Chapter 8

Chapter 8 Chapter 8 Maintaining Mathematical Proficiency (p. 417)

7. −x2 + 4x + 1 = −(−2)2 + 4(−2) + 1

1. Use m = 2 and b = −3 to graph y = 2x − 3.

= −(4) + (−8) + 1 = −4 − 8 + 1

y

2 1

= −12 + 1

−4 −3 −2−1

= −11

2 3 4 x

−2

8. x2 + 8x + 5 = (−2)2 + 8(−2) + 5

−3

y = 2x − 3

= 4 + (−16) + 5 = −12 + 5

−6

= −7

2. Use m = −3 and b = 4 to graph y = −3x + 4.

9. −2x2 − 4x + 3 = −2(−2)2 − 4(−2) + 3

y

= −2(4) − (−8) + 3

y = −3x + 4 4

= −8 + 8 + 3

3

=0+3

2 1

=3

−4 −3 −2−1

1 2 3 4 x

10. −4x2 + 2x − 6 = −4(−2)2 + 2(−2) − 6

−2

= −4(4) + (−4) − 6

1

1

3. Use m = −—2 and b = −2 to graph y = −—2 x − 2. 4

= −16 − 4 − 6 = −20 − 6

y

= −26

3 2 1 −6 −5 −4

11.

−1

1 2 x

x

1

y=

1 y = − 2 x − 2 −3

ax2

y=

y

−4

y=x+5

3

a(2)2

y=

4a

4

a(3)2

9a

y=

a(4)2

16a

5 y = a(5)2 25a

⤻⤻⤻⤻ +3a +5a +7a +9a

The coefficient of each difference is the next consecutive odd integer. So, when x = 6, y = 25a + 11a = 36a.

y

4 3

Chapter 8 Mathematical Practices (p. 418)

2 1 −4 −3 −2−1

y=

a

4. Use m = 1 and b = 5 to graph y = x + 5. 6

2

a(1)2

1. 1 2 x

−2

5. 5x2 − 9 = 5(−2)2 − 9

= 5(4) − 9 = 20 − 9 = 11 6. 3x2 + x − 2 = 3(−2)2 + (−2) − 2

= 3(4) − 2 − 2 = 12 − 2 − 2 = 10 − 2 =8

−4

−2

0

2

4

−(−4)2

−(−2)2

−(0)2

−(2)2

−(4)2

−16

−4

0

−4

−16

(−4, −16)

(−2, −4)

(0, 0)

x y=

−x2

y (x, y)

(2, −4) (4, −16)

y −4 −3 −2

2 3 4 x −4 −6 −8 −10

y = −x2

−12 −14 −16

The graph opens down and the highest point is at the origin.

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Algebra 1 Worked-Out Solutions

455

Chapter 8 2.

−2

−1

0

1

2

2(−2)2

2(−1)2

2(0)2

2(1)2

8

2

0

(−2, 8)

(−1, 2)

(0, 0)

x y = 2x2 y (x, y)

16 14 12 10 8

4.

x

f(x) = 2x2 − 1

f (x)

(x, y)

2(2)2

−2

2(−2)2 − 1

7

(−2, 7)

2

8

−1

2(−1)2 − 1

1

(−1, 1)

(1, 2)

(2, 8)

0

2(0)2 − 1

−1

(0, −1)

1

2(1)2 − 1

1

(1, 1)

2

2(2)2 − 1

7

(2, 7)

y

14 12 10 8

6 4 2

y = 2x2

−4 −3 −2−1

1 2 3 4 x

6 4 2

The graph opens up and the lowest point is at the origin. −4 −3 −2−1

3.

x

f(x) = 2x2 + 1

f (x)

(x, y)

−2

2(−2)2 + 1

9

(−2, 9)

−1

2(−1)2 + 1

3

(−1, 3)

5.

f(x)

(x, y)

+ 4(−8) + 3

3

(−8, 3)

+ 4(−6) + 3

−3

(−6, −3)

+ 4(−4) + 3

−5

(−4, −5)

−2 —12 (−2)2 + 4(−2) + 3

−3

(−2, −3)

3

(0, 3)

2(0)2 + 1

1

(0, 1)

−8

1

2(1)2 + 1

3

(1, 3)

−6 −4

1 —2 (−4)2

16 14 12 10 8

9

(2, 9)

y

1 2 3 4 x

f(x) = —12x2 + 4x + 3

x

0

2(2)2 + 1

f(x) = 2x2 − 1

The graph opens up and the lowest point is at (0, −1).

1 —2 (−8)2 1 —2 (−6)2

2

y

1 —2 (0)2

0

+ 4(0) + 3 y

6 4

−4 −3 −2−1

8

f(x) = 2x2 + 1

6 4 2

1 2 3 4 x

The graph opens up and the lowest point is at (0, 1).

−12

−8

−4

2 4 x −4 −6

1

f(x) = 2 x2 + 4x + 3

The graph opens up and the lowest point is at (−4, −5).

456

Algebra 1 Worked-Out Solutions

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Chapter 8 6.

f (x) = —12x2 − 4x + 3

x

f (x)

(x, y)

0

1 —2 (0)2

− 4(0) + 3

3

(0, 3)

2

1 —2 (2)2

− 4(2) + 3

−3

(2, −3)

4

1 —2 (4)2

− 4(4) + 3

−5

(4, −5)

6

1 —2 (6)2

− 4(6) + 3

−3

(6, −3)

8

—2 (8)2 − 4(8) + 3

3

(8, 3)

1

8.

x

y = −2(x − 1)2 + 1

y

(x, y)

−1

−2(−1 − 1)2 + 1

−7

(−1, −7)

0

−2(0 − 1)2 + 1

−1

(0, −1)

1

−2(1 − 1)2 + 1

1

(1, 1)

2

−2(2 − 1)2 + 1

−1

(2, −1)

3

−2(3 − 1)2 + 1

−7

(3, −7)

y

y

2 −4 −3 −2 −1

1

3 4 x

4 2 −4−2

y = −2(x − 1)2 + 1

2 4 6 8 10 12 x

−4 1

−6

−14

f(x) = 2 x2 − 4x + 3

The graph opens up and the lowest point is at (4, −5). 7.

x

y = −2(x + 1)2 + 1

y

(x, y)

−3

−2(−3 + 1)2 + 1

−7

(−3, −7)

−2

−2(−2 + 1)2 + 1

−1

(−2, −1)

−1

−2(−1 + 1)2 + 1

−1

(−1, 1)

0

−2(0 + 1)2 + 1

−1

(0, −1)

1

−2(1 + 1)2 + 1

−7

(1, −7)

2 −4 −3

−1 −4 −6 −8 −10 −12 −14

The graph opens down and the highest point is at (1, 1). 9. Sample answer: The graphs are all U-shaped. Some open

up, and some open down. Some have the highest point or lowest point at the origin, and some have been translated. Some have been shrunk or stretched. 8.1 Explorations (p. 419) 1. a.

x g(x) = 3x2 g(x)

y

(x, y)

−2

−1

0

1

2

3(−2)2

3(−1)2

3(0)2

3(1)2

3(2)2

12

3

0

3

12

(1, 3)

(2, 12)

(−2, 12) (−1, 3) (0, 0)

1 2 3 4 x

y = −2(x + 1)2 + 1

g(x) = 3x2 11 10 9 8 7 6 5 4 3

The graph opens down and the highest point is at (−1, 1).

y

f(x) = x2

2

−6 −5 −4 −3 −2−1

1 2 3 4 5 6 x

The graph of g is a vertical stretch by a factor of 3 of the graph of f.

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Algebra 1 Worked-Out Solutions

457

Chapter 8 b.

d.

−2

x

−1

0

1

2

g(x) = −5x2 −5(−2)2 −5(−1)2 −5(0)2 −5(1)2 −5(2)2 −20

g(x)

−5

0

−5

−20

(−2, −20) (−1, −5) (0, 0) (1, −5) (2, −20)

(x, y)

6

f(x) =

−6 −5 −4 −3 −2−1

0

2

4

1 2 x g(x) = — 10

(−4)2 — 10

(−2)2 — 10

(0)2 — 10

(2)2 — 10

(4)2 — 10

1.6

0.4

0

0.4

1.6

(0, 0)

(2, 0.4)

(4, 1.6)

1

g(x) (x, y)

1 2 3 4 5 6 x

g(x) = −5x2

1

g(x) = 10 x2

x

g(x) = −0.2x2

g(x)

(x, y)

−4

−0.2(−4)2

−3.2

(−4, −3.2)

−2

−0.2(−2)2

−0.8

(−2, −0.8)

0

−0.2(0)2

0

(0, 0)

2

−0.2(2)2

−0.8

(2, −0.8)

4

−0.2(4)2

−3.2

(4, −3.2)

y

5 4 3

−6 −5 −4 −3

11 10 9 8 7 6 5 4 3

−6 −5 −4 −3 −2−1

The graph of g is a vertical stretch by a factor of 5 and a reflection in the x-axis of the graph of f.

2 1

1

1

1

y

f(x) = x2

2 1

−14 −16 −18

6

1

(−4, 1.6) (−2, 0.4)

x2

−8

1 2 3 4 5 6 x

1 The graph of g is a vertical shrink by a factor of — of the 10 graph of f.

2. Sample answer: They are U-shaped and symmetric. They

either open up with the lowest point at the origin, or they open down with the highest point at the origin. 3. Sample answer: When 0 < a < 1, the graph of f(x) = ax2 is

a vertical shrink of the graph of f(x) = x2. When a > 1, the graph of f(x) = ax2 is a vertical stretch of the graph of f(x) = x2. When −1 < a < 1, the graph of f(x) = ax2 is a vertical shrink and a reflection in the x-axis of the graph of f(x) = x2. When a < −1, the graph of f(x) = ax2 is a vertical stretch and a reflection in the x-axis of the graph of f(x) = x2.

4. The interval that describes the value of a is 0 < a < 1, f(x) =

x2

3 4 5 6 x −2 −3 −4 −5 −6

because the graph is a vertical shrink of the graph of f(x) = x2. 8.1 Monitoring Progress (pp. 421–422) 1. The vertex is (2, −3). The axis of symmetry is x = 2. The

domain is all real numbers. The range is y ≥ −3. When x < 2, y increases as x decreases. When x > 2, y increases as x increases.

g(x) = −0.2x2

The graph of g is a vertical shrink by a factor of 0.2 and a reflection in the x-axis of the graph of f.

458

−2

y

4 2

c.

−4

x

Algebra 1 Worked-Out Solutions

2. The vertex is (−3, 7). The axis of symmetry is x = −3.

The domain is all real numbers. The range is y ≤ 7. When x < −3, y increases as x increases. When x > −3, y increases as x decreases.

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Chapter 8 3.

x g(x) 24 21 18 15

−2

−1

0

1

2

20

5

0

5

20

6.

−2

−1

0

1

2

p(x)

−12

−3

0

−3

−12

y

y −4 −3 −2−1

12 9

−4 −3 −2−1

1 2 3 4 x

Both graphs open up and have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of g is narrower than the graph of f. So, the graph of g is a vertical stretch by a factor of 5 of the graph of f. −3

x h(x)

−1 1 —3

3

0

1

3

0

1 —3

3

7.

x

y

1

1 2 3 4 x

Both graphs open up and have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of h is wider than the graph of f. So, the graph of h is a vertical shrink by a factor of —13 of the graph of f.

−4 −3 −2−1

4

8

−6.4

−1.6

0

−1.6

−6.4

4 6 8 x

q(x) = −0.1x2

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of q opens down and is wider than the graph of f. So, the graph of q is a vertical shrink by a factor of 0.1 and a reflection in the x-axis of the graph of f.

5.

2 1

0

−2 −3 −4 −5 −6 −7 −8

h(x) = 3 x2

−4 −3 −2−1

8 7 6 5 4 3

−4

−8 −6 −4

2 1

n(x)

−8

y

7 6 5 4 3

x

p(x) = −3x2

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of p opens down and is narrower than the graph of f. So, the graph of p is a vertical stretch by a factor of 3 and a reflection in the x-axis of the graph of f.

q(x) 8

1 2 3 4 x

−6 −8 −10 −12 −14 −16

g(x) = 5x2

6

4.

x

−2

−1

0

1

2

6

1.5

0

1.5

6

y

8.

x

−4

−2

0

2

4

g(x)

−4

−1

0

−1

−4

y −8 −6 −4

3

n(x) = 2 x2 1 2 3 4 x

Both graphs open up and have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of n is narrower than the graph of f. So, the graph of n is a vertical stretch by a factor of —32 of the graph of f.

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4 6 8 x −2 −3 −4 −5 −6 −7 −8 1

g(x) = − 4 x2

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of g opens down and is wider than the graph of f. So, the graph of g is a vertical shrink by a factor of —14 and a reflection in the x-axis of the graph of f.

Algebra 1 Worked-Out Solutions

459

Chapter 8 9.

x y

−2

−1

2

1 —2

0

1

2

0

1 —2

2

6.

x

−2

−1

0

1

2

b(x)

10

2.5

0

2.5

10

y

16 14 12 10 8

3

(−2, 2)

(2, 2)

2 1

6

y = 0.5x 2 −2

−1

1 −1

4 2

2 x

(0, 0)

1 2 3 4 x

Both graphs open up and have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of b is narrower than the graph of f. So, the graph of b is a vertical stretch by a factor of 2.5 of the graph of f. 7.

x h(x)

8.1 Exercises (pp. 423−424) Vocabulary and Core Concept Check

8

1. The U-shaped graph of a quadratic function is called a

2. The graph of a quadratic function opens up when a > 0 and

opens down when a < 0.

0

2

4

4

1

0

1

4

y

The domain is all real numbers. The range is y ≥ 4. When x < −2, y increases as x decreases. When x > −2, y increases as x increases.

2 4 6 8 x

Both graphs open up and have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of h is wider than the graph of f. So, the graph of h is a vertical shrink by a factor of —14 of the graph of f.

domain is all real numbers. The range is y ≤ −1. When x < 1, y increases as x increases. When x > 1, y increases as x decreases.

