9-3 graphing quadratic functions
TRANSCRIPT
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9-3 Graphing Quadratic Functions9-3 Graphing Quadratic Functions
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Warm UpFind the axis of symmetry.
1. y = 4x2 – 7 2. y = x2 – 3x + 1
3. y = –2x2 + 4x + 3 4. y = –2x2 + 3x – 1
Find the vertex.
5. y = x2 + 4x + 5 6. y = 3x2 + 2
7. y = 2x2 + 2x – 8
x = 0
x = 1
(–2, 1) (0, 2)
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Recall that a y-intercept is the y-coordinate of the point where a graph intersects the y-axis. The x-coordinate of this point is always 0. For a quadratic function written in the form y = ax2 + bx + c, when x = 0, y = c. So the y-intercept of a quadratic function is c.
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In the previous lesson, you found the axis of symmetry and vertex of a parabola. You can use these characteristics, the y-intercept, and symmetry to graph a quadratic function.
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Additional Example 1: Graphing a Quadratic Function
Graph y = 3x2 – 6x + 1.
Step 1 Find the axis of symmetry.
= 1
The axis of symmetry is x = 1. Simplify.
Use x = . Substitute 3
for a and –6 for b.
Step 2 Find the vertex.y = 3x2 – 6x + 1 = 3(1)2 – 6(1) + 1= 3 – 6 + 1= –2
The vertex is (1, –2).
The x-coordinate of the vertex is 1. Substitute 1 for x.
Simplify.The y-coordinate of the vertex is –2.
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Additional Example 1 Continued
Step 3 Find the y-intercept.
y = 3x2 – 6x + 1
y = 3x2 – 6x + 1
The y-intercept is 1; the graph passes through (0, 1).
Identify c.
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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept.
Since the axis of symmetry is x = 1, choose x-values less than 1.
Let x = –1.
y = 3(–1)2 – 6(–1) + 1
= 3 + 6 + 1 = 10
Let x = –2.
y = 3(–2)2 – 6(–2) + 1
= 12 + 12 + 1 = 25
Substitutex-coordinates.
Simplify.
Two other points are (–1, 10) and (–2, 25).
Additional Example 1 Continued
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Graph y = 3x2 – 6x + 1.Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.
Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
Additional Example 1 Continued
x = 1(–2, 25)
(–1, 10)
(0, 1)(1, –2)
x = 1
(–1, 10)
(0, 1)
(1, –2)
(–2, 25)
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Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry as its reflected point.
Helpful Hint
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Check It Out! Example 1a
Graph the quadratic function.
y = 2x2 + 6x + 2
Step 1 Find the axis of symmetry.
Simplify.
Use x = . Substitute 2
for a and 6 for b.
The axis of symmetry is x .
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Step 2 Find the vertex.
y = 2x2 + 6x + 2
Simplify.
Check It Out! Example 1a Continued
= 4 – 9 + 2
= –2
The x-coordinate of the vertex is
. Substitute for
x.
The y-coordinate of the vertex
is .The vertex is .
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Step 3 Find the y-intercept.
y = 2x2 + 6x + 2
y = 2x2 + 6x + 2
The y-intercept is 2; the graph passes through (0, 2).
Identify c.
Check It Out! Example 1a Continued
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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept.
Let x = –1
y = 2(–1)2 + 6(–1) + 1
= 2 – 6 + 2 = –2
Let x = 1
y = 2(1)2 + 6(1) + 2
= 2 + 6 + 2 = 10
Substitutex-coordinates.
Simplify.
Two other points are (–1, –2) and (1, 10).
Check It Out! Example 1a Continued
Since the axis of symmetry is x = –1 , choose x
values greater than –1 .
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Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.
Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
Check It Out! Example 1a Continued
(–1, –2)
(1, 10)
(–1, –2)
(1, 10)
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Check It Out! Example 1b
Graph the quadratic function.
y + 6x = x2 + 9
Step 1 Find the axis of symmetry.
Simplify.
Use x = . Substitute 1
for a and –6 for b.
The axis of symmetry is x = 3.
= 3
y = x2 – 6x + 9 Rewrite in standard form.
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Step 2 Find the vertex.
Simplify.
Check It Out! Example 1b Continued
= 9 – 18 + 9
= 0
The vertex is (3, 0).
The x-coordinate of the vertex is 3. Substitute 3 for x.
The y-coordinate of the vertex is 0.
y = x2 – 6x + 9
y = 32 – 6(3) + 9
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Step 3 Find the y-intercept.
y = x2 – 6x + 9
y = x2 – 6x + 9
The y-intercept is 9; the graph passes through (0, 9).
Identify c.
Check It Out! Example 1b Continued
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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept.
Since the axis of symmetry is x = 3, choose x-values less than 3.
Let x = 2y = 1(2)2 – 6(2) + 9
= 4 – 12 + 9 = 1
Let x = 1 y = 1(1)2 – 6(1) + 9
= 1 – 6 + 9
= 4
Substitutex-coordinates.
Simplify.
Two other points are (2, 1) and (1, 4).
Check It Out! Example 1b Continued
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Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.
Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.
y = x2 – 6x + 9
Check It Out! Example 1b Continued
x = 3
(3, 0)
(0, 9)
(2, 1)
(1, 4)
(0, 9)
(1, 4)
(2, 1)
x = 3
(3, 0)
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Additional Example 2: Problem-Solving Application
The height in feet of a basketball that is thrown can be modeled by f(x) = –16x2 + 32x, where x is the time in seconds after it is thrown. Find the basketball’s maximum height and the time it takes the basketball to reach this height. Then find how long the basketball is in the air.
