Copyright © 2014, 2010, 2006 Pearson Education, Inc. 1 Chapter 3 Quadratic Functions and Equations.
Transcript of Copyright © 2014, 2010, 2006 Pearson Education, Inc. 1 Chapter 3 Quadratic Functions and Equations.
2Copyright © 2014, 2010, 2006 Pearson Education, Inc.
Quadratic Functions and
Models♦ Learn basic concepts about quadratic functions
and their graphs.♦ Complete the square and apply the vertex
formula.♦ Graph a quadratic function by hand.♦ Solve applications and model data.♦ Use quadratic regression to model data
(optional)
3.1
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Basic Concepts
Recall that a linear function can be written as f(x) = ax + b (or f(x) = mx + b). The formula for a quadratic function is different from that of a linear function because it contains an x2 term.
f(x) = 3x2 + 3x + 5
g(x) = 5 x2
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Quadratic Function
Let a, b, and c be constants with a ≠ 0. A function represented by
f(x) = ax2 + bx + cis a quadratic function.
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Quadratic Function
The graph of a quadratic function is a parabola—a U shaped graph that opens either upward or downward.
A parabola opens upward if a is positive and opens downward if a is negative.
The highest point on a parabola that opens downward and the lowest point on a parabola that opens upward is called the vertex.
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Quadratic Function
The vertical line passing through the vertex is called the axis of symmetry.
The leading coefficient a controls the width of the parabola. Larger values of |a| result in a narrower parabola, and smaller values of |a| result in a wider parabola.
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Example: Analyzing graphs of quadratic functions
Use the graph of the quadratic function to determine the sign of the leading coefficient, its vertex, and the equation of the axis of symmetry. Give intervals where the function is increasing and where it is decreasing. Give the domain and range.
Solution
Leading coefficient: Opens downward, so a is negative.
Vertex: The vertex is (–2, 5).
Axis of symmetry: Vertical line through the vertex with equation x = –2.
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Example: Analyzing graphs of quadratic functions
Increasing for x ≤ –2.
Decreasing for x ≥ –2.
Domain: All real numbers.
Range: | 5y y
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Vertex Form
The parabolic graph of f(x) = a(x – h)2 + kwith a ≠ 0 has vertex (h, k).Its graph opens upward when a > 0 and opens downward when a < 0.
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Completing the Square
We can convert the general formf(x) = ax2 + bx + c to vertex form by completing the square. If a quadratic expression can be written as
then it is a perfect square trinomial and can be factored as
x2 kx
k
2
2
,
x2 kx
k
2
2
x k
2
2
.
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Example: Converting to Vertex Form
Write the formula f(x) = x2 + 6x – 3 in vertex form by completing the square.Solution Given formula
Subtract 3 from each side.
Let k = 10; add (10/2)2 = 25.
Factor perfect square trinomial.
Subtract 12.
2
2
2
2
2
2
6 3
3 6
3
12 ( 3
9
)
( 3) 12
( 3
9
1
6
) 2
y x x
y x x
y x x
y x
y x
f x x Required form.
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Vertex Formula
The vertex of the graph of f(x) = ax2 + bx + c
with a ≠ 0 is the point , .2 2
b bf
a a
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Example: Converting
Use the vertex formula to write f(x) = 3x2 + 12x + 7 in vertex form.
Solution
Vertex: (–2, –5)
1
2
22
2 3
bx
a
22( 2) 3( ) 1 5722( )f
2( ) 3( 2) 5f x x
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Graphing Quadratic Functions
When sketching a parabola, it is important to determine the vertex, the axis of symmetry, and whether the parabola opens upward or downward.
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Example: Graphing quadratic functions by hand
Graph the quadratic functionSolutionThe formula is not in vertex form, but we can find the vertex.
The y-coordinate of the vertex is:
The vertex is at (–1, 5/2). The axis of symmetry is x = –1, and the parabola opens downward because a = –1/2 is negative
21– 2
2h x x x
x b
2a
1
2 12
1
211 1
21
2– 2
5h
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Example: Graphing quadratic functions by hand
Table of Values Graph: 21– 2
2h x x x
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Applications and Models
Sometimes when a quadratic function f is used in applications, the vertex provides important information. The reason is that the y-coordinate of the vertex is the minimum value of f(x) when its graph opens upward and is the maximum value of f(x) when its graph opens downward.
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Example: Maximizing area
A rancher is fencing a rectangular area for cattle using the straight portion of a river as one side of the rectangle. If the rancher has 2400 feet of fence, find the dimensions of the rectangle that give the maximum area for the cattle.SolutionLet W be the width and L be the length of the rectangle. Because the 2400-foot fence does not go along the river, it follows thatW + L + W = 2400 or L = 2400 – 2W
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Example: Maximizing area
Area of the rectangle equals length times width.
This is a parabola that opens downward, and by the vertex formula, the maximum area occurs when
A LW (2400 2W )W 2400W 2W 2
A 2W 2 2400W
W
b
2a
2400
2( 2)600 feet
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Example: Maximizing area
The corresponding length is
L = 2400 – 2W = 2400 – 2(600) = 1200 feet.
The dimensions are 600 feet by 1200 feet.
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Applications and Models
Another application of quadratic functions occurs in projectile motion, such as when a baseball is hit up in the air. If air resistance is ignored, then the formula
s(t) = –16t2 + v0t + h0 calculates the height s of the object above the ground in feet after t seconds. In this formula h0 represents the initial height of the object in feet and v0 represents its initial vertical velocity in feet per second. If the initial velocity is upward, then v0 > 0 and if the initial velocity is downward, then v0 < 0.
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Example: Modeling
A baseball is hit straight up with an initial velocity of v0 = 80 feet per second (or about 55 miles per hour) and leaves the bat with an initial height of h0 = 3 feet, a) Write a formula s(t) that models the height of the baseball after t seconds.b) How high is the baseball after
2 seconds?c) Find the maximum height of
the baseball. Support your answer graphically.
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Example: Modeling
a)
b) Baseball is 99 feet high after 2 seconds.
c) Because a is negative, the vertex is the highest point on the graph, with a t-coordinate of
s(t) 16t 2 v0t h
0
16t 2 80t 3
s(2) 16(2)2 80(2) 3 99
t
b
2a
80
2( 16)2.5