4. The vertex is (−2, 4). The axis of symmetry is x = −2.

8.

x

−4

−2

0

2

4

j(x)

12

3

0

3

12

8

x

−2

−1

0

1

2

g(x)

24

6

0

6

24

2 1 −4 −3 −2−1

g(x) = 6x2 1 2 3 4 x

y

7 6 5 4 3

y

15 12 9

1

h(x) = 4 x2

−8 −6 −4−2

3. The vertex is (1, −1). The axis of symmetry is x = 1. The

−4 −3 −2−1

−2

2 1

Monitoring Progress and Modeling with Mathematics

6

−4

7 6 5 4 3

parabola.

24 21 18

b(x) = 2.5x2

−4 −3 −2−1

The leftmost point on the graph is (−2, 2), and the rightmost point is (2, 2). So, the domain is −2 ≤ x ≤ 2, which represents 4 inches. The lowest point on the graph is (0, 0), and the highest points on the graph are (−2, 2) and (2, 2). So, the range is 0 ≤ y ≤ 2, which represents 2 inches. So, the spotlight is 4 inches wide and 2 inches deep.

5.

y

j(x) = 0.75x2 1 2 3 4 x

Both graphs open up and have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of j is wider than the graph of f. So, the graph of j is a vertical shrink by a factor of 0.75 of the graph of f.

Both graphs open up and have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of g is narrower than the graph of f. So, the graph of g is a vertical stretch by a factor of 6 of the graph of f.

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Algebra 1 Worked-Out Solutions

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Chapter 8 9.

x

−2

−1

0

1

2

m(x)

−8

−2

0

−2

−8

12.

x

−3

−1.5

0

1.5

3

p(x)

−6

−1.5

0

1.5

6

y −4 −3 −2−1

y

1 2 3 4 x

−4 −3 −2

m(x) = −2x2

−3 −4 −5 −6 −7 −8

−2 −3 −4 −5 −6 −7 −8

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of m opens down and is narrower than the graph of f. So, the graph of m is a vertical stretch by a factor of 2 and a reflection in the x-axis of the graph of f. 10.

2 3 4 x

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of p opens down and is wider than the graph of f. So, the graph of p is a vertical shrink by a factor of —23 and a reflection in the x-axis of the graph of f. 13.

x q(x)

−2

−1

0

1

2

−18

−4.5

0

−4.5

−18

2

p(x) = − 3 x2

5

−5

5

y = 4x2 y −4 −3 −2−1

−5

1 2 3 4 x

The graph of y = 4x2 is a reflection in the x-axis of the graph of y = −4x2.

9

q(x) = − 2 x2

−9 −12 −15 −18 −21 −24

14.

y = −0.4x2 −5

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of q opens down and is narrower than the graph of f. So, the graph of q is a vertical stretch by a factor of —92 and a reflection in the x-axis of the graph of f. 11.

5

−5

1 The graph of y = −0.4x2 is a vertical shrink by a factor of — 10 2 of the graph of y = −4x .

15.

x

−5

−2

0

2

5

k(x)

−5

−0.8

0

−0.8

−5

5

−5

5

y=

y −8 −6 −4

5

4 6 8 x −2 −3 −4 −5 −6 −7 −8

k(x) = −0.2x2

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of k opens down and is wider than the graph of f. So, the graph of k is a vertical shrink by a factor of 0.2 and a reflection in the x-axis of the graph of f.

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−0.04x2

−5

The graph of y = −0.04x2 is a vertical shrink by a factor 1 of — of the graph of y = −4x2. 100 16.

2

−15

15

−2

y = −0.004x2

The graph of y = −0.004x2 is a vertical shrink by a factor 1 of — of the graph of y = −4x2. 1000

Algebra 1 Worked-Out Solutions

461

Chapter 8 17. The graph of y = 0.5x2 should be wider than the graph of

y=

x2. 8 7 6 5 4 3

y

decreasing when x > 0. y = x2

23. The point (−2, 3) is in Quadrant II, and part of the graph of

f is in Quadrant II. So, the function f could include the point (−2, 3). f(x) = ax2

y = 0.5x2

3 = a(−2)2

1 2 3 4 x

The graphs have the same vertex and the same axis of symmetry. The graph of y = 0.5x2 is wider than the graph of y = x2. 18. The leftmost point on the graph is (−500, −300), and the

rightmost point is (500, −300). So, the domain is −500 ≤ x ≤ 500, which represents 1000 feet. The lowest points on the graph are (−500, −300) and (500, −300), and the highest point on the graph is (0, 0). So, the range is −300 ≤ y ≤ 0, which represents 300 feet. So, the arch is 1000 feet wide and 300 feet high.

19. a. A negative diameter does not make sense in the context

of the problem. So, the domain is d ≥ 0. Because it is a quadratic function of the form y = ax2, where a > 0, the range is z ≥ 0. b.

d

0

1

2

3

4

z

0

8900

35,600

80,100

142,400

Breaking strength (1000 pounds)

Manila Rope z 160 140 120 100 80 60 40 20 0

increasing when x < 0. 22. The function f is decreasing when x < 0, and the function g is

2 1 −4 −3 −2−1

21. The function f is increasing when x > 0, and the function g is

3 = a(4) a(4) 3 = ____ __ 4

4

3=a __ 4

So, the value of a is —34 when the graph passes through (−2, 3). 24. The x-intercept of the graph of y = ax2 is always 0. At

an x-intercept, y = 0, so ax2 = 0. By the definition of a quadratic function, a ≠ 0. So, by the Zero-Product Property, x2 = 0, which means x = 0.

25. Sample answer: The vertex of a parabola that opens up is

the minimum point, so its y-coordinate is the minimum value of y. The graph passes through (6, −3), so 2 is not the minimum value of y. 26. The graph of f(x) = ax2 is sometimes narrower than the

graph of g(x) = x2 when a > 0. Sample answer: The graph of f will be narrower than the graph of g when a > 1, but it will be wider when 0 < a < 1.

27. The graph of f(x) = ax2 is always narrower than the graph of

z = 8900d 2

g(x) = x2 when ∣ a ∣ > 1. Sample answer: When ∣ a ∣ > 1, the graph of f will be a vertical stretch of the graph of g, so it will be narrower.

28. The graph of f(x) = ax2 is always wider than the graph of 0 1 2 3 4 5 6 7 8 d

Diameter (inches)

c. A manila rope that has 4 times the breaking strength of

another manila rope does not have 4 times the diameter. Sample answer: The relationship is quadratic. So, a rope with 4 times the diameter will have 42 = 16 times the breaking strength.

g(x) = x2 when 0 < ∣ a ∣ < 1. Sample answer: When 0 < ∣ a ∣ < 1, the graph of f will be a vertical shrink of the graph of g, so it will be wider.

29. The graph of f(x) = ax2 is never wider than the graph of

g(x) = dx2 when ∣ a ∣ > ∣ d ∣. Sample answer: When ∣ a ∣ > ∣ d ∣, the graph of f will be a vertical stretch of the graph of g, so it will be narrower, not wider.

20. a. Both graphs open up and have the same vertex, (0, 0), and

axis of symmetry, x = 0, but the graph of g is narrower than the graph of f. So, the graph of g is a vertical stretch of the graph of f, which means that a > 1. b. The graphs have the same vertex, (0, 0), and axis of

symmetry, x = 0, but the graph of g opens down and is wider than the graph of f. So, the graph of g is a vertical shrink and a reflection in the x-axis of the graph of f, which means that −1 < a < 0.

462

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Chapter 8 30.

8.2 Explorations (p. 425)

(0, 4)

1. a. (−6, −4)

x

−2

−1

0

1

2

4

1

0

1

4

−2

−1

0

1

2

6

3

2

3

6

f(x)

(6, −4)

y = ax2

x

−4 = a(6)2

g(x)

−4 = a(36)

g(x) = x2 + 2

−4 a(36) 36 36 1 __ − =a 9

— = _____

10 9 8 7 6 5

So, a quadratic equation whose graph models the shape is 1 y = −—9x2 . 31. a. The leftmost point on the graph is (−4, 3.2), and the

1 −6 −5 −4 −3 −2−1

b. Your friend is incorrect. Sample answer: A faster

=9+5

1 2 3 4 5 6 x

−2

b.

x

−2

−1

0

1

2

8

2

0

2

8

−2

−1

0

1

2

6

0

−2

0

8

f(x)

Maintaining Mathematical Proficiency 32. n2 + 5 = (3)2 + 5

f(x) = x2

4 3

rightmost point is (4, 3.2). So, the domain is −4 ≤ x ≤ 4, which represents 8 centimeters. So, the mouth of the glass is 8 centimeters wide. rotational speed would increase the depth. The diagram shown has a depth of 3.2 centimeters. A model of y = 0.1x2 would only have a depth of 1.6 centimeters, so it would have a slower rotational speed.

y

x g(x)

= 14 f(x) = 2x2

33. 3x2 − 9 = 3(−2)2 − 9

10 9 8 7 6 5 4 3 2

= 3(4) − 9 = 12 − 9 =3 34. −4n2 + 11 = −4(3)2 + 11

= −4(9) + 11 = −36 + 11 = −25

−6 −5 −4 −3 −2

y

g(x) = 2x2 − 2

2 3 4 5 6 x

35. n + 2x2 = 3 + 2(−2)2

= 3 + 2(4) =3+8

Sample answer: In g(x) = ax2 + c, c causes a vertical shift in the graph of f(x) = ax2.

= 11

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Algebra 1 Worked-Out Solutions

463

Chapter 8 2. a.

x

−2

−1

0

1

2

y

−3

−6

−7

−6

−3

3 2 1 −6 −5 −4 −3

6

−2 −3

−6

−4 −5

−2

The graph appears to intersect the x-axis where x ≈ 2.6 and x ≈ −2.6. The x-intercepts of the graph occur where y = 0.

4

−6

6

+7

7 = x2

—

−4

—

±√x2

±√7 = —

5. The graph is narrower than y = x2, so it is a vertical stretch.

±√7 = x

—

So, — the x-intercepts of the graph are √ 7 , or about 2.6 and −√7 , or about −2.6. b.

An example of an equation with c < 0 is h(x) = x2 − 2. So, graph h(x) = x2 − 2 and f(x) = x2 on the same graph to verify that the graph of h is a vertical translation 2 units down of the graph of f.

7

0 = x2 − 7 +7

6

y = x2 − 7

−8

y=

4. Sample answer: An example of an equation with c > 0 is

1 2 3 4 5 6 x

−1

x2 −

translated up. When c < 0, the graph is translated down.

g(x) = x2 + 3. So, graph g(x) = x2 + 3 and f(x) = x2 on the same graph to verify that the graph of g is a vertical translation 3 units up of the graph of f.

y

4

3. It causes a vertical shift of c units. When c > 0, the graph is

x

−2

−1

0

−1

2

y

−3

0

1

0

−3

y

4 3

y = −x2 + 1

2

The vertex is 1 unit above the origin, so the graph is a vertical translation 1 unit up of the graph of y = x2. So, a > 1 and c = 1. 8.2 Monitoring Progress (pp. 426–428) 1. g(x) = x2 − 5

x

−2

−1

0

1

2

g(x)

−1

−4

−5

−4

−1

2 1

2 3 4 5 6 x

−6 −5 −4 −3 −2 −2 −3

−4 −3

−4 −5 −6 −7 −8

−1

y

1

3 4 x

−2 −3

−6

The graph appears to intersect the x-axis where x = −1 and x = 1. To verify, let y = 0, and solve for x. y = −x2 + 1 0 = −x2 + 1

g(x) = x2 − 5

Both graphs open up and have the same axis of symmetry, x = 0. The vertex of the graph of g, (0, −5), is below the vertex of the graph of f, (0, 0). So, the graph of g is a vertical translation 5 units down of the graph of f.

0 = 12 − x2 0 = (1 + x)(1 − x) 1+x= −1

0

or

1−x=0

−1

+x +x

x = −1

1=x

So, the x-intercepts of the graph are −1 and 1.

464

Algebra 1 Worked-Out Solutions

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Chapter 8 5. a. The graph of g is a vertical translation 3 units up of the

2. h(x) = x2 + 3

x h(x) 8

−2

−1

0

1

2

7

4

3

4

7

graph of f.

y

7 6 5 4

−2

−1

0

1

2

f(x)

11

2

−1

2

11

g(x) = f(x) + 3

14

5

2

5

14

g(x) = f(x) + 3 14 12 10 8

h(x) = x2 + 3

2 1 −4 −3 −2−1

1 2 3 4 x

4

f(x) = 3x2 − 1 −4 −3 −2−1

3. g(x) = 2x2 − 5

x g(x) 10 8

y

6

Both graphs open up and have the same axis of symmetry, x = 0. The vertex of the graph of h, (0, 3), is above the vertex of the graph of f, (0, 0). So, the graph of h is a vertical translation 3 units up of the graph of f.

1 2 3 4 x

b. g(x) = f(x) + 3

−2

−1

0

1

2

3

−3

−5

−3

3

= (3x − 1) + 3 = 3x + (−1 + 3) = 3x + 2 So, g(x) = 3x + 2.

y

6

6. Sample answer: Negative values of t would represent times

4 2

before the egg was dropped, which have no meaning in the context of this problem.

−4 −3 −2

1 2 3 4 x −6

g(x) =

2x2

7.

−5

1 4

4. h(x) = −__x2 + 4

x h(x)

0

2

4

60

0

3

4

3

0

30

3 2 1

1

1.5

2

2.5

100

96

84

64

36

0

f(t)

−2

1

0.5

120

f(t) = −16t 2 + 100

90

h(x) = − 4 x2 + 4

0

Disregard negative values of t. So, the egg hits the ground 2.5 seconds after it is dropped.

−4

y

−2

t f(t)

Both graphs open up and have the same axis of symmetry, x = 0. The graph of g is narrower, and its vertex, (0, −5), is below the vertex of the graph of f, (0, 0). So, the graph of g is a vertical stretch by a factor of 2 and a vertical translation 5 units down of the graph of f.

