Download Solve each equation by graphing. 1. x + 3x − 10 = 0 SOLUTION

9-2 Solving Quadratic Equations by Graphing
Solve each equation by graphing.
2
1. x + 3x − 10 = 0
SOLUTION: 2
Graph the related function f (x) = x + 3x − 10.
The x-intercepts appear to be at 2 and –5, so the solutions are 2 and –5.
Check:
2
2. 2x − 8x = 0
SOLUTION: 2
Graph the related function f (x) = 2x − 8x.
The x-intercepts appear to be at 0 and 4, so the solutions are 0 and 4.
Check:
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Page 1
9-2 Solving Quadratic Equations by Graphing
2
2. 2x − 8x = 0
SOLUTION: 2
Graph the related function f (x) = 2x − 8x.
The x-intercepts appear to be at 0 and 4, so the solutions are 0 and 4.
Check:
2
3. x + 4x = −4
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 4x + 4.
Notice that the vertex of the parabola is the only x-intercept, therefore there is only one solution, –2.
Check:
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2
4. x + 12 = −8x
Page 2
9-2 Solving Quadratic Equations by Graphing
2
3. x + 4x = −4
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 4x + 4.
Notice that the vertex of the parabola is the only x-intercept, therefore there is only one solution, –2.
Check:
2
4. x + 12 = −8x
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 8x + 12.
The x-intercepts appear to be at –6 and –2, so the solutions are –6 and –2.
Check:
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Page 3
9-2 Solving Quadratic Equations by Graphing
2
4. x + 12 = −8x
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 8x + 12.
The x-intercepts appear to be at –6 and –2, so the solutions are –6 and –2.
Check:
Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest
tenth.
2
5. −x − 5x + 1 = 0
SOLUTION: 2
Graph the related function f (x) = −x − 5x + 1. The x-intercepts are located between –6 and –5 and between 0 and 1. Make a table using an increment of 0.1 for
the x-values located between –6 and –5 and between 0 and 1.
–5.6 –5.5 –5.4 –5.3
–
–
–
–
eSolutions Manual
2.36- Powered
1.75 by Cognero
1.16 0.59
x
y
–5.2
–
0.04
–5.1
0.49
Page 4
x
0.1
0.2
0.3
0.4
0.5
0.6
9-2 Solving Quadratic Equations by Graphing
Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest
tenth.
2
5. −x − 5x + 1 = 0
SOLUTION: 2
Graph the related function f (x) = −x − 5x + 1. The x-intercepts are located between –6 and –5 and between 0 and 1. Make a table using an increment of 0.1 for
the x-values located between –6 and –5 and between 0 and 1.
x
y
–5.6
–
2.36
x
y
0.1
0.49
–5.5
–
1.75
–5.4
–
1.16
–5.3
–
0.59
–5.2
–
0.04
–5.1
0.49
0.2
0.3
0.4
0.5
0.6
–
–
–
–
–
0.04 0.59
1.16
1.75
2.36
For each table, the function value that is closest to zero when the sign change is –0.04. Thus, the roots are
approximately –5.2 and 0.2.
6. −9 = x
2
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 9.
This graph has no x-intercepts. Therefore, this equation has no real number solutions. The solution set is {∅}.
Check: There are no factors of 9 that have a sum of 0, so the expression is not factorable. Thus, the equation has no
real number solutions.
2
7. x = 25
SOLUTION: eSolutions
Manual - Powered by Cognero
Write the equation in standard form.
Page 5
This graph has no x-intercepts. Therefore, this equation has no real number solutions. The solution set is {∅}.
9-2 Solving
Quadratic
Graphing
Check: There
are noEquations
factors of 9by
that
have a sum of 0, so the expression is not factorable. Thus, the equation has no
real number solutions.
2
7. x = 25
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 25.
The x-intercepts appear to be at –5 and 5, so the solutions are –5 and 5.
Check:
2
8. x − 8x = −9
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 8x + 9. The x-intercepts are located between 1 and 2 and between 6 and 7. Make a table using an increment of 0.1 for the
x-values located between 1 and 2 and between 6 and 7.
x
y
1.1
1.41
1.2
0.84
1.3
0.29
1.4
–0.24
1.5
–0.75
1.6
–1.24
x
6.3
6.4
6.5
6.6
6.7
6.8
y
–1.71
–1.24
–0.75
–0.24
0.29
0.84
For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the roots are
approximately 6.6 and 1.4.
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9. SCIENCE FAIR Ricky built a model rocket. Its flight can be modeled by the equation shown, where h is the
height of the rocket in feet after t seconds. About how long was Ricky’s rocket in the air?
Page 6
9-2 Solving Quadratic Equations by Graphing
2
8. x − 8x = −9
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 8x + 9. The x-intercepts are located between 1 and 2 and between 6 and 7. Make a table using an increment of 0.1 for the
x-values located between 1 and 2 and between 6 and 7.
x
y
1.1
1.41
1.2
0.84
1.3
0.29
1.4
–0.24
1.5
–0.75
1.6
–1.24
x
6.3
6.4
6.5
6.6
6.7
6.8
y
–1.71
–1.24
–0.75
–0.24
0.29
0.84
For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the roots are
approximately 6.6 and 1.4.
9. SCIENCE FAIR Ricky built a model rocket. Its flight can be modeled by the equation shown, where h is the
height of the rocket in feet after t seconds. About how long was Ricky’s rocket in the air?
SOLUTION: 2
You will need to find the roots of -16t + 135t = 0. Use a graphing calculator to graph the related function f (x) = 2
16t + 135t.Use the WINDOW option to adjust the viewing window. Select the zero option from the 2nd [CALC]
menu to find the roots.
The positive x-intercept is about 8.4.
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Manual
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by the
Cognero
So, the
rocket
was in
air for
about 8.4 seconds.
Solve each equation by graphing.
Page 7
x
6.3
6.4
6.5
6.6
6.7
6.8
y
–1.71
–1.24
–0.75
–0.24
0.29
0.84
9-2 Solving
Equations
by that
Graphing
For eachQuadratic
table, the function
value
is closest to zero when the sign changes is –0.24. Thus, the roots are
approximately 6.6 and 1.4.
9. SCIENCE FAIR Ricky built a model rocket. Its flight can be modeled by the equation shown, where h is the
height of the rocket in feet after t seconds. About how long was Ricky’s rocket in the air?