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Additional Example 2 Continued
11 Understand the Problem
The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the ground.
• The function f(x) = –16x2 + 32x models the height of the basketball after x seconds.
List the important information:
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22 Make a Plan
Find the vertex of the graph because the maximum height of the basketball and the time it takes to reach it are the coordinates of the vertex. The basketball will hit the ground when its height is 0, so find the zeros of the function. You can do this by graphing.
Additional Example 2 Continued
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Solve33
Step 1 Find the axis of symmetry.
Use x = . Substitute
–16 for a and 32 for b.
Simplify.
The axis of symmetry is x = 1.
Additional Example 2 Continued
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Step 2 Find the vertex.
f(x) = –16x2 + 32x
= –16(1)2 + 32(1)
= –16(1) + 32
= –16 + 32
= 16
The vertex is (1, 16).
The x-coordinate of the vertex is 1. Substitute 1 for x.
Simplify.
The y-coordinate of the vertex is 16.
Additional Example 2 Continued
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Step 3 Find the y-intercept.
Identify c.f(x) = –16x2 + 32x + 0
The y-intercept is 0; the graph passes through (0, 0).
Additional Example 2 Continued
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Additional Example 2 Continued
Step 4 Find another point on the same side of the axis of symmetry as the point containing the y-intercept.
Since the axis of symmetry is x = 1, choose an x-value that is less than 1.
Let x = 0.5
f(x) = –16(0.5)2 + 32(0.5)= –4 + 16
= 12Another point is (0.5, 12).
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Step 5 Graph the axis of symmetry, the vertex, and the point containing the y-intercept, and the other point. Then reflect the points across the axis of symmetry. Connect the points with a smooth curve.
Additional Example 2 Continued
(0, 0)
(1, 16)
(2, 0)
(0.5, 12) (1.5, 12)
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The vertex is (1, 16). So at 1 second, the basketball has reached its maximum height of 16 feet. The graph shows the zeros of the function are 0 and 2. At 0 seconds the basketball has not yet been thrown, and at 2 seconds it reaches the ground. The basketball is in the air for 2 seconds.
Additional Example 2 Continued
(0, 0)
(1, 16)
(2, 0)
(0.5, 12) (1.5, 12)
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Look Back44
Check by substituting (1, 16) and (2, 0) into the function.
Additional Example 2 Continued
16 = –16(1)2 + 32(1)
0 = –16(2)2 + 32(2)
16 16 16 –16 + 32
0 0 0 –64 + 64
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Check It Out! Example 2
As Molly dives into her pool, her height in feet above the water can be modeled by the function f(x) = –16x2 + 24x, where x is the time in seconds after she begins diving. Find the maximum height of her dive and the time it takes Molly to reach this height. Then find how long it takes her to reach the pool.
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11 Understand the Problem
The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the pool.
Check It Out! Example 2 Continued
List the important information:
• The function f(x) = –16x2 + 24x models the height of the dive after x seconds.
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22 Make a Plan
Find the vertex of the graph because the maximum height of the dive and the time it takes to reach it are the coordinates of the vertex. The diver will hit the water when its height is 0, so find the zeros of the function. You can do this by graphing.
Check It Out! Example 2 Continued
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Solve33
Step 1 Find the axis of symmetry.
Use x = . Substitute
–16 for a and 24 for b.
Simplify.
The axis of symmetry is x = 0.75.
Check It Out! Example 2 Continued
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Step 2 Find the vertex.
f(x) = –16x2 + 24x
= –16(0.75)2 + 24(0.75)
= –16(0.5625) + 18
= –9 + 18
= 9
The vertex is (0.75, 9).
Simplify.
The y-coordinate of the vertex is 9.
The x-coordinate of the vertex is 0.75. Substitute 0.75 for x.
Check It Out! Example 2 Continued
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Step 3 Find the y-intercept.
Identify c.f(x) = –16x2 + 24x + 0
The y-intercept is 0; the graph passes through (0, 0).
Check It Out! Example 2 Continued
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Step 4 Find another point on the same side of the axis of symmetry as the point containing the y-intercept.
Since the axis of symmetry is x = 0.75, choose an x-value that is less than 0.75.
Let x = 0.5
f(x) = –16(0.5)2 + 24(0.5)
= –4 + 12
= 8 Another point is (0.5, 8).
Substitute 0.5 for x.
Simplify.
Check It Out! Example 2 Continued
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Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and the other point. Then reflect the points across the axis of symmetry. Connect the points with a smooth curve.
(1.5, 0)
(0.75, 9)
(0, 0)
(0.5, 8) (1, 8)
Check It Out! Example 2 Continued
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The vertex is (0.75, 9). So at 0.75 seconds, Molly's dive has reached its maximum height of 9 feet. The graph shows the zeros of the function are 0 and 1.5. At 0 seconds the dive has not begun, and at 1.5 seconds she reaches the pool. Molly reaches the pool in 1.5 seconds.
(1.5, 0)
(0.75, 9)
(0, 0)
(0.5, 8) (1, 8)
Check It Out! Example 2 Continued
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Look Back44
Check by substituting (0.75, 9) and (1.5, 0) into the function.
Check It Out! Example 2 Continued
9 = –16(0.75)2 + 24(0.75)
0 = –16(1.5)2 + 24(1.5)
9 9 9 –9 + 18
0 0 0 –36 + 36
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Lesson Quiz1. Graph y = –2x2 – 8x + 4.
2. The height in feet of a fireworks shell can be modeled by h(t) = –16t2 + 224t, where t is the time in seconds after it is fired. Find the maximum height of the shell, the time it takes to reach its maximum height, and length of time the shell is in the air.
784 ft; 7 s; 14 s