−8 −6

x

0

(2.5, 0) 0

1

2

3

4

t

8.2 Exercises (pp. 429–430) 2

6 8 x

−2 −3 −4

The graphs have the same axis of symmetry, x = 0. The graph of h opens down and is wider than the graph of f, and its vertx, (0, 4), is above the vertex of the graph of f, 1 (0, 0). So, the graph of h is a vertical shrink by a factor of —, 4 a reflection in the x-axis, and a vertical translation 4 units up of the graph of f. Copyright © Big Ideas Learning, LLC All rights reserved.

Vocabulary and Core Concept Check 1. The graph of y = ax2 + c has a vertex of (0, c) and an axis of

symmetry of x = 0.

2. When c > 0, the graph of f(x) = ax2 + c is a vertical

translation c units up of the graph of f(x) = ax2. When c < 0, the graph of f(x) = ax2 + c is a vertical translation ∣ c ∣ units down of the graph of f(x) = ax2.

Algebra 1 Worked-Out Solutions

465

Chapter 8 Monitoring Progress and Modeling with Mathematics 3. g(x) =

x2

+6

6. q(x) = x2 − 1

x

x

−2

−1

0

1

2

g(x)

10

7

6

7

10

q(x) 6

y

16 14 12 10 8

−2

−1

0

1

2

3

0

−1

0

3

y

5 4 3 2 1

q(x) = x2 − 1

−4 −3 −2−1

g(x) = x2 + 6

4 2 −4 −3 −2−1

1 2 3 4 x

−2

1 2 3 4 x

Both graphs open up and have the same axis of symmetry, x = 0. The vertex of the graph of g, (0, 6), is above the vertex of the graph of f, (0, 0). So, the graph of g is a vertical translation 6 units up of the graph of f. 4. h(x) = x2 + 8

x

−2

−1

0

1

2

h(x)

12

9

8

9

12

Both graphs open up and have the same axis of symmetry, x = 0. The vertex of the graph of q, (0, −1), is below the vertex of the graph of f, (0, 0). So, the graph of q is a vertical translation 1 unit down of the graph of f. 7. g(x) = −x2 + 3

x

−2

−1

0

1

2

g(x)

−1

2

3

2

−1

4 2 1 −4 −3

4 2 −4 −3 −2−1

−1

1

3 4 x

−2 −3 −4

h(x) = x2 + 8

6

1 2 3 4 x

Both graphs open up and have the same axis of symmetry, x = 0. The vertex of the graph of h, (0, 8), is above the vertex of the graph of f, (0, 0). So, the graph of h is a vertical translation 8 units up of the graph of f.

The graphs have the same vertex, (0, 0), and axis of symmetry, x = 0, but the graph of g opens down, and the vertex of the graph of g, (0, 3), is above the vertex of the graph of f, (0, 0). So, the graph of g is a reflection in the x-axis and a vertical translation 3 units up of the graph of f. 8. h(x) = −x2 − 7

5. p(x) = x2 − 3

p(x)

g(x) = −x2 + 3

y

16 14 12 10

x

y

−2

−1

0

1

2

x

−2

−1

0

1

2

1

−2

−3

−2

1

h(x)

−11

−8

−7

−8

−11

y 4

y

−4 −3 −2−1 −4 −6

3 2 1 −4 −3 −2

1 2 3 4 x

−4

p(x) = x2 − 3

Both graphs open up and have the same axis of symmetry, x = 0. The vertex of the graph of p, (0, −3), is below the vertex of the graph of f, (0, 0). So, the graph of p is a vertical translation 3 units down of the graph of f.

466

Algebra 1 Worked-Out Solutions

1 2 3 4 x

h(x) = −x2 − 7

−10 −12 −14 −16

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of h opens down, and the vertex of the graph of h, (0, −7), is below the vertex of the graph of f, (0, 0). So, the graph of h is a reflection in the x-axis and a vertical translation 7 units down of the graph of f.

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 1 3

9. s(x) = 2x2 − 4

x s(x) 4 3

11. p(x) = −—x2 − 2

−2

−1

0

1

2

4

−2

−4

−2

4

−3

x

−5

p(x)

y

3

−2

1 −3— 3

−5

1 2 3 4 x

s(x) = 2x2 − 4

Both graphs open up and have the same axis of symmetry, x = 0, but the graph of s is narrower than the graph of f. Also, the vertex of the graph of s, (0, −4), is below the vertex of the graph of f, (0, 0). So, the graph of s is a vertical stretch by a factor of 2 and a vertical translation 4 units down of the graph of f. 10. t(x) = −3x2 + 1

x

−2

−1

0

1

2

t(x)

−11

−2

1

−2

−11

y 2 3 4 x

t(x) = −3x2 + 1

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of t opens down and is narrower than the graph of f. Also, the vertex of the graph of t, (0, 1), is above the vertex of the graph of f, (0, 0). So, the graph of t is a vertical stretch by a factor of 3, a reflection in the x-axis, and a vertical translation 1 unit up of the graph of f.

Copyright © Big Ideas Learning, LLC All rights reserved.

p(x) = − 3 x2 − 2

−3 −4 −5 −6 −7 −8

1 2 3 4 x

−4 −6 −8 −10 −12 −14

2

1

−4 −3 −2

−4 −3 −2

0

y −4 −3 −2−1

2 1

2

−2 1 −3— 3

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of p opens down and is wider than the graph of f. Also, the vertex of the graph of p, (0, −2), is below the vertex of the graph of f, (0, 0). So, 1 the graph of p is a vertical shrink by a factor of —, a reflection 3 in the x-axis, and a vertical translation 2 units down of the graph of f. 1 2

12. q(x) = —x2 + 6

x

−4

−2

0

2

4

q(x)

14

8

6

8

14

16 14 12 10 8 4 2 −8 −6 −4−2

y

1

q(x) = 2 x2 + 6 2 4 6 8 x

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of q is wider than the graph of f. Also, the vertex of the graph of q, (0, 6), is above the vertex of the graph of f, (0, 0). So, the graph of q is a vertical 1 shrink by a factor of — and a vertical translation 6 units up of 2 the graph of f.

Algebra 1 Worked-Out Solutions

467

Chapter 8 13. The graph of g is a vertical translation 2 units up of the graph

of f.

15. The graph of g is a vertical translation 3 units down of the

graph of f. −2

−1

0

1

2

+4

16

7

4

7

16

g(x) = f(x) + 2

18

9

6

9

18

x f(x) =

3x2

g(x) = f(x) + 2 24 21 18 15 12 9

x

−4

−2

0

2

4

1 x2 − 6 f(x) = −__ 4

−10

−7

−6

−7

−10

g(x) = f(x) − 3

−13

−10

−9

−10

−13

y

y −4 −3 −2−1

−4 −3 −2−1

1 2 3 4 x

g(x) = f(x) + 2

(

= 3x2 + (4 + 2) = 3x2 + 6 So, g(x) = 3x2 + 6. 14. The graph of g is a vertical translation 4 units down of the

graph of f. x 1 x2 + 1 f(x) = __ 2

g(x) = f(x) − 4 f(x) =

1 2 2x

−2

0

2

4

9

3

1

3

9

of f.

5

−1

−3

−1

5

x

+1

y

16. The graph of g is a vertical translation 7 units up of the graph

−4

−2

0

2

4

−5

59

11

−5

11

59

g(x) = f(x) + 7

66

18

2

18

66

f(x) =

4x2

g(x) = f(x) + 7

1

−4

3 4 x

g(x) = f(x) − 4

g(x) = f(x) − 4

(

)

1 = −—x2 − 6 − 3 4 1 = −—x2 + (−6 − 3) 4 1 = −—x2 − 9 4 1 So, g(x) = −—x2 − 9. 4

−4

4 2 −1

g(x) = f(x) − 3

g(x) = f(x) − 3

g(x) = (3x2 + 4) + 2

−4 −3

1

f(x) = −4 x2 − 6

−8 −10 −12 −14 −16

f(x) = 3x2 + 4

12 10 8 6

1 2 3 4 x

−4

)

1 = —x2 + 1 − 4 2 1 = —x2 + (1 − 4) 2 1 = —x2 − 3 2 1 So, g(x) = —x2 − 3. 2

70 60 50 40 30 20 10 −4 −3 −2−1

y

f(x) = 4x2 − 5 1 2 3 4 x

g(x) = f(x) + 7 = (4x2 − 5) + 7 = 4x2 + (−5 + 7) = 4x2 + 2 So, g(x) = 4x2 + 2. 17. The graph of y = 3x2 + 2 is narrower, so it should be a

stretch not a shrink. The graph of y = 3x2 + 2 is a vertical stretch by a factor of 3 and a translation 2 units up of the graph of y = x2.

468

Algebra 1 Worked-Out Solutions

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Chapter 8 18. The function g is not graphed correctly.

21. f(x) = −x2 + 25

x

−2

−1

0

1

2

g(x)

−6

−9

−10

−9

−6

x f(x)

f(x) = x2

−4

−2

0

2

4

9

21

25

21

9

f(x)

y

25 20 15 10 5

f(x) = −x 2 + 25 10

(−5, 0)

(5, 0)

−8

−8 −6 −4

4 6 8 x

−10

g(x) = x2 − 10

−15

8 x

4

Both graphs open up and have the same axis of symmetry. However, the vertex of the graph of g, (0, −10), is 10 units below the vertex of the graph of f, (0, 0). 19. y = x2 − 1

x

−2

−1

0

1

2

y

3

0

−1

0

3

The graph intersects the x-axis at (−5, 0) and (5, 0). So, the zeros are −5 and 5. 22. f(x) = −x2 + 49

x

−8

−4

0

4

8

f(x)

−15

33

49

33

−15

60

f(x)

f(x) = −x 2 + 49

y

40

4 20 2

(−7, 0)

(−1, 0) −4

−8

(1, 0)

−2

2 −2

4 x

y = x2 − 1

The graph intersects the x-axis at (−7, 0) and (7, 0). So, the zeros are −7 and 7.

20. y = x2 − 36

23. f(x) = 4x2 − 16

x

x

−6

−3

0

3

6

f(x)

y

0

−27

−36

−27

0

(−2, 0) −3

y

9 −3

−2

−1

0

1

2

0

−12

−16

−12

0

f(x) −1

1

(2, 0) 3

x

−4

27

−9

8 x

4 −20

The graph intersects the x-axis at (−1, 0) and (1, 0). So, the zeros are −1 and 1.

(−6, 0)

(7, 0)

−4

−8

(6, 0) 3

9

x

f(x) = 4x 2 − 16 −45

y = x 2 − 36

The graph crosses the x-axis at (−2, 0) and (2, 0). So, the zeros are −2 and 2.

The graph intersects the x-axis at (−6, 0) and (6, 0). So, the zeros are −6 and 6.

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Algebra 1 Worked-Out Solutions

469

Chapter 8 24. f(x) = 3x2 − 27

27. a. f(t) = −16t2 + 144

x

−2

−1

0

1

2

f(x)

−15

−24

−27

−24

−15

t f(t)

(−3, 0)

(3, 0)

−2

1

2

3

144

128

80

0

f(t)

f(x) −4

0

2

160 4 x

f(t) = −16t 2 + 144

120

−10

80 −30

40

f(x) = 3x 2 − 27

0

The graph crosses the x-axis at (−3, 0) and (3, 0). So, the zeros are −3 and 3.

−1

1 −__ 2

0

1 __

1

f(x)

−9

0

3

0

−9

2

( 12 , 0(

−1

1

t

3 seconds to hit the ground. When k < 0, the water balloon will take less than 3 seconds to hit the ground.

x

0

0.5

1

1.5

y

36

32

20

0

y

y = −16x 2 + 36

−8

30

(

) ( )

1 1 The graph crosses the x-axis at −—, 0 and —, 0 . So, the 2 2 1 1 zeros are −— and —. 2 2 26. f(x) = −8x2 + 98

x

−4

−2

0

2

4

f(x)

−30

66

98

66

−30

20 10 0

(1.5, 0) 0

1

2

3

4 x

Disregard the negative value(s). So, the x-intercept is 1.5. This means the apple hits the ground after 1.5 seconds. Let x = 0. y = −16x2 + 36

f(x)

y = −16(0)2 + 36

80

= 0 + 36

40

−4

4

2 x

−4

(−72 , 0(

3

b. When k > 0, the water balloon will take more than

f(x) = −12x 2 + 3

(−12 , 0( −2

2

28. y = −16x2 + 36

f(x) 4

1

Find the positive zero of the function. When t = 3, f(t) = 0. So, the zero is 3. So, the water balloon hits the ground 3 seconds after it is dropped.

25. f(x) = −12x2 + 3

x

0

= 36

( 72 , 0(

−2

2

So, the y-intercept is 36. This means the apple fell from a height of 36 feet.

4 x

−40

29. Sample answer: f(x) = −8x 2 + 98

(

8 7 6 5 4

) ( )

7 7 The graph crosses the x-axis at −—, 0 and —, 0 . So, the 2 2 7 7 zeros are −— and —. 2 2

y

2 1 −4 −3 −2−1

470

Algebra 1 Worked-Out Solutions

1 2 3 4 x

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Chapter 8 30. Sample answer:

A = x2

35.

192 = x2 + 48

y

− 192

3

−4 −3

−1

1

3 4 x

− 12

or x − 12 =

− 12

+ 12

0 + 12

x = 12

Disregard the negative value. So, the solution is x = 12. The graph of A = x2 + 48 confirms that the area is 192 square feet when x = 12.

2 3 4 x −2 −3 −4 −5 −6 −7

x

0

4

8

12

A

48

64

112

192

Patio Area

Area (square feet)

32. Sample answer: y −4 −3 −2−1

1 2 3 4 x

−4 −8 −10 −12 −14 −16

A 200 175 150 125 100 75 50 25 0

(12, 192)

A = x2 + 48 0 2 4 6 8 10 12 14 16 x

Side length (feet)

So, the original patio is 12 feet by 12 feet.