SOLUTION: 2
You will need to find the roots of -16t + 135t = 0. Use a graphing calculator to graph the related function f (x) = 2
16t + 135t.Use the WINDOW option to adjust the viewing window. Select the zero option from the 2nd [CALC]
menu to find the roots.
The positive x-intercept is about 8.4.
So, the rocket was in the air for about 8.4 seconds.
Solve each equation by graphing.
2
10. x + 7x + 14 = 0
SOLUTION: 2
Graph the related function f (x) = x + 7x + 14.
This graph has no x-intercepts. Therefore, this equation has no real number solutions. The solution set is {∅}.
2
11. x + 2x − 24 = 0
SOLUTION: 2
Graph the related function f (x) = x + 2x − 24.
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Page 8
This graph has no x-intercepts. Therefore, this equation has no real number solutions. The solution set is {∅}.
9-2 Solving
Quadratic Equations by Graphing
2
11. x + 2x − 24 = 0
SOLUTION: 2
Graph the related function f (x) = x + 2x − 24.
The x-intercepts appear to be at –6 and 4, so the solutions are –6 and 4.
Check:
2
12. x − 16x + 64 = 0
SOLUTION: 2
Graph the related function f (x) = x − 16x + 64.
Notice that the vertex of the parabola is the only x-intercept, therefore there is only one solution, 8.
Check:
2
13. x −Manual
5x + 12
= 0 by Cognero
eSolutions
- Powered
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SOLUTION: 2
Graph the related function f (x) = x − 5x + 12.
9-2 Solving Quadratic Equations by Graphing
2
13. x − 5x + 12 = 0
SOLUTION: 2
Graph the related function f (x) = x − 5x + 12.
This graph has no x-intercepts. Therefore, this equation has no real number solutions. The solution set is {∅}.
2
14. x + 14x = −49
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 14x + 49.
Notice that the vertex of the parabola is the only x-intercept, therefore there is only one solution, –7.
Check:
2
15. x = 2x − 1
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 2x + 1.
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Page 10
9-2 Solving Quadratic Equations by Graphing
2
15. x = 2x − 1
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 2x + 1.
Notice that the vertex of the parabola is the only x-intercept, therefore there is only one solution, 1.
Check:
2
16. x − 10x = −16
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 10x + 16.
The x-intercepts appear to be at 2 and 8, so the solutions are 2 and 8.
Check:
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Page 11
9-2 Solving Quadratic Equations by Graphing
2
16. x − 10x = −16
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 10x + 16.
The x-intercepts appear to be at 2 and 8, so the solutions are 2 and 8.
Check:
2
17. −2x − 8x = 13
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = –2x – 8x – 13.
This graph has no x-intercepts. Therefore, this equation has no real number solutions. The solution set is {∅}.
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2
18. 2x − 16x = −30
Page 12
9-2 Solving Quadratic Equations by Graphing
2
17. −2x − 8x = 13
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = –2x – 8x – 13.
This graph has no x-intercepts. Therefore, this equation has no real number solutions. The solution set is {∅}.
2
18. 2x − 16x = −30
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = 2x – 16x + 30.
The x-intercepts appear to be at 3 and 5, so the solutions are 3 and 5.
Check:
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2
19. 2x = −24x − 72
SOLUTION: Page 13
This graph
has no x-intercepts.
Therefore,
this equation has no real number solutions. The solution set is {∅}.
9-2 Solving
Quadratic
Equations by
Graphing
2
18. 2x − 16x = −30
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = 2x – 16x + 30.
The x-intercepts appear to be at 3 and 5, so the solutions are 3 and 5.
Check:
2
19. 2x = −24x − 72
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = 2x + 24x + 72.
Notice that the vertex of the parabola is the only x-intercept, therefore there is only one solution, –6.
Check:
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Page 14
9-2 Solving Quadratic Equations by Graphing
2
19. 2x = −24x − 72
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = 2x + 24x + 72.
Notice that the vertex of the parabola is the only x-intercept, therefore there is only one solution, –6.
Check:
2
20. −3x + 2x = 15
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = –3x + 2x – 15.
This graph has no x-intercepts. Therefore, this equation has no real number solutions. The solution set is {∅}.
2
21. x = −2x + 80
SOLUTION: Write the equation in standard form.
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Manual
- Powered
by Cognero
Graph
the related
function
f (x)
2
= x + 2x – 80.
Page 15
This graph
has no x-intercepts.
Therefore,
this equation has no real number solutions. The solution set is {∅}.
9-2 Solving
Quadratic
Equations by
Graphing
2
21. x = −2x + 80
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 2x – 80.
The x-intercepts appear to be at –10 and 8, so the solutions are –10 and 8.
Check:
Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest
tenth.
2
22. x + 2x − 9 = 0
SOLUTION: 2
Graph the related function f (x) = x + 2x – 9.
The x-intercepts are located between –5 and –4 and between 2 and 3. Make a table using an increment of 0.1 for
the x-values located between –5 and –4 and between 2 and 3.
x
y
–4.4
1.56
x
2.0
–4.3
0.89
–4.2
0.24
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–4.1
–0.39
–4.0
–1
2.1
2.2
2.3
2.4
Page 16
9-2 Solving Quadratic Equations by Graphing
Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest
tenth.
2
22. x + 2x − 9 = 0
SOLUTION: 2
Graph the related function f (x) = x + 2x – 9.
The x-intercepts are located between –5 and –4 and between 2 and 3. Make a table using an increment of 0.1 for
the x-values located between –5 and –4 and between 2 and 3.
x
y
–4.4
1.56
–4.3
0.89
–4.2
0.24
–4.1
–0.39
–4.0
–1
x
2.0
2.1
2.2
2.3
2.4
y
–1
–0.39
0.24
0.89
1.56
For each table, the function value that is closest to zero when the sign changes is 0.24. Thus, the roots are
approximately –4.2 and 2.2.
2
23. x − 4x = 20
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 4x – 20.
The x-intercepts are located between –2 and –3 and between 6 and 7. Make a table using an increment of 0.1 for
the x-values located between –2 and –3 and between 6 and 7.
x
y
–3.0
1
–2.9
0.01
–2.8
–0.96
–2.7
–1.91
–2.6
–2.84
x
6.6
6.7
6.8
6.9
7.0
y
–2.84
–1.91
–0.96
0.01
1
eSolutions Manual - Powered by Cognero
For each table, the function value that is closest to zero when the sign changes is 0.01. Thus, the roots are
approximately –2.9 and 6.9.
