = number of $2
price increases15 + 2x = 0 200 10x = 0
25.3781)25.6(10200)25.6(215)25.6( !!f
2x = -15Vertex is:
25.3781,25.6
So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = $27.50
f(x) = income
200 = 10x
215!x x!20
25.6:4
252
240
215
!!!
xsymmetryofaxis
Maximumincome:
VERTEX
27.50 137.50
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A taxi service operates between two airports transporting200 passengers a day. The charge is $15.00. The ownerestimates that 10 passengers will be lost for each $2increase in the fare. What charge would be most profitablefor the service? What is the maximum income?
Jeff Bivin -- LZHS
Income = Price Quantity
f(x) = ( 15 + 2x ) ( 200 10x )
Define the variable
x = number of $2
price increasesf(x) = 3000 150x + 400x 20x2
f(x) = 20x2 + 250x + 3000
VERTEX
a
bx2
!
)20(2250
!x
25.6!x
f(6.25) = 20(6.25)2 + 250(6.25) + 3000
f(6.25) = 3781.25 Vertex is: 25.3781,25.6
So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = $27.50
f(x) = income
Maximum income = f(x) = $3781.25