33. a. T(x) = h(x) − g(x)

= (−16x2 + 256) − (−16x2 + 300) = −16x2 + 256 + 16x2 − 300 = (−16x2 + 16x2) + (256 − 300) = 0 − 44 = −44 So, T(x) = −44. This means that your ball is 44 feet below your friend’s ball at any point while they are falling. b. Your friend’s ball is 44 feet high when your ball hits the

ground.

36. As c increases, the graph of f is translated up. So, it gets

closer to A. 37. One way to find the zeros of the function f(x) = −16t2 + 400

is to graph it and determine the x-intercepts. Another way is to set the function equal to zero, factor −16t2 + 400, and apply the Zero-Product Property. 38. a. Because 4.8 is the greatest value of c, Waterfall 1 drops

water from highest point. b. Because 3.5 is the greatest value of ∣ a ∣, Waterfall 2

follows the narrowest path. c.

5

Falling Balls

Height (feet)

0

x = −12

y −4 −3 −2

− 144

0 = (x + 12)(x − 12) x + 12 =

31. Sample answer:

waterfall 1

g(x) = −16x2 + 300 h(x) = −16x2 + 256

(4, 44) 0

x2

0 = x2 − 122

−2 −3 −4

y 320 280 240 200 160 120 80 40 0

− 192

0=

2 1

1

2

3

4

waterfall 2 0

0

waterfall 3

1.5

The graph with the greatest x-intercept is Waterfall 1. So, Waterfall 1 sends water the farthest.

5 x

Time (seconds)

34. Your friend is incorrect. Sample answer: Changing a causes

a vertical stretch or shrink, which does not change the vertex. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

471

Chapter 8 8.3 Explorations (p. 431)

39. a. h = −16t2 + s0

1. a. y = 2x2 − 8x

Acorn 1: h = −16t2 + 45 Acorn 2: h = −16t2 + 32 So, an equation that represents the height of the first acorn is h = −16t2 + 45, and an equation for the height of the second acorn is h = −16t2 + 32.

x

0

1

2

3

4

y

0

−6

−8

−6

0

8

b. The graph of the height of the second acorn,

y

h = −16t2 + 32, is a vertical translation 13 units down of the graph of the height of the first acorn, h = −16t2 + 45. −2−1

40. Sample answer:

Area under the curve ≈ 2

of a triangle whose base ⋅ Area and height are on the x- and

y-axis 1 ≈ 2 —bh 2 ≈ bh ≈ 2(4) ≈8

1 2 3

−4 −6 −8

5 6 x

y = 2x2 − 8x

y = 2x2 − 8x + 6

( )

x

0

1

2

3

4

y

6

0

−2

0

6

y

So, the area under the curve is approximately 8 square units. 6

41. y = 0.012x2

4 2

= 0.012(40)2

−2−1

= 0.012(1600)

−4 −6 −8

= 19.2 The outer edges of the antenna are located 40 feet from the center. Substituting this into y = 0.012x2 indicates they are 19.2 feet above the x-axis. To be 25 feet above the x-axis, they must be vertically translated up 25 − 19.2 = 5.8 feet. So, the vertex of the cross section will be at (0, 5.8).

−(−3) 2(4)

4+3 12 − 3

7 9

−[−3 + 2(4)] 4(−3)

−(−3 + 8) −12

but not the same y-coordinate. c. The graph appears to intersect the x-axis when x = 0 and

x = 4.

or

x−4=0 +4+4 x=4

The x-intercepts are 0 and 4.

44. — = —— = — = —

b + 2a ab

b. The vertex of each graph has the same x-coordinate, x = 2,

2x = 0 2x 0 —=— 2 2 x=0

3 8

4 − (−3) 3(4) + (−3)

y = 2x2 − 8x + 6

0 = 2x(x − 4)

43. −— = — = —

a−b 3a + b

4 5 6 x

0 = 2x2 − 8x

Maintaining Mathematical Proficiency 1 1 4 a 42. — = — = — = −— 3 4b 4(−3) −3 b 2a

2

d. Sample answer: The value of the x-coordinate of the

−5 −12

5 12

45. −— = —— = — = — = —

vertex is the average of the values of the x-intercepts. 2. a. 0 = ax2 + bx

0 = x(ax + b) x=0

or

ax + b = 0 −b

−b

ax = − b −b a b x = −— a

ax a

—=—

b The solutions are x = 0 and x = −—. a

472

Algebra 1 Worked-Out Solutions

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Chapter 8 b. The x-intercepts of the graph of y = ax2 + bx are 0

b and −—. a

c.

y=

x

ax2

( ) ( ) − 2( 13 ) 1 2 = 3( ) − 9 3

1 1 f — =3 — 3 3

+ bx

y = a(0)2 + b(0)

—

(

=0+0 =0

6 6 b 2a 2(1) 2 The axis of symmetry is x = −3.

( ) ( ) ( ) ( ) ( )

b. g(x) = x2 + 6x + 5

g(−3) = (−3)2 + 6(−3) + 5 = 9 − 18 + 5 = −9 + 5 = −4 The vertex is (−3, −4).

b a

3. Find the average of 0 and −—.

−7

b

7 =—=7 3. a. x = −— = −______ 2a −1 1 2 −—

( 2)

b ÷ 2 = −— ÷ 2 a b 1 = −— — a 2 b = −— 2a

⋅

So, the vertex of the graph of y = ax2 + bx occurs when b x = −—. Also, the vertex of the graph of y = ax2 + bx + c 2a b occurs when x = −—. 2a b 2a

4. Sample answer: Find the x-coordinate using x = −—, then

use the function to find the y-coordinate.

)

2. a. x = −— = −— = −— = −3

−b 2 −b y=a — +b — a a 2 2 (−1) b −b2 =a — + — a2 a 2 2 b b = a —2 − — a a 2 2 b b =—−— a a =0

b 0 + −— a

—

1 2 =—−— 3 3 1 = −— 3 1 1 The vertex is —, − — . 3 3

y=0

[ ( )]

2

—

y=0

0 b −— a

b. f(x) = 3x2 − 2x

5. f(x) = x2 − 4x + 3

b (−4) 4 x = −— = −— = — = 2 2a 2(1) 2 f(2) = 22 − 4(2) + 3

The axis of symmetry is x = 7. 1 b. h(x) = −—x2 + 7x − 4 2 1 h(7) = −—(7)2 + 7(7) − 4 2 1 = −—(49) + 49 − 4 2 49 = −— + 49 − 4 2 49 =—−4 2 49 8 =—−— 2 2 41 =— 2 41 The vertex is 7, — . 2

( )

=4−8+3 = −4 + 3 = −1 So, the vertex of the graph of f(x) = x2 − 4x + 3 is at (2, −1). 8.3 Monitoring Progress (pp. 432–435) (−2) 2 1 b 1. a. x = −— = −— = — = — 2a 2(3) 6 3 1 The axis of symmetry is x = —. 3

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Algebra 1 Worked-Out Solutions

473

Chapter 8 4 b 2a 2(2) f(x) = 2x2 + 4x + 1

4 4

4. x = −— = −— = −— = −1

f(−1) = 2(−1)2 + 4(−1) + 1 = 2(1) − 4 + 1 =2−4+1 = −2 + 1 = −1

g(x) = 8x2 − 8x + 6

( ) ( ) − 8( 12 ) + 6 1 = 8( ) − 4 + 6 4

1 1 g — =8 — 2 2

2

—

= −2 + 6

4 3 2 1

h(x) = 2x2 + 4x + 1

=4 The minimum value is 4.

−4 −3 −2

1 2 3 4 x

The domain is all real numbers. The range is y ≥ −1. (−8) 8 b 5. x = −— = −— = — = 4 2a 2(1) 2 x2

− 8x + 7

k(4) = 42 − 8(4) + 7 = 16 − 32 + 7 = −16 + 7 = −9 y 9 6 3 −4−2

2 4 6 8 10 12 x

−6 −9 −12

1 1 1 4 4 4 the parabola opens down, and the function has a maximum

8. For h(x) = −—x2 + 3x + 1, a = −— and −— < 0. So,

−2

value.

⋅ ⋅

b −3 2 −6 3 =— x = −— = −______ =—=6 2a −1 1 1 2 −— −— 2 4 2 1 h(x) = −—x2 + 3x + 1 4 1 h(6) = −—(6)2 + 3(6) + 1 4 1 = −—(36) + 18 + 1 4 = −9 + 18 + 1

( )

=9+1 = 10 The maximum value is 10.

k(x) = x2 − 8x + 7

The domain is all real numbers. The range is y ≥ −9. b 2a

(−10) 2(−5)

10 −10 p(x) = −5x2 − 10x − 2

6. x = −— = −— = — = −1

p(−1) = −5(−1)2 − 10(−1) −2 = −5(1) + 10 − 2 = −5 + 10 − 2 =5−2=3 4

−1

y

b 2a

(−0.46) 2(0.00016)

0.46 0.00032

9. x = −— = −— = — = 1437.5

f(x) = 0.00016x2 − 0.46x + 507 f(1437.5) = 0.00016(1437.5)2 − 0.46(1437.5) + 507 = 0.00016(2,066,406.25) − 661.25 + 507 = 330.625 − 661.25 + 507 = −330.625 + 507 = 176.375 The cable is about 176 feet above the water at its lowest point. 10. From the graphs, you can see that the first water balloon is at

1 2 3 4 x

−4 −6 −8 −10 −12

p(x) = −5x2 − 10x − 2

The domain is all real numbers. The range is y ≤ 3.

474

(−8) b 1 8 x = −— = −— = — = — 2a 2(8) 16 2

=2−4+6

5

−4 −3

opens up, and the function has a minimum value.

—

y

k(x) =

7. For g(x) = 8x2 − 8x + 6, a = 8 and 8 > 0. So, the parabola

Algebra 1 Worked-Out Solutions

a height of 0 after about 5 seconds, while the second water balloon is at a height of 0 after about 5.7 seconds. So, the second water balloon is in the air longer. 11. From the graphs, you can see that the first water balloon

reaches its maximum height after about 2.5 seconds, while the second water balloon reaches its maximum height after about 2.8 seconds. So, the first water balloon reaches its maximum height faster. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 8.3 Exercises (pp. 436–438) Vocabulary and Core Concept Check 1. Sample answer: If the leading coefficient is positive, the

graph has a minimum value. If the leading coefficient is negative, the graph has a maximum value. 2. The question that is different is “What is the axis of

symmetry of the graph of the function?” 8 b −8 x = −— = −— = — = 2 2a 2(−2) −4 So, the axis of symmetry is x = 2. f(x) = −2x2 + 8x + 24 f(2) =

−2(2)2

+ 8(2) + 24

= −2(4) + 16 + 24

b. y = 3x2 + 2x

( ) ( ) ()

1 2 1 y = 3 −— + 2 −— 3 3 1 2 =3 — −— 9 3 1 2 =—−— 3 3 1 = −— 3 1 1 The vertex is −—, −— . 3 3

(

)

(−18) b 18 2a 2(−9) −18 The axis of symmetry is x = −1.

9. a. x = −— = −— = — = −1

b. y = −9x2 − 18x − 1

= −8 + 16 + 24

y = −9(−1)2 − 18(−1) − 1

= 8 + 24

= −9(1) + 18 − 1

= 32 So, the maximum value is 32, which is the answer to the other three questions.

= −9 + 18 − 1 =9−1 =8

Monitoring Progress and Modeling with Mathematics 3. The vertex is (2, −1). The axis of symmetry is x = 2. The

y-intercept of the graph is 1. 4. The vertex is (−3, 2). The axis of symmetry is x = −3. The

y-intercept of the graph is −1.

5. The vertex is (−2, 0). The axis of symmetry is x = −2. The

y-intercept of the graph is −3.

6. The vertex is (−1, 1). The axis of symmetry is x = −1. The

y-intercept of the graph is 5. (−4) 4 b 7. a. x = −— = −— = — = 1 2a 2(2) 4 The axis of symmetry is x = 1. b. f(x) = 2x2 − 4x

24 b −24 2a 2(−6) −12 The axis of symmetry is x = 2.

10. a. x = −— = −— = — = 2 b. f(x) = −6x2 + 24x − 20

f(2) = −6(2)2 + 24(2) − 20 = −6(4) + 48 − 20 = −24 + 48 − 20 = 24 − 20 =4 The vertex is (2, 4).

= 2(1) − 4 =2−4

⋅ ⋅

5 4 — (−4) _____ b 5 4 =— _____ =5 = 11. a. x = −— = − 2a 4 5 1 2 2 — — — 5 4 5 The axis of symmetry is x = 5.

()

f(1) = 2(1)2 − 4(1)

2 5 2 f(5) = —(5)2 − 4(5) + 14 5 2 = —(25) − 20 + 14 5 = 10 − 20 + 14

b. f(x) = —x2 − 4x + 14

= −2 The vertex is (1, −2). 2 2(3)

−2 6

1 3 1 The axis of symmetry is x = −—. 3 b 2a

The vertex is (−1, 8).

8. a. x = −— = −— = — = −—

= −10 + 14 =4 The vertex is (5, 4).

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Algebra 1 Worked-Out Solutions

475

Chapter 8

⋅ ⋅

2 −9 — b −6 9 3 =— ______ _______ =6 12. a. x = −— = − = 3 2 −1 2a 3 −— — 2 −— 2 3 4 The axis of symmetry is x = 6. 3 b. y = −—x2 + 9x − 18 4 3 y = −—(6)2 + 9(6) − 18 4 3 = −—(36) + 54 − 18 4 = −27 + 54 − 18

( )

= 27 − 18

24 8

24 2(4)

14. x = −— = −— = −3

So, the axis of symmetry is x = −3. y = 4x2 + 24x + 13 y = 4(−3)2 + 24(−3) + 13 = 4(9) − 72 + 13 = 36 − 72 + 13 = −36 + 13 = −23 So, the vertex is (−3, −23). The y-intercept is 13. So, the points (0, 13) and (−6, 13) lie on the graph.