Page 17
x
2.0
2.1
2.2
2.3
2.4
y
–1
–0.39
0.24
0.89
1.56
For each table, the function value that is closest to zero when the sign changes is 0.24. Thus, the roots are
9-2 Solving
Quadratic
approximately
–4.2 Equations
and 2.2. by Graphing
2
23. x − 4x = 20
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x – 4x – 20.
The x-intercepts are located between –2 and –3 and between 6 and 7. Make a table using an increment of 0.1 for
the x-values located between –2 and –3 and between 6 and 7.
x
y
–3.0
1
–2.9
0.01
–2.8
–0.96
–2.7
–1.91
–2.6
–2.84
x
6.6
6.7
6.8
6.9
7.0
y
–2.84
–1.91
–0.96
0.01
1
For each table, the function value that is closest to zero when the sign changes is 0.01. Thus, the roots are
approximately –2.9 and 6.9.
2
24. x + 3x = 18
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 3x – 18.
The x-intercepts appear to be at –6 and 3, so the solutions are –6 and 3.
Check:
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Page 18
x
6.6
6.7
6.8
6.9
7.0
y
–2.84
–1.91
–0.96
0.01
1
9-2 Solving
Equations
by that
Graphing
For eachQuadratic
table, the function
value
is closest to zero when the sign changes is 0.01. Thus, the roots are
approximately –2.9 and 6.9.
2
24. x + 3x = 18
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 3x – 18.
The x-intercepts appear to be at –6 and 3, so the solutions are –6 and 3.
Check:
2
25. 2x − 9x = −8
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = 2x – 9x + 8. The x-intercepts are located between 1 and 2 and between 3 and 4. Make a table using an increment of 0.1 for the
Page 19
x-values located between 1 and 2 and between 3 and 4.
eSolutions Manual - Powered by Cognero
x
1.1
1.2
1.3
1.4
1.5
9-2 Solving Quadratic Equations by Graphing
2
25. 2x − 9x = −8
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = 2x – 9x + 8. The x-intercepts are located between 1 and 2 and between 3 and 4. Make a table using an increment of 0.1 for the
x-values located between 1 and 2 and between 3 and 4.
x
y
1.1
0.52
1.2
0.08
1.3
–0.32
1.4
–0.68
1.5
–1
x 3.1
3.2
3.3 3.4 3.5
y –0.68 –0.32 0.08 0.52 1
For each table, the function value that is closest to zero when the sign changes is 0.08. Thus, the roots are
approximately 1.2 and 3.3.
2
26. 3x = −2x + 7
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = 3x + 2x – 7.
The x-intercepts are located between –2 and –1 and between 1 and 2. Make a table using an increment of 0.1 for
the x-values located between –2 and –1 and between 1 and 2.
x
y
–2.0
1
–1.9
0.03
–1.8
–0.88
–1.7
–1.73
–1.6
–2.52
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Page 20
x
y
1.1
–1.17
1.2
–0.28
1.3
0.67
1.4
1.68
1.5
2.75
x 3.1
3.2
3.3 3.4 3.5
y –0.68 –0.32 0.08 0.52 1
9-2 Solving
Equations
by that
Graphing
For eachQuadratic
table, the function
value
is closest to zero when the sign changes is 0.08. Thus, the roots are
approximately 1.2 and 3.3.
2
26. 3x = −2x + 7
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = 3x + 2x – 7.
The x-intercepts are located between –2 and –1 and between 1 and 2. Make a table using an increment of 0.1 for
the x-values located between –2 and –1 and between 1 and 2.
x
y
–2.0
1
–1.9
0.03
–1.8
–0.88
–1.7
–1.73
–1.6
–2.52
x
1.1
1.2
1.3
1.4
1.5
y
–1.17
–0.28
0.67
1.68
2.75
For the first table, the function value that is closest to zero when the sign changes is 0.03. For the second table, the
function value that is closest to zero when the sign changes is –0.28. Thus, the roots are approximately –1.9 and 1.2.
27. 5x = 25 − x
2
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 5x – 25.
The x-intercepts are located between –9 and –8 and between 3 and 4. Make a table using an increment of 0.1 for
the x-values located between –8 and –7 and between 3 and 4.
x
y
–8.4
3.56
–8.3
2.39
–8.2
1.24
–8.1
0.11
–8.0
–1
x
3.0
3.1
3.2
3.3
3.4
y
–1
0.11
1.24
2.39
3.56
For each table, the function value that is closest to zero when the sign changes is 0.11. Thus, the roots are
approximately –8.1 and 3.1.
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Page 21
x
1.1
1.2
1.3
1.4
1.5
y
–1.17
–0.28
0.67
1.68
2.75
9-2 Solving
Quadratic
Equations
by Graphing
For the first
table, the
function value
that is closest to zero when the sign changes is 0.03. For the second table, the
function value that is closest to zero when the sign changes is –0.28. Thus, the roots are approximately –1.9 and 1.2.
27. 5x = 25 − x
2
SOLUTION: Write the equation in standard form.
2
Graph the related function f (x) = x + 5x – 25.
The x-intercepts are located between –9 and –8 and between 3 and 4. Make a table using an increment of 0.1 for
the x-values located between –8 and –7 and between 3 and 4.
x
y
–8.4
3.56
–8.3
2.39
–8.2
1.24
–8.1
0.11
–8.0
–1
x
3.0
3.1
3.2
3.3
3.4
y
–1
0.11
1.24
2.39
3.56
For each table, the function value that is closest to zero when the sign changes is 0.11. Thus, the roots are
approximately –8.1 and 3.1.
2
28. SOFTBALL The equation h = −16t + 47t + 3 models the height h, in feet, of a ball that Sofia hits after t seconds.
How long is the ball in the air?
SOLUTION: You need to find the roots of the equation
2
function f (x) = -16t + 47t + 3.
. Use a graphing calculator to graph the related
The positive x-intercept of the graph is at 3. Therefore, the ball is in the air for 3 seconds.
2
29. RIDES The Terror Tower launches riders straight up and returns straight down. The equation h = –16t + 122t
models the height h, in feet, of the riders from their starting position after t seconds. How long is it until the riders
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Page 22
return
to the
bottom?
SOLUTION: 2
The positive x-intercept of the graph is at 3. Therefore,
the ball is
in the air by
for Graphing
3 seconds.
9-2 Solving
Quadratic
Equations
2
29. RIDES The Terror Tower launches riders straight up and returns straight down. The equation h = –16t + 122t
models the height h, in feet, of the riders from their starting position after t seconds. How long is it until the riders
return to the bottom?
SOLUTION: 2
You will need to find the roots of the equation –16t + 122t = 0. Use a graphing calculator to graph the related
2
function f (x) = –16t + 122t.
The positive x-intercept of the graph is approximately 7.6.
Therefore, the riders return to the bottom in about 7.6 seconds.
Use factoring to determine how many times the graph of each function intersects the x-axis. Identify each
zero.
2
30. y = x − 8x + 16
SOLUTION: Because there is only one root, the graph of the function intersects the x-axis one time. The root is 4.
2
31. y = x + 4x + 4
SOLUTION: Because there is only one root, the graph intersects the x-axis one time. The root is –2.
2
32. y = x + 2x − 24
SOLUTION: The graph of the function intersects the x-axis two times. The roots are –6 and 4.
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2
33. y = x + 12x + 32
Page 23
9-2 Solving Quadratic Equations by Graphing
Because there is only one root, the graph intersects the x-axis one time. The root is –2.
2
32. y = x + 2x − 24
SOLUTION: The graph of the function intersects the x-axis two times. The roots are –6 and 4.
2
33. y = x + 12x + 32
SOLUTION: The graph of the function intersects the x-axis two times. The roots are –4 and –8.
34. NUMBER THEORY Use a quadratic equation to find two numbers that have a sum of 9 and a product of 20.
SOLUTION: Let x and y be the numbers. Then x + y = 9 and xy = 20. Use substitution to combine the system into one equation. y = 9 – x Substitute into the second equation and solve for x. 4+5=9
4 • 5 = 20
The two numbers are 4 and 5.
35. NUMBER THEORY Use a quadratic equation to find two numbers that have a sum of 1 and a product of −12.
SOLUTION: Let x and y be the numbers. Then x + y = 1 and xy = –12.
Use substitution to solve the system. y=1–x
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4+5=9
9-2 Solving
Quadratic Equations by Graphing
4 • 5 = 20
The two numbers are 4 and 5.
35. NUMBER THEORY Use a quadratic equation to find two numbers that have a sum of 1 and a product of −12.
SOLUTION: Let x and y be the numbers. Then x + y = 1 and xy = –12.
Use substitution to solve the system. y=1–x
4 + (–3) = 1
4 • (–3) = –12
The two numbers are –3 and 4.
2
36. CCSS MODELING The height of a golf ball in the air can be modeled by the equation h = −16t + 72t, where h is
the height in feet of the ball after t seconds.
a. How long was the ball in the air?
b. What is the ball’s maximum height?
c. When will the ball reach its maximum height?
SOLUTION: 2
a. Use a graphing calculator to graph the related function f (x) = 16t +72t. Use the WINDOW option to adjust the
viewing window. Select the Zero option from the 2nd [CALC] menu to find the positive x-intercept.
The positive x-intercept is 4.5. So, the ball was in the air for 4.5 seconds.
b. Select the Maximum option from the 2nd [CALC] menu to find the vertex of the parabola.
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Page 25
The vertex of the graph is about (2.2, 81). Since the y-value of the vertex is 81, the maximum height of the ball is 81
feet.
4 + (–3) = 1
9-2 Solving
Equations by Graphing
4 • (–3) Quadratic
= –12
The two numbers are –3 and 4.
2
36. CCSS MODELING The height of a golf ball in the air can be modeled by the equation h = −16t + 72t, where h is
the height in feet of the ball after t seconds.
a. How long was the ball in the air?
b. What is the ball’s maximum height?
c. When will the ball reach its maximum height?
SOLUTION: 2
a. Use a graphing calculator to graph the related function f (x) = 16t +72t. Use the WINDOW option to adjust the
viewing window. Select the Zero option from the 2nd [CALC] menu to find the positive x-intercept.
The positive x-intercept is 4.5. So, the ball was in the air for 4.5 seconds.
b. Select the Maximum option from the 2nd [CALC] menu to find the vertex of the parabola.
The vertex of the graph is about (2.2, 81). Since the y-value of the vertex is 81, the maximum height of the ball is 81
feet.
c. The x-value of the vertex is about 2.2. So, the ball reached its maximum height at about 2.2 seconds.
2
37. SKIING Stefanie is in a freestyle aerial competition. The equation h = −16t + 30t + 10 models Stefanie’s height h,
in feet, t seconds after leaving the ramp.
a. How long is Stefanie in the air?
b. When will Stefanie reach a height of 15 feet?
c. To earn bonus points in the competition, you must reach a height of 20 feet. Will Stefanie earn bonus points?
SOLUTION: 2
a. Use a graphing calculator to graph the related function f (x) = -16t + 30t + 10. Use the WINDOW option to
adjust the viewing window. Select the zero option from the 2nd [CALC] menu to find the roots.
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Page 26
The vertex of the graph is about (2.2, 81). Since the y-value of the vertex is 81, the maximum height of the ball is 81
feet.
9-2 Solving
Quadratic Equations by Graphing
c. The x-value of the vertex is about 2.2. So, the ball reached its maximum height at about 2.2 seconds.
2
37. SKIING Stefanie is in a freestyle aerial competition. The equation h = −16t + 30t + 10 models Stefanie’s height h,
in feet, t seconds after leaving the ramp.
a. How long is Stefanie in the air?
b. When will Stefanie reach a height of 15 feet?
c. To earn bonus points in the competition, you must reach a height of 20 feet. Will Stefanie earn bonus points?
SOLUTION: 2
a. Use a graphing calculator to graph the related function f (x) = -16t + 30t + 10. Use the WINDOW option to
adjust the viewing window. Select the zero option from the 2nd [CALC] menu to find the roots.
The positive x-intercept is about 2.2. So, Stefanie is in the air for about 2.2 seconds.
b. You will need to find the roots of the equation -16t2 + 30t + 10 = 15. Use a graphing calculator to graph the
2
related function f (x) = -16t + 30t - 5. Select the zero option from the 2nd [CALC] menu to find the roots.
The roots are about 0.2 and 1.7. So, Stefanie will reach a height of 15 feet in about 0.2 seconds and 1.7 seconds.
c. Graph the related function f (x) = -16t2 + 30t + 10. Select the maximum option from the 2nd [CALC] menu to
find the maximum of the graph.
The maximum of the graph is about 24, so she will reach a height of 20 feet on the way up and down. Therefore,
Stefanie will earn bonus points. 38. MULTIPLE REPRESENTATIONS In this problem, you will explore how to further interpret the relationship
between quadratic functions and graphs.