=9 The vertex is (6, 9). b 2a

12 2(2)

y

12 4

13. x = −— = −— = −— = −3

So, the axis of symmetry is x = −3.

−7 −6

−4 −3 −2

f(x) = 2x2 + 12x + 4 f(−3) =

2(−3)2

y = 4x2 + 24x + 13

= 18 − 36 + 4

The domain is all real numbers. The range is y ≥ −23.

= −18 + 4 = −14 So, the vertex is (−3, −14). The y-intercept is 4. So, the points (0, 4) and (−6, 4) lie on the graph. y

−4 −3 −2−1

(−16) b 16 2a 2(−8) −16 So, the axis of symmetry is x = −1.

15. x = −— = −— = — = −1

y = −8x2 − 16x − 9 y = −8(−1)2 − 16(−1) − 9 = −8(1) + 16 − 9

6 3 −7 −6

−15 −20 −25

+ 12(−3) + 4

= 2(9) − 36 + 4

1 x

= −8 + 16 − 9 1 x

−9 −12 −15

f(x) = 2x2 + 12x + 4

The domain is all real numbers. The range is y ≥ −14.

=8−9 = −1 So, the vertex is (−1, −1). The y-intercept is −9. So, the points (0, −9) and (−2, −9) lie on the graph. y −4 −3 −2

1 2 3 4 x

−8 −10 −12 −14 −16

y = −8x2 − 16x − 9

The domain is all real numbers. The range is y ≤ −1.

476

Algebra 1 Worked-Out Solutions

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Chapter 8 20 b −20 2a 2(−5) −10 So, the axis of symmetry is x = 2.

16. x = −— = −— = — = 2

f(x) = −5x2 + 20x − 7 f(2) =

−5(2)2

+ 20(2) − 7

= −5(4) + 40 − 7 = −20 + 40 − 7 = 20 − 7 = 13 So, the vertex is (2, 13). The y-intercept is −7. So, the points (0, −7) and (4, −7) lie on the graph. 15 12 9

y

y = −5x2 + 20x − 7

6 3 −2−1

(−3)

b

3

18. x = −— = −______ = — = −3 2a 1 −1 2 −—

( 2)

So, the axis of symmetry is x = −3. 1 f(x) = −—x2 − 3x − 4 2 1 f(−3) = −—(−3)2 − 3(−3) − 4 2 1 = −—(9) + 9 − 4 2 9 = −— + 9 − 4 2 9 =—−4 2 1 =— 2 1 So, the vertex is −3, — . 2 The y-intercept is −4. So, the points (0, −4) and (−6, −4) lie on the graph.

(

)

1 2 3 4 5 6 x

−6

y

1

−9

−7 −6 −5

−3 −2

1 x

The domain is all real numbers. The range is y ≤ 13.

()

() () 2 81 = ( ) − 27 + 5 3 4 27 = — − 27 + 5 2 27 = −— + 5 2 17 = −— 2

19. There should be two negatives in the substitution, one from

the formula and one because b is −12.

20. The formula for the axis of symmetry has a negative sign.

4 b −4 The axis of symmetry is x = −— = −— = — = −2. 2a 2(1) 2

(

)

The y-intercept is 5. So, the points (0, 5) and (9, 5) lie on the graph.

f(x) = x2 + 4x − 3 f(−2) = (−2)2 + 4(−2) + 3 = 4 − 8 + 3 = −4 + 3 = −1 So, the vertex is (−2, −1). The y-intercept is 3, so the points (0, 3) and (−4, 3) lie on the graph.

y

4 2

−4 −6 −8 −10

1 The domain is all real numbers. The range is y ≤ —. 2

The axis of symmetry is x = 2.

9 17 So, the vertex is —, −— . 2 2

−4−2

1

f(x) = − 2 x2 − 3x − 4

(−12) 12 b x = −— = −— = — = 2 2a 2(3) 6

— —

6

−4 −5 −6 −7

⋅ ⋅

9 3 — (−6) 6 — b 9 4 = __ 2 =— 17. x = −— = −_____ = _____ 4 3 1 2 2a 2 — — 2 — 3 4 3 9 So, the axis of symmetry is x = —. 2 2 y = —x2 − 6x + 5 3 2 9 2 9 y=— — −6 — +5 3 2 2

y 2 4 6

5 4 3

10 12 x

2

y = 3 x2 − 6x + 5

17 The domain is all real numbers. The range is y ≥ −—. 2 Copyright © Big Ideas Learning, LLC All rights reserved.

−6 −5 −4

f(x) =

x2

1 2 x

+ 4x + 3 −2

Algebra 1 Worked-Out Solutions

477

Chapter 8 21. For y = 3x2 − 18x + 15, a = 3 and 3 > 0. So, the parabola

opens up, and the function has a minimum value. (−18) 18 b x = −— = −— = — = 3 2a 2(3) 6 y = 3x2 − 18x + 15 y = 3(3)2 − 18(3) + 15 = 3(9) − 54 + 15 = 27 − 54 + 15 = −27 + 15 = −12 The minimum value is −12. 22. For f(x) = −5x2 + 10x + 7, a = −5 and −5 < 0. So, the

parabola opens down, and the function has a maximum value. 10 b −10 x = −— = −— = — = 1 2a 2(−5) −10 f(x) = −5x2 + 10x + 7 f(1) = −5(1)2 + 10(1) + 7 = −5(1) + 10 + 7 = −5 + 10 + 7 = 5 + 7 = 12 The maximum value is 12. 23. For f(x) = −4x2 + 4x − 2, a = −4 and −4 < 0. So, the

parabola opens down, and the function has a maximum value. 4 −4 1 b = −— = — =— x = −— 2a 2(−4) −8 2 f(x) = −4x2 + 4x − 2

()

( ) + 4( 21 ) − 2 1 = −4( ) + 2 − 2 4

1 1 f — = −4 — 2 2

2

—

—

= −1 + 2 − 2 = 1 − 2 = −1 The maximum value is −1. 24. For y = 2x2 − 10x + 13, a = 2 and 2 > 0. So, the parabola

opens up, and the function has a minimum value. (−10) 10 5 x = −— = — = — 2(2) 4 2 y = 2x2 − 10x + 13

() () ( )

5 2 5 y = 2 — − 10 — + 13 2 2 25 = 2 — − 25 + 13 4 25 = — − 25 + 13 2 25 = −— + 13 2 1 =— 2 1. The minimum value is — 2

478

Algebra 1 Worked-Out Solutions

1 1 1 2 2 2 parabola opens down, and the function has a maximum value.

25. For y = −—x2 − 11x + 6, a = −— and −— < 0. So, the

(−11) b 11 x = −— = −______ = — = −11 2a 1 −1 2 −— 2 1 2 y = −—x − 11x + 6 2 1 y = −—(−11)2 − 11(−11) + 6 2 1 = −—(121) + 121 + 6 2 121 = −— + 121 + 6 2 121 =—+6 2 1 = 66— 2 1 The maximum value is 66—. 2

( )

1 1 1 5 5 5 opens up, and the function has a minimum value.

26. For f(x) = —x2 − 5x + 27, a = — and — > 0. So, the parabola

⋅ ⋅

(−5) 5 _____ b 25 = 5 5 =— x = −— = −_____ = __ 2 2 2a 1 2 — — 5 2 — 5 5 5 1 f(x) = —x2 − 5x + 27 5 25 1 25 2 25 f  — = — — − 5 — + 27 2 5 2 2 1 625 125 = — — − — + 27 5 4 2 125 125 = — − — + 27 4 2 125 = −— + 27 4 17 1 = −—, or −4— 4 4 1 The minimum value is −4—. 4

()

( ) ( ) ( ) ( )

b −128 128 2a 2(−16) −32 The firework explodes 4 seconds after it is launched.

27. a. t = −— = −— = — = 4

b. h = −16t2 + 128t

h = −16(4)2 + 128(4) = −16(16) + 512 = −256 + 512 = 256 The firework explodes at a height of 256 feet. 16 b −16 1 2a 2(−16) −32 2 The horse reaches its maximum height 0.5 second after it jumps.

28. a. t = −— = −— = — = —, or 0.5

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Chapter 8 b.

32. Your friend is correct. Sample answer: If the two points

h(t) = −16t2 + 16t h(0.5) = −16(0.5)2 + 16(0.5) = −16(0.25) + 8 = −4 + 8 =4 The horse reaches a maximum height of 4 feet. So, it can clear the fence by 4 − 3.5 = 0.5 foot. h(t) =

+ 16t

0=

−16t2

+ 16t

34. The maximum feature of a graphing calculator gives

0 = −16t(t − 1) −16t = 0 −16t

or

t−1=0 +1

0

= −16 −16

+1

t=1

t=0 The horse is on the ground when t = 0, then it is in the air, then it is on the ground again 1 second after it jumped. So, the horse is in the air for 1 − 0 = 1 second.

⋅ ⋅

(−1) b 200 1 200 = — = 200 29. a. x = −— = −______ = _________ 2a 1 1 1 — 200 2 — 200 400 The lowest point of the cable is 200 feet from each tower. 1 b. y = —x2 − x + 150 400 1 y = —(200)2 − 200 + 150 400 1 = —(40,000) − 200 + 150 400 = 100 − 200 + 150

( )

33. The minimum feature of a graphing calculator gives an

x-value of about −1.414216 and a y-value of −4. So, the approximate coordinates of the vertex are (−1.41, −4).

c. Let h = 0.

−16t2

have the same y-coordinate, then a parabola can be drawn by selecting a vertex with its x-coordinate halfway between the x-coordinates of the two points, and with a different y-coordinate. If the two points have different y-coordinates, a parabola can be drawn by making one of the points the vertex and drawing the parabola through the other point.

= −100 + 150 = 50

an x-value of about 0.38709677 and a y-value of about −0.0709677. So, the approximate coordinates of the vertex are (0.39, −0.07). 35. The maximum feature of a graphing calculator gives

an x-value of about 0.47746377 and a y-value of about 0.71619724. So, the approximate coordinates of the vertex are (0.48, 0.72). 36. The minimum feature of a graphing calculator gives

an x-value of about 5.8480362 and a y-value of about −6.54988. So, the approximate coordinates of the vertex are (5.85, −6.55). 37. Graph y = −0.006x2 + 1.5x on a graphing calculator. 125 1st aircraft hangar

0

a. You can see that first aircraft hangar has a maximum

height that is less than 100 feet, while the second aircraft hangar has a maximum height that is greater than 100 feet. So, the second aircraft hangar is taller.

The lowest point of the cable is 50 feet above the water. So, the road is also 50 feet above the water. c. Because the point that is midway between the towers

occurs when x = 200, the least value of x is 0, and the greatest value of x is 400. So, the domain is 0 ≤ x ≤ 400. The y-intercept is c = 150, which is the greatest value of y, and the least value is y = 50. So, the range is 50 ≤ y ≤ 150. 0 b 2a 2a So, the axis of symmetry is x = 0 when b = 0. You cannot find an axis of symmetry when a = 0. Sample answer: If b you substitute 0 for a in −—, then you are dividing by 0, 2a which is undefined. Also, if a = 0, the equation is linear, not quadratic, so it has no axis of symmetry.

250

0

b. You can see that the first aircraft hangar has a width of

250 feet, while the second aircraft hangar has a width of less than 250 feet. So, the first aircraft hangar is wider. 38. a.

15,000

30. x = −— = −— = 0

0

0

20

The maximum feature of a graphing calculator gives an x-value of 5 and a y-value of 10,800. So, they should decrease the price by 5($6) = $30 and sell it for $120 − $30 = $90.

31. Another point on the parabola is (0, 8). Sample answer:

Because the axis of symmetry is x = 3, the point (6, 8) is 6 − 3 = 3 units to the right of the axis of symmetry. So, the point on the other side of the axis is also 3 units away, where x = 3 − 3 = 0. So, (0, 8) would also lie on the graph. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

479

Chapter 8 b. Second revenue model: R(n) = (120 − 4n)(80 + 5n).

b. A = −2x2 + 8x + 24

A = −2(2)2 + 8(2) + 24

15,000

= −2(4) + 16 + 24 = −8 + 16 + 24 = 8 + 24 0

= 32

20

0

The maximum feature of a graphing calculator gives a y-value of 10,800 for the first model and 10,580 for this model. So, the first model results in a greater maximum monthly revenue of $10,800.

The maximum area is 32 square feet. 41. g(x) = x2 + 4x + 1 b 4 = −4 = −2 x = −___ = −____ —

2a

2(1)

g(x) = x2 + 4x + 1

39.

g(−2) = (−2)2 + 4(−2) + 1

x

=4−8+1

6

x

= −4 + 1 = −3

(6 − x)

So, the vertex is (−2, −3). (1.5x − x)

Area = length

⋅ width + length ⋅ width

The y-intercept is 1. So, the points (0, 1) and (−4, 1) lie on the graph.

A = 6(1.5x − x) + x(6 − x) A = 6(0.5x) + x(6) − x(x)

y 4

A = 3x + 6x − x2

2

A = −x2 + 9x b 2a

9 2(−1)

−9 −2

−6

a. x = −— = −— = — = 4.5

The value of x that maximizes the area of the figure is 4.5. b. A = −x2 + 9x

A = −(4.5)2 + 9(4.5) = −20.25 + 40.5 = 20.25

−4

2 x

g(x) = x 2 + 4x + 1

−4

h(x) = x2 − 4x + 1 (−4) 4 b x = −___ = −_____ = __ =2 2a 2(1) 2

The maximum area of the figure is 20.25 square inches.

So, the axis of symmetry is x = 2.

1 2 1 A = —(x + 2)[12 + (12 − 4x)] 2 1 = —(x + 2)(24 − 4x) 2 1 = —[x(24) + x(−4x) + 2(24) + 2(−4x)] 2 1 = —(24x − 4x2 + 48 − 8x) 2 1 = —(−4x2 + 16x + 48) 2 1 1 1 = —(−4x2) + —(16x) + —(48) 2 2 2

h(x) = x2 − 4x + 1

40. Area = —h(b1 + b2)

= −2x2 + 8x + 24 8 b −8 2a 2(−2) −4 The value of x that maximizes the area of the figure is 2.

a. x = −— = −— = — = 2

480

2

So, the axis of symmetry is x = −2.