2
a. GRAPHICAL Graph y = x .
b. ANALYTICAL Cognero
the vertex and two other points on the graph.
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c. GRAPHICAL Graph y = x2 + 2, y = x2 + 4, and y = x2 + 6 on the same coordinate plane as the previous graph.
d. ANALYTICAL Name the vertex and two points from each of these graphs that have the same x-coordinates as
9-2 Solving
Quadratic
Equations
by Graphing
The maximum
of the
graph is about
24, so she will reach a height of 20 feet on the way up and down. Therefore,
Stefanie will earn bonus points. 38. MULTIPLE REPRESENTATIONS In this problem, you will explore how to further interpret the relationship
between quadratic functions and graphs.
2
a. GRAPHICAL Graph y = x .
b. ANALYTICAL Name the vertex and two other points on the graph.
c. GRAPHICAL Graph y = x2 + 2, y = x2 + 4, and y = x2 + 6 on the same coordinate plane as the previous graph.
d. ANALYTICAL Name the vertex and two points from each of these graphs that have the same x-coordinates as
the first graph.
e. ANALYTICAL What conclusion can you draw from this?
SOLUTION: a.
b.The vertex is at (0, 0). Two other points that can be seen on the graph include (1, 1) and (−1, 1).
c.
2
d. The vertex of y = x + 2 is (0, 2). Two other points that can be seen on the graph include (1, 3) and (−1, 3).
2
The vertex of y = x + 4 is (0, 4). Two other points that can be seen on the graph include (1, 5) and (−1, 5).
2
The vertex of y = x + 6 is (0, 6). Two other points that can be seen on the graph include (1, 7) and (−1, 7).
e. x = −1
x=1
2
y =1
y =1
y=x +2
2
y =3
y =3
2
y =5
y =5
y=x
y=x +4
y
y=x +6 =
7
2
y =7
2
2
2
2
The graphs of y = x + 2, y = x + 4, and y = x + 6 can be obtained by moving the graph of y = x straight up 2
units, 4 units, and 6 units, respectively.
GRAPHING CALCULATOR Solve each equation by graphing.
3
2
39. x − 3x − 6x + 8 = 0
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Page 28
SOLUTION: 3
2
Use a graphing calculator to graph f (x) = x − 3x − 6x + 8 and find the roots.
2
y=x +6 =
y =7
7
2
2
2
2
9-2 Solving
Quadratic
The graphs
of y = xEquations
+ 2, y = x by
+ Graphing
4, and y = x + 6 can be obtained by moving the graph of y = x straight up 2
units, 4 units, and 6 units, respectively.
GRAPHING CALCULATOR Solve each equation by graphing.
3
2
39. x − 3x − 6x + 8 = 0
SOLUTION: 3
2
3
2
Use a graphing calculator to graph f (x) = x − 3x − 6x + 8 and find the roots.
Let Y1 = x3 − 3x2 − 6x + 8. Use the WINDOW option to adjust the viewing window. Select the Zero option from
the 2nd [CALC] menu to find the roots. [-10, 10] scl: 1 by [-8, 12] scl: 1
[-10, 10] scl: 1 by [-8, 12] scl: 1
[-10, 10] scl: 1 by [-8, 12] scl: 1
The roots are –2, 1, and 4.
3
2
40. x − 8x + 15x = 0
SOLUTION: Use a graphing calculator to graph g(x) = x − 8x + 15x and find the roots.
Let Y1 = x3 − 8x2 + 15x. Use the WINDOW option to adjust the viewing window. Select the Zero option from
the 2nd [CALC] menu to find the roots. eSolutions Manual - Powered by Cognero
Page 29
[-10, 10] scl: 1 by [-8, 12] scl: 1
9-2 Solving
Quadratic
by Graphing
The roots
are –2, 1,Equations
and 4.
3
2
40. x − 8x + 15x = 0
SOLUTION: 3
2
Use a graphing calculator to graph g(x) = x − 8x + 15x and find the roots.
Let Y1 = x3 − 8x2 + 15x. Use the WINDOW option to adjust the viewing window. Select the Zero option from
the 2nd [CALC] menu to find the roots. The roots are 0, 3 and 5.
41. CCSS STRUCTURE Iku and Zachary are finding the number of real zeros of the function graphed. Iku says that
the function has no real zeros because there are no x-intercepts. Zachary says that the function has one real zero
because the graph has a y-intercept. Is either of them correct? Explain your reasoning.
SOLUTION: Iku; sample answer: The zeros of a quadratic function are the x-intercepts of the graph. Since the graph does not
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Manual
- Powered
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Page 30
intersect
the
x-axis,bythere
are no x-intercepts and no real zeros.
42. OPEN ENDED Describe a real-world situation in which a thrown object travels in the air. Write an equation that
9-2 Solving
Quadratic Equations by Graphing
The roots are 0, 3 and 5.
41. CCSS STRUCTURE Iku and Zachary are finding the number of real zeros of the function graphed. Iku says that
the function has no real zeros because there are no x-intercepts. Zachary says that the function has one real zero
because the graph has a y-intercept. Is either of them correct? Explain your reasoning.
SOLUTION: Iku; sample answer: The zeros of a quadratic function are the x-intercepts of the graph. Since the graph does not
intersect the x-axis, there are no x-intercepts and no real zeros.
42. OPEN ENDED Describe a real-world situation in which a thrown object travels in the air. Write an equation that
models the height of the object with respect to time, and determine how long the object travels in the air.
SOLUTION: 2
A tennis ball being hit in the air; an equation is h = −16t + 25t + 2.
Use a graphing calculator to find the zeros. [-2, 3] scl: 0.5 by [-5, 15] scl: 2
[-2, 3] scl: 0.5 by [-5, 15] scl: 2
Time can not be negative, so -0.76 is not a possible solution. Thus, the ball is in the air for about 1.6 seconds.
43. REASONING The graph shown is that of a quadratic inequality. Analyze the graph, and determine whether the
y-value of a solution of the inequality is sometimes, always, or never greater than 2. Explain.
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SOLUTION: Page 31
[-2, 3] scl: 0.5 by [-5, 15] scl: 2
9-2 Solving
Quadratic Equations by Graphing
Time can not be negative, so -0.76 is not a possible solution. Thus, the ball is in the air for about 1.6 seconds.
43. REASONING The graph shown is that of a quadratic inequality. Analyze the graph, and determine whether the
y-value of a solution of the inequality is sometimes, always, or never greater than 2. Explain.
SOLUTION: Within the shaded region of the graph, some y-values are less than 2 and some y-values are greater than 2. So, the
y-value of a solution of the inequality is sometimes greater than 2.
44. CHALLENGE Write a quadratic equation that has the roots described.
a. one double root
b. one rational (nonintegral) root and one integral root
c. two distinct integral roots that are additive opposites.
SOLUTION: a. A double root occurs when the curves vertex is on the x-axis. Consider x2 + 8x + 16 = 0 or (x + 4)2 = 0. There
is a double root at x = −4. [-8, 2] scl: 1 by [-2.5, 2.5] scl: 0.5
2
x + 8x + 16 = 0
b. To have one rational root and one integral root, find two binomials where one equals a integer and there other a
rational number. Consider 2x2 − 23x + 45 = 0 or (2x − 5) (x − 9) = 0. The zeros are 9 and .
[-5, 15] scl: 1 by [-10, 10] scl: 1
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9-2 Solving Quadratic Equations by Graphing
[-5, 15] scl: 1 by [-10, 10] scl: 1
[-5, 15] scl: 1 by [-10, 10] scl: 1
c. For the roots to be additive opposites, the quadratic must not have a middle terms. Also, the quadratic must be
a difference of perfect square. Consider x2 − 4 = 0. The roots are −2 and 2 which are additive opposites.
[-10, 10] scl: 1 by [-10, 10] scl: 1
[-10, 10] scl: 1 by [-10, 10] scl: 1
2
45. CHALLENGE Find the roots of x = 2.25 without using a calculator. Explain your strategy.
SOLUTION: The square root of 2.25 is less than ±2. Therefore, make a table of values for x from −2.0 to 2.0. Use increments of
0.1.
x
y x
y
−2.0 4.00 0.1 0.01
−1.9 3.61 0.2 0.04
−1.8 3.24 0.3 0.09
−1.7 2.89 0.4 0.16
−1.6 2.56 0.5 0.25
−1.5 2.25 0.6 0.36
−1.4 1.96 0.7 0.49
−1.3 1.69 0.8 0.64
−1.2 1.44 0.9 0.81
−1.1 1.21 1.0 1.00
−1.0
1.00
1.1 by 1.21
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−0.9 0.81 1.2 1.44
−0.8 0.64 1.3 1.69
9-2 Solving Quadratic Equations by Graphing
[-10, 10] scl: 1 by [-10, 10] scl: 1
2
45. CHALLENGE Find the roots of x = 2.25 without using a calculator. Explain your strategy.
SOLUTION: The square root of 2.25 is less than ±2. Therefore, make a table of values for x from −2.0 to 2.0. Use increments of
0.1.
x
y x
y
−2.0 4.00 0.1 0.01
−1.9 3.61 0.2 0.04
−1.8 3.24 0.3 0.09
−1.7 2.89 0.4 0.16
−1.6 2.56 0.5 0.25
−1.5 2.25 0.6 0.36
−1.4 1.96 0.7 0.49
−1.3 1.69 0.8 0.64
−1.2 1.44 0.9 0.81
−1.1 1.21 1.0 1.00
−1.0 1.00 1.1 1.21
−0.9 0.81 1.2 1.44
−0.8 0.64 1.3 1.69
−0.7 0.49 1.4 1.96
−0.6 0.36 1.5 2.25
−0.5 0.25 1.6 2.56
−0.4 0.16 1.7 2.89
−0.3 0.09 1.8 3.24
−0.2 0.04 1.9 3.61
-0.1 0.01 2.0 4.00
2
The roots of x = 2.25 are 1.5 and –1.5.
46. WRITING IN MATH Explain how to approximate the roots of a quadratic equation when the roots are not
integers.
SOLUTION: First graph the related function. Then determine between which two integers the graph crosses the x-axis. Make a
table going by tenths of the values between the integers. Locate where the function value changes signs. The xvalue for which the function value is closest to zero is the best approximation of the root of the equation.
2
Consider the quadratic equation x – 3.35 = 0. The zeros are between – 2 and – 1 and 1 and 2. eSolutions Manual
- Powered
x
y by Cognero
x
y
0.65 1
–2
– 2.35
0.26 1.1
– 1.9
– 2.14
Page 34
−0.3 0.09 1.8 3.24
−0.2 0.04 1.9 3.61
-0.1 0.01 2.0 4.00
9-2 Solving Quadratic Equations by Graphing
2
The roots of x = 2.25 are 1.5 and –1.5.
46. WRITING IN MATH Explain how to approximate the roots of a quadratic equation when the roots are not
integers.
SOLUTION: First graph the related function. Then determine between which two integers the graph crosses the x-axis. Make a
table going by tenths of the values between the integers. Locate where the function value changes signs. The xvalue for which the function value is closest to zero is the best approximation of the root of the equation.
2
Consider the quadratic equation x – 3.35 = 0. The zeros are between – 2 and – 1 and 1 and 2. x
y
x
y
0.65 1
–2
– 2.35
0.26 1.1
– 1.9
– 2.14
1.2
– 1.8
– 0.11 – 1.91
1.3
– 1.7
– 0.46 – 1.66
1.4
– 1.6
– 0.79 – 1.39
1.5
– 1.5
– 1.1 – 1.1
1.6
– 1.4
– 1.39 – 0.79
1.7
– 1.3
– 1.66 – 0.46
1.8
– 1.2
– 1.91 – 0.11
1.9
0.26
– 1.1
– 2.14 2
0.65
–1
– 2.35 The zeros are about – 1.8 and 1.8.
47. Adrahan earned 50 out of 80 points on a test. What percentage did Adrahan score on the test?
A 62.5%
B 6.25%
C 1.6%
D 16%
SOLUTION: The correct choice is A.
48. Ernesto needs to loosen a bolt. He needs a wrench that is smaller than a
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wrench. Which of the following sizes should Ernesto use?
F
inch
-inch wrench, but larger than a
-inch
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9-2 Solving Quadratic Equations by Graphing
The correct choice is A.
48. Ernesto needs to loosen a bolt. He needs a wrench that is smaller than a
-inch wrench, but larger than a
-inch
wrench. Which of the following sizes should Ernesto use?
F
inch
G
inch
H
J
inch
inch
SOLUTION: Choices F, G, and J are all smaller than both
and .
Choice H is the only one smaller than
, but larger than
.
The correct choice is H.
49. EXTENDED RESPONSE Two boats leave a dock. One boat travels 4 miles east and then 5 miles north. The
second boat travels 12 miles south and 9 miles west. Draw a diagram that represents the paths traveled by the boats.