Algebra 1 Worked-Out Solutions

h(2) = 22 − 4(2) + 1 =4−8+1 = −4 + 1 = −3 So, the vertex is (2, −3). The y-intercept is 1. So, the points (0, 1) and (4, 1) lie on the graph. y

2

−2

4

6 x

−2 −4

h(x) = x 2 − 4x + 1

The graph of g is a reflection in the y-axis of the graph of h. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 42. a. When x = 0, y = 1.5. So, the initial height of the arrow is

1.5 meters. b. The maximum height of the arrow appears to be about

1.6 meters.

47. 3y = 2x − 14

3y _______ ___ = 2x − 14 3

3

2 x − ___ 14 y = __

c. When the value of x is about 90, the value of y is 0. So,

the arrow hits the ground after traveling about 90 meters.

3

3 1 14 2 Graph y = −—x2 + 4x and y = —x − — on a graphing 8 3 3 calculator.

43. The parabola opens down. Sample answer: Because (3, 2)

and (9, 2) have the same y-coordinate, any point with an x-coordinate between 3 and 9 lies on the part of the parabola between these two points that passes through the vertex. Because 7 is greater than 2, the vertex must be above these 2 points, so the parabola opens down.

( )

b 44. f −— represents the maximum or minimum value of the 2a b function. Sample answer: Because −— is the x-coordinate 2a b of the vertex, f −— is the y-coordinate of the vertex, 2a which is also the maximum value or minimum value of the function.

( )

40

Intersection Y=14

0 X=28 0

Use the intersect feature of a graphing calculator to find that the T-shirt lands in the bleachers at a height of 14 feet. 48. Sample answer:

Sketch a graph of y = x2 and a tangent line through (1, 1). y 4

45. Use the points (1, 6) and (3, 6) and the equation

y = ax2 + bx to write a system of linear equations. 6 = a + b Equation 1 6 = 9a + 3b Equation 2

40

y = x2 −4

2

(1, 1)

−2

2

4 x

(0, −1)

Solve the system using substitution. a=6−b 6 = 9(6 − b) + 3b 6 = 54 − 9b + 3b −48 = −6b 8=b a = 6 − 8 = −2 So, a = −2 and b = 8. Therefore, a function of the form y = ax2 + bx whose graph contains the points (1, 6) and (3, 6) is y = −2x2 + 8x. 46. It is not possible to identify characteristics of parabola A

with only two points. The vertex of parabola B is (3, −4) because 3 is halfway between 1 and 5, which are the x-coordinates of points that have the same y-coordinates. Parabola B opens up because −4, the y-coordinate of the vertex, is the minimum value.

Select another point on the tangent line, such as (0, −1). Substitute (1, 1) and (0, −1) into the slope formula to calculate the slope of the tangent line. y2 − y1 1 − (−1) _____ 2 , or 2 m=— = — = 1 + 1 = __ 1 1 x2 − x1 1−0 So, the slope of the tangent line is 2. 49. Perimeter = ℓ + 2w

k = ℓ + 2w k − 2w = ℓ + 2w − 2w k − 2w = ℓ

⋅

Area = length width

⋅

A = (k − 2w) w A = k(w) − 2w(w) A = kw − 2w2 A = −2w2 + kw k k b k The axis of symmetry is w = −— = −— = −— = — 2a 2(−2) −4 4 A = −2w2 + kw 2

() ()

k + k __ k = −2 __ 4 4 k2 k2 2k2 k2 k2 + — = −__ + — = __ = −2 ___ 8 8 16 4 8

( )

k2 So, the maximum area of the outside enclosure is — square 8 feet. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

481

Chapter 8 Maintaining Mathematical Proficiency

4.

50. The function q is of the form y = f(x − h), where h = −6.

x p(x)

So, the graph of q is a horizontal translation 6 units left of the graph of f.

−2

−1

0

1

2

10

4

2

4

10

16 14 12 10 8 6

51. The function h is of the form y = −af(x), where a = 0.5.

So, the graph of h is a vertical shrink by a factor of 0.5 and a reflection in the x-axis of the graph of f. 52. The function g is of the form y = f(x − h) + k, where h = 2

y

4

and k = 5. So, the graph of g is horizontal translation 2 units right and a vertical translation 5 units up of the graph of f.

p(x) = 2x2 + 2 −4 −3 −2−1

53. The function p is of the form y = af(x − h), where a = 3 and

1 2 3 4 x

The graph of p is a vertical stretch by a factor of 2 and a vertical translation 2 units up of the graph of f.

h = −1. So, the graph of p is a vertical stretch by a factor of 3 and a horizontal translation 1 unit left of the graph of f.

5.

8.1–8.3 What Did You Learn? (p. 439)

x

−2

−1

0

1

2

0

−12

−16

−12

0

r(x)

1. Sample answer: Because the highest point has a y-coordinate

of 0, the height is the opposite of the y-coordinate of the lowest points. The width is the absolute value of the difference of the x-coordinates of the endpoints.

6 3 −4 −3

2. Sample answer: The t-intercept of the graph is the total time

before the water balloon hits the ground.

y

−1

1

3 4 x

−6 −9

3. Sample answer: Use the definition of vertex to identify

( )

b f −— as the y-coordinate of the vertex. Then use the 2a definition of maximum value/minimum value to recognize that this is also the maximum or minimum value of the function.

1. The vertex is (1, 4). The axis of symmetry is x = 1. The

domain is all real numbers. The range is y ≤ 4. When x < 1, y increases as x increases. When x > 1, y increases as x decreases.

2. The vertex is (−2, 5). The axis of symmetry is x = −2.

The domain is all real numbers. The range is y ≥ 5. When x < −2, y increases as x decreases. When x > −2, y increases as x increases.

3.

The graph of r is a vertical stretch by a factor of 4 and a vertical translation 16 units down of the graph of f. 6.

8.1–8.3 Quiz (p. 440)

x

−2

−1

0

1

2

h(x)

−4

−1

0

−1

−4

r(x) = 4x2 − 16

−18

x

−2

−1

0

1

2

b(x)

32

8

0

8

32

32 28 24 20 16

y

12 8

−4 −3 −2−1

b(x) = 8x2 1 2 3 4 x

The graph of b is a vertical stretch by a factor of 8 of the graph of f.

y −4 −3 −2−1 −2 −3 −4 −5 −6 −7 −8

1 2 3 4 x

h(x) = −x2

The graph of h is a reflection in the x-axis of the graph of f.

482

Algebra 1 Worked-Out Solutions

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 7.

x

−4

−2

0

2

4

g(x)

6.4

1.6

0

1.6

6.4

10. The graph of g is a vertical translation 9 units down of the

graph of f. x

8 7 6 5 4 3

y

f(x) =

g(x) = f(x) − 9

−1

0

1

2

0

9

12

9

0

−9

0

3

0

−9

y

f(x) = −3x2 + 12

2 1

g(x) =

−4 −3 −2−1

2 2 5x

8 6

1 2 3 4 x

4

The graph of g is a vertical shrink by a factor of —25 of the graph of f. 8.

−2 + 12

−3x2

−4 −3

3 4 x −4

x

−2

−1

0

1

2

m(x)

−6

−4.5

−4

−4.5

−6

g(x) = f(x) − 9

g(x) = f(x) − 9 g(x) = (−3x2 + 12) − 9

y −4 −3 −2−1

= −3x2 + (12 − 9)

1 2 3 4 x

−2 −3

= −3x2 + 3

1

m(x) = − 2 x2 − 4

So, g(x) = −3x2 + 3.

−5 −6 −7 −8

11. The graph of g is a vertical translation 6 units down of the

graph of f. −4

−2

0

2

4

f(x) = —12x2 − 2

6

0

−2

0

6

g(x) = f(x) − 6

0

−6

−8

−6

0

x

The graph of m is a vertical shrink by a factor of —12 , a reflection in the x-axis, and a vertical translation 4 units down of the graph of f. 9. The graph of g is a vertical translation 2 units up of the graph

of f.

1

f(x) = 2 x2 − 2

−2

−1

0

1

2

8 6

f(x) = 2x2 + 1

9

3

1

3

9

4 2

g(x) = f(x) + 2

11

5

3

5

11

x

−8 −6

6 8 x −4

g(x) = f(x) + 2 16 14 12 10 8 6

y

g(x) = f(x) − 6

g(x) = f(x) − 6

(

)

= —12 x2 − 2 − 6 = —12x2 + (−2 − 6) f(x) = 2x2 + 1

−4 −3 −2−1

y

1 2 3 4 x

= —12x2 − 8 So, g(x) = —12x2 − 8.

g(x) = f(x) + 2 g(x) = (2x2 + 1) + 2 = 2x2 + (1 + 2) = 2x2 + 3 So, g(x) = 2x2 + 3. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

483

Chapter 8 12. The graph of g is a vertical translation 1 unit up of the graph

of f.

b 12 −12 = −3. 14. The axis of symmetry is x = −___ = −____ = ____ 2a

2(2)

4

f(x) = 2x2 + 12x + 5 −1

−0.5

0

0.5

1

f(x) = 5x2 − 3

2

−1.75

−3

−1.75

2

= 2(9) − 36 + 5

g(x) = f(x) + 1

3

−0.75

−2

−0.75

3

= 18 − 36 + 5

x

f(−3) = 2(−3)2 + 12(−3) + 5

= −18 + 5 = −13

g(x) = f(x) + 1 4 3

So, the vertex is (−3, −13).

y

The y-intercept is 5. So, the points (0, 5) and (−6, 5) lie on the graph.

2 1

y

−2

−1

1

2 x 6 3

f(x) = 5x2 − 3

−4

−7 −6

−4 −3 −2−1

g(x) = f(x) + 1 g(x) = (5x2 − 3) + 1 f(x) = 2x2 + 12x + 5

= 5x2 + (−3 + 1) = 5x2 − 2 (−4) b 1. 4 = −__ 13. The axis of symmetry is x = −___ = −— = ___ 2a

2(−4)

−8

2

f(x) = −4x2 − 4x + 7

( )

( )

2

( )

1 +7 − 4 −__ 2

()

1 +2+7 = −4 __ 4

b 4 −4 = −2. 15. The axis of symmetry is x = −___ = −____ = ___ 2a

=8

y = (−2)2 + 4(−2) − 5 =4−8−5 = −4 − 5 = −9

)

1, 8 . So, the vertex is −__ 2 The y-intercept is 7. So, the points (0, 7) and (−1, 7) lie on the graph.

6

−6

−4 −3 −2−1

2 x

−4

f(x) = −4x2 − 4x + 7

4 2 −1

y

4 2

y

−4 −3

2

The y-intercept is −5. So, the points (0, −5) and (−4, −5) lie on the graph.

=1+7

6

2(1)

y = x2 + 4x − 5

So, the vertex is (−2, −9).

= −1 + 2 + 7

(

−9 −12 −15

The domain is all real numbers. The range is y ≥ −13.

So, g(x) = 5x2 − 2.

1 = −4 −__ 1 f −__ 2 2

1 x

y = x2 + 4x − 5 2 3 4 x

−10

The domain is all real numbers. The range is y ≥ −9.

−4 −6 −8

The domain is all real numbers. The range is y ≤ 8.

484

Algebra 1 Worked-Out Solutions

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Chapter 8 6 b −6 = 1. 16. The axis of symmetry is x = −___ = −______ = ___ 2a

2(−3)

−6

19. For y = −x2 + 4x + 12, a = −1 and −1 < 0. So, the

= −3(1) + 6 + 9

parabola opens down, and the function has a maximum value. b 4 = ___ −4 = 2 x = −___ = −______ 2a 2(−1) −2 y = −x2 + 4x + 12

= −3 + 6 + 9

y = −(2)2 + 4(2) + 12

y = −3x2 + 6x + 9 y = −3(1)2 + 6(1) + 9

=3+9

= −4 + 8 + 12

= 12

= 4 + 12 = 16

So, the vertex is (1, 12). The y-intercept is 9. So, the points (0, 9) and (2, 9) lie on the graph.

12 10 8

y

The maximum value is 16. 20. For y = 2x2 + 8x + 3, a = 2 and 2 > 0. So, the parabola

opens up, and the function has a minimum value. 8 8 b x = −___ = −____ = −__ = −2 2a 2(2) 4 y = 2x2 + 8x + 3

y = −3x2 + 6x + 9

y = 2(−2)2 + 8(−2) + 3 2 −3 −2

= 2(4) − 16 + 3 1 2

4 5 x

= 8 − 16 + 3

−4

= −8 + 3 = −5

The domain is all real numbers. The range is y ≤ 12. 17. For f(x) = 5x2 + 10x − 3, a = 5 and 5 > 0, So, the parabola

opens up, and the function has a minimum value. 10 b −10 = −1 x = −___ = −____ = ____ 2a 2(5) 10 f(x) = 5x2 + 10x − 3 f(−1) = 5(−1)2 + 10(−1) − 3 = 5(1) − 10 − 3 = 5 − 10 − 3 = −5 − 3 = −8

The minimum value is −5. 21.

t

0

0.5

1

1.5

2

y

0

4

16

36

64

y 80

(2, 64)

60 40

y = 16t 2

20

The minimum value is −8. 1 1 1 18. For f(x) = −—x2 + 2x + 16, a = −— and −— < 0. So, the 2 2 2 parabola opens down, and the function has a maximum value. 2 b −2 = 2 x = −___ = −— = ___ 2a −1 1 2 −—2 1 __ 2 f(x) = − x + 2x + 16 2 __ f(2) = − 1 (2)2 + 2(2) + 16 2 __ = − 1 (4) + 4 + 16 2 = −2 + 4 + 16

( )

= 2 + 16 = 18 The maximum value is 18.

0

0

1

2

4 t

3

The point (2, 64) lies on the graph. So, it takes 2 seconds for the coconut to fall 64 feet. 22. a.

t

0

0.25

0.5

0.75

1

1.25

y

25

24

21

16

9

0

y

y = −16t 2 + 25

24 18 12 6 0

(1.25, 0) 0

0.5

1

1.5

2 t

The positive t-intercept is 1.25. So, the pinecone hits the ground after 1.25 seconds. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

485

Chapter 8 b.

b. y = −16t2 + 36

t

0

0.5

1

1.5

y

36

32

20

0

−2 −1

x

0

1

2

3

4

f(x) = 2x2

8

2

0

2

8

18

32

g(x) = 2(x − 2)2

32

18

8

2

0

2

8

y 40

y = −16t 2 + 36

8 7 6 5

30 20

4 3

10 0

y

f(x) = 2x2 0

0.5

1

1.5

2 t

−3 −2−1

The positive t-intercept of the graph that represents the height of the second pinecone is 1.5. So, the second pinecone hits the ground after 1.5 seconds, which means the first pinecone hits the ground in the least amount of time. 32 b −32 = 1 23. t = −___ = −_______ = ____

g(x) = 2(x − 2)2

2 1

(1.5, 0)

1 2 3 4 5 x

Sample answer: The value of h causes a horizontal translation of the graph of y = ax2. 2. a.

2a 2(−16) −32 h(t) = −16t2 + 32t + 2

x

−4 −3 −2

−1

0

1

2

f(x) = −x2

−16 −9 −4

−1

0

−1

−4

g(x) = −(x + 2)2 −4 −1

h(1) = −16(1)2 + 32(1) + 2 = −16(1) + 32 + 2

−5 −4

= 16 + 2

−2

1 2 3 x

g(x) = −(x + 2)2

= 18

b.

8.4 Explorations (p. 441) −2 −1

f(x) = −x2 −4 −5 −6 −7 −8

Because the midpoint of the graph occurs when x = 1, the domain is 0 ≤ t ≤ 2. The highest point is the vertex (1, 18), and the lowest points are (0, 0) and (2, 0). So, the range is 0 ≤ h ≤ 18. The maximum height of the softball is 18 feet.

x

x

−4 −3 −2 −1

0

−32 −18 −8 −2

0

0

1

2

3

4

f(x) = −2x2

g(x) = −2(x + 2)2 −8 −2

f(x) = x2

4

1

0

1

4

9

16

g(x) = (x − 2)2

16

9

4

1

0

1

4

y

g(x) = −2(x + 2)2

4 3

f(x) = x2 −3 −2−1

2 1

0

1

2

−2 −8

−2 −8 −18 −32

y −5 −4 −3 −2−1

8

−1 −4 −9 −16

y

= −16 + 32 + 2

1. a.

0

−2 −3 −4 −5 −6 −7 −8

1 2 3 x

f(x) = −2x2

g(x) = (x − 2)2 1 2 3 4 5 x

Sample answer: The value of h causes a horizontal translation of the graph of y = ax2. 3. When h > 0, the graph of f(x) = a(x − h)2 is a horizontal

translation h units to the right of the graph of f(x) = ax2. When h < 0, the graph of f(x) = a(x − h)2 is a horizontal translation ∣ h ∣ units to the left of the graph of f(x) = ax2.

486

Algebra 1 Worked-Out Solutions

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Chapter 8 4. a. The graph of y = (x − 3)2 is a horizontal translation

3 units right of the graph of y =

x2.

5

4. g(x) = 2(x + 5)2

Because h = −5, graph the axis of symmetry x = −5 and plot the vertex (−5, 0). When x = −7:

When x = −6:

g(−7) = 2(−7 + 5)2 −1

8 −1

b. The graph of y = (x + 3)2 is a horizontal translation

3 units left of the graph of y =

x2.

g(−6) = 2(−6 + 5)2

= 2(−2)2

= 2(−1)2

= 2(4)

= 2(1)

=8

=2

So, plot (−7, 8) and (−6, 2), and the reflected points (−3, 8) and (−4, 2).

5

g(x) = 2(x + 5)2

−8

16 14 12 10 8 6

1 −1

c. The graph of y = −(x − 3)2 is a horizontal translation

3 units right and a reflection in the x-axis of the graph of y = x2.

y

4 2 −8 −7 −6 −5 −4 −3 −2−1

x

1 −1

8

−5

8.4 Monitoring Progress (pp. 442–445) 1.

f(x) = 5x f(−x) = 5(−x) = −5x = −f(x) Because f(−x) = −f(x), the function is odd.

2.

g(x) = 2x g(−x) =

2−x

1 = __ 2x

1 Because g(−x) = —x and −g(x) = −2x, you can conclude 2 that g(−x) ≠ g(x) and g(−x) ≠ −g(x). So, the function is neither even nor odd. 3.

Both graphs open up. The graph of g is narrower than the graph of f(x) = x2. The axis of symmetry, x = −5, and the vertex, (−5, 0), of the graph of g are 5 units left of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of g is a horizontal translation 5 units left and a vertical stretch by a factor of 2 of the graph of f. 5. h(x) = −(x − 2)2

Because h = 2, graph the axis of symmetry x = 2 and plot the vertex (2, 0). When x = 0:

When x = 1:

h(0) = −(0 − 2)2

h(1) = −(1 − 2)2

= −(−2)2

= −(−1)2

= −4

= −1

So, plot (0, −4) and (1, −1), and the reflected points (4, −4) and (3, −1). y −2−1 −2 −3 −4

2

4 5 6 x

h(x) = −(x − 2)2

h(x) = 2x2 + 3 h(−x) = 2(−x)2 + 3 = 2x2 + 3 = h(x) Because h(−x) = h(x), the function is even.

Copyright © Big Ideas Learning, LLC All rights reserved.

The graph of h opens down, while the graph of f(x) = x2 opens up. Also, the axis of symmetry, x = 2, and the vertex, (2, 0), of the graph of h are 2 units right of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of h is a reflection in the x-axis and a horizontal translation 2 units right of the graph of f.

Algebra 1 Worked-Out Solutions

487

Chapter 8 6. g(x) = 3(x − 1)2 + 6

8. The function f is of the form y = g(x) + k, where k = −3.

Because h = 1, graph the axis of symmetry x = 1. Because x = 1 and k = 6, plot the vertex (1, 6). When x = −1:

When x = 0:

g(−1) = 3(−1 − 1)2 + 6

g(0) = 3(0 − 1)2 + 6

= 3(−2)2 + 6

= 3(−1)2 + 6

= 3(4) + 6

= 3(1) + 6

= 12 + 6

=3+6

= 18

=9

So, the graph of f is a vertical translation 3 units down of the graph of g. To graph f, subtract 3 from the y-coordinates of the points on the graph of g. y −6 −5 −4 −3

f(x) = g(x) − 3

−7 −8

So, plot (−1, 18) and (0, 9), and the reflected points (3, 18) and (2, 9). 24

1 2 x

9. f(x) = a(x − h)2 + k

y

f(x) = a(x − 3)2 + 6 0 = a(6 − 3)2 + 6 0 = a(3)2 + 6 0 = a(9) + 6

9 6 3 −3 −2−1

g(x) = 3(x −

1)2

−6

+6

1 2 3 4 5 x

Both graphs open up. The graph of g is narrower than the graph of f(x) = x2. The vertex, (1, 6), is 1 unit right and 6 units up of the vertex, (0, 0), of the graph of f. So, the graph of g is a vertical stretch by a factor of 3 and a translation 1 unit right and 6 units up of the graph of f. 1

7. h(x) = —2 (x + 4)2 − 2

Because h = −4, graph the axis of symmetry x = −4. Because h = −4 and k = −2, plot the vertex (−4, −2). When x = −8:

When x = −6:

h(−8) = —12(−8 + 4)2 − 2

h(−6) = —12(−6 + 4)2 − 2

= —12(−4)2 − 2 =

1 —2 (16)

= —12 (−2)2 − 2

=8−2

=2−2

=6

=0

So, plot (−8, 6) and (−6, 0), and the reflected points (0, 6) and (−2, 0). y 12 10 8 1

h(x) = 2 (x + 4)2 − 2 2 4 x −4

Both graphs open up. The graph of h is wider than the graph of f(x) = x2. The vertex, (−4, −2), is 4 units left and 2 units down of the vertex, (0, 0), of the graph of f. So, the graph of h is a vertical shrink by a factor of —12 and a translation 4 units left and 2 units down of the graph of f.

488

When x = 4.5: 2 (4.5 − 3)2 + 6 f(4.5) = −__ 3 2 (1.5)2 + 6 = −__ 3 2 (2.25) + 6 = −__ 3 = −1.5 + 6 = 4.5 So, plot (4.5, 4.5) and (6, 0), and the reflected points (1.5, 4.5) and (0, 0). 8 7 6 5

y

4 3

6

−10 −8

−6 = 9a −6 ___ 9a —= 9 9 2 __ − =a 3 2 So, f(x) = −—3 (x − 3)2 + 6 models the path of a stream of water. Because h = 3, graph the axis of symmetry x = 3. Plot the vertex (3, 6).

= —12 (4) − 2

−2

−6

Algebra 1 Worked-Out Solutions

2 1 1 2 3 4 5 6 7 8 x

8.4 Exercises (pp. 446–448) Vocabulary and Core Concept Check 1. Sample answer: The graph of an even function is symmetric

about the y-axis. The graph of an odd function is symmetric about the origin.

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 2. A function whose graph has a vertex of (1, 2) will be of the

form y = a(x − + k, where h = 1 and k = 2. Sample answer: One possible function is y = −5(x − 1)2 + 2. h)2

10.

1

f(−x) = −—2(−x) = —12x

3. The graph of g is a horizontal translation h units right if h

is positive or ∣ h ∣ units left if h is negative, and a vertical translation k units up if k is positive or ∣ k ∣ units down if k is negative of the graph of f.

= −f(x) Because f(−x) = −f(x), the function is odd. 11.

4. The function that does not belong with the other three is

= 2x2 + 7x + 3 Because n(x) = 2x2 + 7x + 3 and −n(x) = −2x2 − 7x − 3, you can conclude that n(−x) ≠ n(x) and n(−x)≠ −n(x). So, the function is neither odd nor even.

Monitoring Progress and Modeling with Mathematics f(x) = 4x + 3 f(−x) = 4(−x) + 3 = −4x + 3 Because f(x) = 4x + 3 and −f(x) = −4x − 3, you can conclude that f(−x) ≠ f(x) and f(−x) ≠ −f(x). So, the function is neither odd nor even. 6.

g(x) = 3x2 g(−x) = 3(−x)2 = 3x2 = g(x) Because g(−x) = g(x), the function is even.

7.

h(x) = 5x + 2 h(−x) = 5−x + 2 1 +2 = __ 5x Because h(x) = 5x + 2 and −h(x) = −5x − 2, you can conclude that h(−x) ≠ h(x) and h(−x) ≠ −h(x). So, the function is neither odd nor even.

8.

m(x) = 2x2 − 7x m(−x) = 2(−x)2 − 7(−x) = 2x2 + 7x Because m(x) = 2x2 − 7 and −m(x) = −2x2 + 7x, you can conclude that m(−x) ≠ m(x) and m(−x) ≠ −m(x). So, the function is neither odd nor even.

9.

p(x) = −x2 + 8 p(−x) = −(−x)2 + 8

n(x) = 2x2 − 7x + 3 n(−x) = 2(−x)2 − 7(−x) + 3

f(x) = 2(x + 0)2. It is the only function that is not a horizontal translation of the parent function g(x) = x2.

5.

1

f(x) = −—2x

12.

r(x) = −6x2 + 5 r(−x) = −6(−x)2 + 5 = −6x2 + 5 = r(x) Because r(−x) = r(x), the function is even.

13. The graph is symmetric about the y-axis. So, the function is even. 14. The graph is symmetric about neither the y-axis nor the

origin. So, the function is neither even nor odd. 15. The graph is symmetric about neither the y-axis nor the

origin. So, the function is neither even nor odd. 16. The graph is symmetric about the y-axis. So, the function is even. 17. The graph is symmetric about the origin. So, the function is odd. 18. The graph is symmetric about neither the y-axis nor the

origin. So, the function is neither even nor odd. 19. For f(x) = 3(x + 1)2, because h = −1, the axis of symmetry

is x = −1, and the vertex is (−1, 0). 1

20. For f(x) = —4 (x − 6)2, because h = 6, the axis of symmetry is

x = 6, and the vertex is (6, 0). 1

21. For y = −—8 (x − 4)2, because h = 4, the axis of symmetry is

x = 4, and the vertex is (4, 0).

22. For y = −5(x + 9)2, because h = −9, the axis of symmetry

is x = −9, and the vertex is (−9, 0).

= −x2 + 8 = p(x) Because p(−x) = p(x), the function is even.

Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

489

Chapter 8 1

23. g(x) = 2(x + 3)2

25. r(x) = —4 (x + 10)2

Because h = −3, graph the axis of symmetry x = −3 and the vertex (−3, 0).

Because h = −10, graph the axis of symmetry x = −10 and the vertex (−10, 0).

When x = −5:

When x = −4:

When x = −14:

When x = −12:

g(−5) = 2(−5 + 3)2

g(−4) = 2(−4 + 3)2

r(−14) = —14(−14 + 10)2

r(−12) = —14(−12 + 10)2

= —14(−4)2

= —14(−2)2

= 2(−2)2

= 2(−1)2

= 2(4)

= 2(1)

=8

=2

=

= —14(4)

=4

So, plot (−5, 8) and (−4, 2), and the reflected points (−1, 8) and (−2, 2). 8 7 6 5

1 —4 (16)

=1

So, plot (−14, 4) and (−12, 1), and the reflected points (−6, 4) and (−8, 1). y

y

10 8 6

4 3 2 1

g(x) = 2(x + 3)2 −7 −6 −5 −4 −3 −2−1

−21

1 x

Both graphs open up. The graph of g is narrower than the graph of f(x) = x2. The axis of symmetry, x = −3, and the vertex, (−3, 0), of the graph of g are 3 units left of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of g is a horizontal translation 3 units left and a vertical stretch by a factor of 2 of the graph of f.