How far apart are the boats in miles?
SOLUTION: Boat 1 is a total of 13 miles east of Boat 2 and a total of 17 miles north of Boat 2.
The boats are about 21.4 miles apart.
50. The formula
represents the distance s in meters that a free-falling object will fall on a planet or moon in a
given time t in seconds. Solve the formula for a, the acceleration due to gravity.
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A
B a = 2s − t2
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9-2 Solving Quadratic Equations by Graphing
The boats are about 21.4 miles apart.
50. The formula
represents the distance s in meters that a free-falling object will fall on a planet or moon in a
given time t in seconds. Solve the formula for a, the acceleration due to gravity.
A
B a = 2s − t2
C
D
SOLUTION: The correct choice is D.
Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of each
function. Identify the vertex as a maximum or minimum. Then graph the function.
51. y = 3x
2
SOLUTION: The axis of symmetry is x = 0.
The coordinate of the vertex is (0, 0).
Because the equation is in standard form and a is positive, the graph opens up and the vertex is a minimum.
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9-2 Solving Quadratic Equations by Graphing
The correct choice is D.
Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of each
function. Identify the vertex as a maximum or minimum. Then graph the function.
51. y = 3x
2
SOLUTION: The axis of symmetry is x = 0.
The coordinate of the vertex is (0, 0).
Because the equation is in standard form and a is positive, the graph opens up and the vertex is a minimum.
2
52. y = −4x − 5
SOLUTION: The axis of symmetry is x = 0.
The coordinate of the vertex is (0, –5).
Because the equation is in standard form and a is negative, the graph opens down and the vertex is a maximum.
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9-2 Solving Quadratic Equations by Graphing
2
52. y = −4x − 5
SOLUTION: The axis of symmetry is x = 0.
The coordinate of the vertex is (0, –5).
Because the equation is in standard form and a is negative, the graph opens down and the vertex is a maximum.
2
53. y = −x + 4x − 7
SOLUTION: The axis of symmetry is x = 2.
The coordinate of the vertex is (2, –3).
Because the equation is in standard form and a is negative, the graph opens down and the vertex is a maximum.
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9-2 Solving Quadratic Equations by Graphing
2
53. y = −x + 4x − 7
SOLUTION: The axis of symmetry is x = 2.
The coordinate of the vertex is (2, –3).
Because the equation is in standard form and a is negative, the graph opens down and the vertex is a maximum.
2
54. y = x − 6x − 8
SOLUTION: The axis of symmetry is x = 3.
The coordinate of the vertex is (3, –17).
Because the equation is in standard form and a is positive, the graph opens up and the vertex is a minimum.
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2
Page 40
9-2 Solving Quadratic Equations by Graphing
2
54. y = x − 6x − 8
SOLUTION: The axis of symmetry is x = 3.
The coordinate of the vertex is (3, –17).
Because the equation is in standard form and a is positive, the graph opens up and the vertex is a minimum.
2
55. y = 3x + 2x + 1
SOLUTION: The axis of symmetry is x =
The coordinate of the vertex is
.
.
Because the equation is in standard form and a is positive, the graph opens up and the vertex is a minimum.
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9-2 Solving Quadratic Equations by Graphing
2
55. y = 3x + 2x + 1
SOLUTION: The axis of symmetry is x =
The coordinate of the vertex is
.
.
Because the equation is in standard form and a is positive, the graph opens up and the vertex is a minimum.
2
56. y = −4x − 8x + 5
SOLUTION: The axis of symmetry is x = –1.
The coordinate of the vertex is (–1, 9).
Because the equation is in standard form and a is negative, the graph opens down and the vertex is a maximum.
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9-2 Solving Quadratic Equations by Graphing
2
56. y = −4x − 8x + 5
SOLUTION: The axis of symmetry is x = –1.
The coordinate of the vertex is (–1, 9).
Because the equation is in standard form and a is negative, the graph opens down and the vertex is a maximum.
Solve each equation. Check the solutions.
2
57. 2x = 32
SOLUTION: Solve using the Zero Product Property. The solutions are –4 and 4.
Check:
and
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9-2 Solving Quadratic Equations by Graphing
Solve each equation. Check the solutions.
2
57. 2x = 32
SOLUTION: Solve using the Zero Product Property. The solutions are –4 and 4.
Check:
and
2
58. (x − 4) = 25
SOLUTION: Solve for x using the Zero Product Property.
The solutions are –1 and 9.
Check:
and
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and
9-2 Solving Quadratic Equations by Graphing
2
58. (x − 4) = 25
SOLUTION: Solve for x using the Zero Product Property.
The solutions are –1 and 9.
Check:
and
2
59. 4x − 4x + 1 = 16
SOLUTION: Solve for x using the Zero Product Property.
The solutions are
.
Check:
and
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Page 45
and
9-2 Solving Quadratic Equations by Graphing
2
59. 4x − 4x + 1 = 16
SOLUTION: Solve for x using the Zero Product Property.
The solutions are
.
Check:
and
2
60. 2x + 16x = −32
SOLUTION: Solve for x using the Zero Product Property.
The solution is –4.
Check:
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Page 46
and
9-2 Solving Quadratic Equations by Graphing
2
60. 2x + 16x = −32
SOLUTION: Solve for x using the Zero Product Property.
The solution is –4.
Check:
2
61. (x + 3) = 5
SOLUTION: and The solutions are
Check:
.
and
2
eSolutions
Manual
by Cognero
12x =- Powered
62. 4x −
−9
SOLUTION: Page 47
and
9-2 Solving Quadratic Equations by Graphing
2
62. 4x − 12x = −9
SOLUTION: To solve for x, use the Zero Product Property.
The solution is
.
Check:
Find each sum or difference.
2
2
63. (3n − 3) + (4 + 4n )
SOLUTION: 2
2
64. (2d − 7d − 3) − (4d + 7)
SOLUTION: 3
2
4
2
65. (2b − 4b + 4) − (3b + 5b − 9)
SOLUTION: 2
4
2
3
66. (8 − 4h + 6h ) + (5h − 3 + 2h )
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SOLUTION: Page 48
65. (2b − 4b + 4) − (3b + 5b − 9)
SOLUTION: 9-2 Solving Quadratic Equations by Graphing
2
4
2
3
66. (8 − 4h + 6h ) + (5h − 3 + 2h )
SOLUTION: 67. GEOMETRY Supplementary angles are two angles with measures that have a sum of 180°. For the supplementary
angles in the figure, the measure of the larger angle is 24° greater than the measure of the smaller angle. Write and solve a system of equations to find these measures.
SOLUTION: Write an equation in point-slope form for the line that passes through each point with the given slope.
68. (2, 5), m = 3
SOLUTION: 69. (−3, 6), m = −7
SOLUTION: 70. (−1, −2), m = −
SOLUTION: Graph each function.
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2
71. y = x + 5
SOLUTION: Page 49
SOLUTION: 9-2 Solving Quadratic Equations by Graphing
Graph each function.
2
71. y = x + 5
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, 5). a > 0, so the graph opens upward and the vertex is a minimum.
The y-intercept is (0, 5).
Find a couple of points and reflect them across the line of symmetry.
2
If x = –1, then y = (–1) + 5 or 6. Plot the point (–1, 6).
Reflect (–1, 6) over the axis of symmetry to obtain the point (1, 6).
2
If x = –2, then y = (–2) + 9 or 9. Plot the point (–2, 9).
Reflect (–2, 9) over the axis of symmetry to obtain the point (2, 9).
Connect the points with a smooth curve.
2
72. y = x − 8
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, –8). a > 0, so the graph opens upward and the vertex is a minimum.
The y-intercept is (0, –8).
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Find a couple of points and reflect them across the line of symmetry.
2
If x = –1, then y = (–1) − 8 or –7. Plot the point (–1, –7).
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9-2 Solving Quadratic Equations by Graphing
2
72. y = x − 8
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, –8). a > 0, so the graph opens upward and the vertex is a minimum.
The y-intercept is (0, –8).
Find a couple of points and reflect them across the line of symmetry.
2
If x = –1, then y = (–1) − 8 or –7. Plot the point (–1, –7).
Reflect (–1, –7) over the axis of symmetry to obtain the point (1, –7).
2
If x = –2, then y = (–2) − 8 or –4. Plot the point (–2, –4).
Reflect (–2, –4) over the axis of symmetry to obtain the point (2, –4).
Connect the points with a smooth curve.
2
73. y = 2x − 7
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, –7). a > 0, so the graph opens upward and the vertex is a minimum.
The y-intercept is (0, –7).
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a couple
of points
and reflect
them across the line of symmetry.
If x = –1, then y = 2(−1) − 7 or –5. Plot the point (–1, –5).
Reflect (–1, –5) over the axis of symmetry to obtain the point (1, –5).
2
Page 51
9-2 Solving Quadratic Equations by Graphing
2
73. y = 2x − 7
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, –7). a > 0, so the graph opens upward and the vertex is a minimum.
The y-intercept is (0, –7).
Find a couple of points and reflect them across the line of symmetry.
2
If x = –1, then y = 2(−1) − 7 or –5. Plot the point (–1, –5).
Reflect (–1, –5) over the axis of symmetry to obtain the point (1, –5).
2
If x = –2, then y = 2(−2) − 7 or 1. Plot the point (–2, 1).
Reflect (–2, 1) over the axis of symmetry to obtain the point (2, 1).
Connect the points with a smooth curve.
2
74. y = −x + 2
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, 2). a < 0, so the graph opens downward and the vertex is a maximum.
The y-intercept is (0, 2).
Find a couple of points and reflect them across the line of symmetry.
2
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If x = –1, then y = −(−1) + 2 or 1. Plot the point (–1, 1).
Reflect (–1, 1) over the axis of symmetry to obtain the point (1, 1).
2
If x = –2, then y = −(−2) + 2 or –2. Plot the point (–2, –2).
Page 52
9-2 Solving Quadratic Equations by Graphing
2
74. y = −x + 2
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, 2). a < 0, so the graph opens downward and the vertex is a maximum.
The y-intercept is (0, 2).
Find a couple of points and reflect them across the line of symmetry.
2
If x = –1, then y = −(−1) + 2 or 1. Plot the point (–1, 1).
Reflect (–1, 1) over the axis of symmetry to obtain the point (1, 1).
2
If x = –2, then y = −(−2) + 2 or –2. Plot the point (–2, –2).
Reflect (–2, –2) over the axis of symmetry to obtain the point (2, –2).
Connect the points with a smooth curve.
2
75. y = −0.5x − 3
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, –3). a < 0, so the graph opens downward and the vertex is a maximum.
The y-intercept is (0, –3).
Find a couple of points and reflect them across the line of symmetry.
− 3 or –3.5. Plot the point (–1, –3.5).
Reflect (–1, –3.5) over the axis of symmetry to obtain the point (1, –3.5).
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−1)2
If x Manual
= –1, then
y = by
−0.5(
2
If x = –2, then y = −0.5( −2) − 3 or –5. Plot the point (–2, –5).
Page 53
9-2 Solving Quadratic Equations by Graphing
2
75. y = −0.5x − 3
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, –3). a < 0, so the graph opens downward and the vertex is a maximum.
The y-intercept is (0, –3).
Find a couple of points and reflect them across the line of symmetry.
2
If x = –1, then y = −0.5( −1) − 3 or –3.5. Plot the point (–1, –3.5).
Reflect (–1, –3.5) over the axis of symmetry to obtain the point (1, –3.5).
2
If x = –2, then y = −0.5( −2) − 3 or –5. Plot the point (–2, –5).
Reflect (–2, –5) over the axis of symmetry to obtain the point (2, –5).
Connect the points with a smooth curve.
2
76. y = (−x) + 1
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, 1). a > 0, so the graph opens upward and the vertex is a minimum.
The y-intercept is (0, 1).
Find a couple of points and reflect them across the line of symmetry.
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2
If x = –1, then y = (−(–1)) + 1 or 2. Plot the point (–1, 2).
Reflect (–1, 2) over the axis of symmetry to obtain the point (1, 2).
2
Page 54
9-2 Solving Quadratic Equations by Graphing
2
76. y = (−x) + 1
SOLUTION: The axis of symmetry is x = 0.
The vertex is (0, 1). a > 0, so the graph opens upward and the vertex is a minimum.
The y-intercept is (0, 1).
Find a couple of points and reflect them across the line of symmetry.
2
If x = –1, then y = (−(–1)) + 1 or 2. Plot the point (–1, 2).
Reflect (–1, 2) over the axis of symmetry to obtain the point (1, 2).
2
If x = –2, then y = (−(–2)) + 1 or 5. Plot the point (–2, 5).
Reflect (–2, 5) over the axis of symmetry to obtain the point (2, 5).
Connect the points with a smooth curve.
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Page 55