When x = −1:

When x = 0:

p(−1) = 3(−1 −

1)2

p(0) = 3(0 −

1)2

= 3(−2)2

= 3(−1)2

= 3(4)

= 3(1)

= 12

=3

So, plot (−1, 12) and (0, 3), and the reflected points (3, 12) and (2, 3).

−3

x

Both graphs open up. The graph of r is wider than the graph of f(x) = x2. The axis of symmetry, x = −10, and the vertex, (−10, 0), of the graph of r are 10 units left of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of r is a horizontal translation 10 units left and a vertical shrink by a factor of —14 of the graph of f. Because h = 6, graph the axis of symmetry x = 6 and the vertex (6, 0). When x = 0:

When x = 3:

n(0) = —13 (0 − 6)2

n(3) = —13 (3 − 6)2

= —13 (−6)2 =

= —13(−3)2

1 —3 (36)

= —13(9)

= 12

=3

So, plot (0, 12) and (3, 3), and the reflected points (12, 12) and (9, 3). y

y

p(x) = 3(x − 1)2 2 1 2 3 4 5 x

Both graphs open up. The graph of p is narrower than the graph of f(x) = x2. The axis of symmetry, x = 1, and the vertex, (1, 0), of the graph of p are 1 unit right of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of p is a horizontal translation 1 unit right and a vertical stretch by a factor of 3 of the graph of f.

490

−9

1

Because h = 1, graph the axis of symmetry x = 1 and the vertex (1, 0).

−3 −2−1

−15

26. n(x) = —3 (x − 6)2

24. p(x) = 3(x − 1)2

16 14

4 2

1

r(x) = 4 (x + 10)2

Algebra 1 Worked-Out Solutions

10 8 6 4 2

1

n(x) = 3(x − 6)2 2 4 6 8 10 12 14 16 x

Both graphs open up. The graph of n is wider than the graph of f(x) = x2. The axis of symmetry, x = 6, and the vertex, (6, 0), of the graph of n are 6 units right of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of n is a horizontal translation 6 units right and a vertical shrink by a factor of —13 of the graph of f. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 1

27. d(x) = —5 (x − 5)2

29. If f(−x) = f(x), then the function is even, not odd.

Because h = 5, graph the axis of symmetry x = 5 and the vertex (5, 0). When x = 0:

When x = 2:

d(0) = —15(0 − 5)2

d(2) = —15 (2 − 5)2

= —15(−5)2

= —15 (−3)2

= —15(25)

= —15 (9)

=5

= 1.8

So, plot (0, 5) and (2, 1.8), and the reflected points (10, 5) and (8, 1.8). 16 14 12

y

d(x) =

1 5(x

f(x) = x2 + 3 f(−x) = (−x)2 + 3 = x2 + 3 = f(x) Because f(−x) = f(x), the function is even. 30. The value of h is the x-coordinate of the vertex, not the

y-coordinate. y = −(x + 8)2 Because h = −8, the vertex is (−8, 0). 31. For y = −6(x + 4)2 − 3, because h = −4 and k = −3,

−

the vertex is (−4, −3), and the axis of symmetry is x = −4.

5)2

32. For f(x) = 3(x − 3)2 + 6, because h = 3 and k = 6, the

vertex is (3, 6), and the axis of symmetry is x = 3.

4 2 −2

33. For f(x) = −4(x + 3)2 + 1, because h = −3 and k = 1, the

vertex is (−3, 1), and the axis of symmetry is x = −3.

2 4 6 8 10 12 14 x

Both graphs open up. The graph of d is wider than the graph of f(x) = x2. The axis of symmetry, x = 5, and the vertex, (5, 0), of the graph of d are 5 units right of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of d is a horizontal translation 5 units right and a vertical shrink by a factor of —15 of the graph of f. 28. q(x) = 6(x + 2)2

Because h = −2, graph the axis of symmetry x = −2 and the vertex (−2, 0). When x = −4:

When x = −3:

q(−4) = 6(−4 + 2)2

q(−3) = 6(−3 + 2)2

=

= 6(−1)2

6(−2)2

= 6(4)

= 6(1)

= 24

=6

34. For y = −(x − 6)2 − 5, because h = 6 and k = −5, the

vertex is (6, −5), and the axis of symmetry is x = 6.

35. C; For y = −(x + 1)2 − 3, because a = −1, the graph opens

down, because h = −1 and k = −3, the vertex is (−1, −3). 1

1

36. A; For y = −—2 (x − 1)2 + 3, because a = −—2 , the graph

opens down and is wider than the graph of f(x) = x2. Also, because h = 1 and k = 3, the vertex is (1, 3). 1

1

37. D; For y = —3 (x − 1) + 3, because a = —3 , the graph opens up

and is wider than the graph of f(x) = and k = 3, the vertex is (1, 3).

x2. Also,

because h = 1

38. B; For y = 2(x + 1)2 − 3, because a = 2, the graph opens

up and is narrower than the graph of f(x) = x2. Also, because h = −1 and k = −3, the vertex is (−1, −3).

So, plot (−4, 24) and (−3, 6), and the reflected points (0, 24) and (−1, 6). 24 21

y

9

q(x) = 6(x + 2)2 −6 −5 −4 −3 −2−1

6 3 1 2 x

Both graphs open up. The graph of q is narrower than the graph of f(x) = x2. The axis of symmetry, x = −2, and the vertex, (−2, 0), of the graph of q are 2 units left of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of q is a horizontal translation 2 units left and a vertical stretch by a factor of 6 of the graph of f. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

491

Chapter 8 39. h(x) = (x − 2)2 + 4

41. r(x) = 4(x − 1)2 − 5

Because h = 2 and k = 4, plot the vertex (2, 4), and graph the axis of symmetry x = 2.

Because h = 1 and k = −5, plot the vertex (1, −5), and graph the axis of symmetry x = 1.

When x = 0:

When x = 1:

When x = −1:

When x = 0:

h(0) = (0 − 2)2 + 4

h(1) = (1 − 2)2 + 4

r(−1) = 4(−1 − 1)2 − 5

r(0) = 4(0 − 1)2 − 5

= 22 + 4

= (−1)2 + 4

= 4(−2)2 − 5

= 4(−1)2 − 5

=4+4

=1+4

= 4(4) − 5

= 4(1) − 5

=8

=5

= 16 − 5

=4−5

= 11

= −1

Plot (0, 8) and (1, 5), and the reflected points (4, 8) and (3, 5). 16 14

So, plot (−1, 11) and (0, −1), and the reflected points (3, 11) and (2, −1).

y

18 15 12 9

8 6 4 2 −2−1

y

h(x) = (x − 2)2 + 4

r(x) = 4(x − 1)2 − 5

1 2 3 4 5 6 x −3 −2−1

Both graphs open up. The vertex, (2, 4), of the graph of h is 2 units right and 4 units up from the vertex, (0, 0), of the graph of f. So, the graph of h is a translation 2 units right and 4 units up of the graph of f. 40. g(x) = (x + 1)2 − 7

Because h = −1 and k = −7, plot the vertex (−1, −7), and graph the axis of symmetry x = −1. When x = −5:

When x = −3:

g(−5) = (−5 + 1)2 − 7

g(−3) = (−3 + 1)2 − 7

= (−4)2 − 7

= (−2)2 − 7

= 16 − 7

=4−7

=9

= −3

y

Both graphs open up. The graph of r is narrower than the graph of f(x) = x2. The vertex, (1, −5), is 1 unit right and 5 units down from the vertex, (0, 0), of the graph of f. So, the graph of r is a vertical stretch by a factor of 4 and a translation 1 unit right and 5 units down of the graph of f. 42. n(x) = −(x + 4)2 + 2

Because h = −4 and k = 2, plot the vertex (−4, 2) and graph the axis of symmetry x = −4. When x = −6: n(−6) = −(−6 +

−2−1

+2

n(−5) = −(−5 + 4)2 + 2

= −(−2)2 + 2

= −(−1)2 + 2

= −4 + 2

= −1 + 2

= −2

=1

2

1 2 3 x

−4

g(x) = (x + 1)2 − 7

When x = −5: 4)2

So, plot (−6, −2) and (−5, 1), and the reflected points (−2, −2), and (−3, 1).

4 2 −5 −4

3 4 5 x

−6

So, plot (−5, 9) and (−3, −3), and the reflected points (3, 9) and (1, −3). 8 6

1

−8

Both graphs open up. The vertex, (−1, −7), of the graph of g is 1 unit left and 7 units down from the vertex, (0, 0), of the graph of f. So, the graph of g is a translation 1 unit left and 7 units down of the graph of f.

−8 −7

−4 −3

−1

y x

−4

n(x) = −(x + 4)2 + 2

The graph of n opens down, while the graph of f(x) = x2 opens up. The vertex, (−4, 2), of the graph of n is 4 units left and 2 units up from the vertex, (0, 0), of the graph of f. So, the graph of n is a reflection in the x-axis followed by a translation 4 units left and 2 units up of the graph of f.

492

Algebra 1 Worked-Out Solutions

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 1

43. g(x) = −—3 (x + 3)2 − 2

45. A; The graph of f(x) = (x − 2)2 + 1 has the vertex (2, 1),

Because h = −3 and k = −2, plot the vertex (−3, −2) and graph the axis of symmetry x = −3. When x = −9: When x = −6: 1

g(−9) = −—3(−9 + 3)2 − 2

1

g(−6) = −—3 (−6 + 3)2 − 2

1

= −—3 (−3)2 − 2

1

= −—3 (9) − 2

= −—3 (−6)2 − 2 = −—3 (36) − 2

1 1

= −12 − 2 = −3 − 2 = −14 = −5 So, plot (−9, −14) and (−6, −5), and the reflected points (3, −14) and (0, −5). y −10 −8 −6 −4−2

2 4 6 x

−6 −8 −10 −12 −14 −16

The graph of g opens down, while the graph of f(x) = x2 opens up. Also, the graph of g is wider than the graph of f, and the vertex, (−3, −2), of the graph of g is 3 units left and 2 units down from the vertex, (0, 0), of the graph of f. So, the graph of g is a reflection in the x-axis, a vertical shrink by a factor of —13, and a translation 3 units left and 2 units down of the graph of f. 1 —2 (x

− 2)2 − 4 Because h = 2 and k = −4, plot the vertex (2, −4) and graph the axis of symmetry x = 2. When x = −2: When x = 0: r(−2) = —12(−2 − 2)2 − 4 r(0) = —12 (0 − 2)2 − 4 = —12 (−4)2 − 4 =

and g(x) = f(x + 2) is of the form g(x) = f(x − h), where h = −2. So, the graph of g is a horizontal translation 2 units left of the graph of f, and the vertex of the graph of g is (2 − 2, 1) = (0, 1).

47. B; The graph of f(x) = (x − 2)2 + 1 has the vertex (2, 1),

and g(x) = f(x) + 2 is of the form g(x) = f(x) + k, where k = 2. So, the graph of g is a vertical translation 2 units up of the graph of f, and the vertex of the graph of g is (2, 1 + 2) = (2, 3).

g(x) = f(x) − 3 is of the form g(x) = f(x) + k, where k = −3. So, the graph of g is a vertical translation 3 units down of the graph of f, and the vertex of the graph of g is (2, 1 − 3) = (2, −2).

1

1 —2 (16)

46. C; The graph of f(x) = (x − 2)2 + 1 has the vertex (2, 1),

48. D; The graph of f(x) = (x − 2)2 + 1 has vertex (2, 1), and

g(x) = − 3 (x + 3)2 − 2

44. r(x) =

and g(x) = f(x − 1) is of the form g(x) = f(x − h), where h = 1. So, the graph of g is a horizontal translation 1 unit right of the graph of f, and the vertex of the graph of g is (2 + 1, 1) = (3, 1).

= —12 (−2)2 − 4 = —12 (4) − 4

−4

=8−4

=2−4

=4

= −2

49. For f(x) = 2(x − 1)2 + 1, because h = 1 and k = 1, the

vertex of f is (1, 1) and the axis of symmetry is x = 1. When x = −1: f(−1) = 2(−1 − =

2(−2)2

+1

+1

= 2(−1)2 + 1 = 2(1) + 1

=8+1

=2+1

=9

=3

So, the points (−1, 9) and (0, 3), and the reflected points (3, 9) and (2, 3) lie on the graph of f. The function g is of the form y = f(x − h), where h = −3. So, the graph of g is a horizontal translation 3 units left of the graph of f. To graph g, subtract 3 from the x-coordinates of the points on the graph of f. So, the points (−4, 9), (−3, 3), (−2, 1), (−1, 3), and (0, 9) lie on the graph of g. y 14 12 10 8

y

g(x) = f(x + 3)

4 −6 −5 −4 −3 −2−1 −6 −4−2 −4 −6

2

f(0) = 2(0 − 1)2 + 1

= 2(4) + 1

So, plot (−2, 4) and (0, −2), and the reflected points (6, 4) and (4, −2). 10 8 6

When x = 0: 1)2

4 2 1 2 x

6 8 10 x

1

r(x) = 2(x − 2)2 − 4

Both graphs open up. The graph of r is wider than the graph of f(x) = x2, and the vertex, (2, −4), of the graph of r is 2 units right and 4 units down from the vertex, (0, 0), of the graph of f. So, the graph of r is a vertical shrink by a factor of —12 , and a translation 2 units right and 4 units down of the graph of f. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

493

Chapter 8 50. For f(x) = −(x + 1)2 + 2, because h = −1 and k = 2, the

vertex of f is (−1, 2) and the axis of symmetry is x = −1. When x = −5:

When x = −3:

f(−5) = −(−5 + =

−(−4)2

1)2

+2

+2

=

vertex of f is (3, −1), and the axis of symmetry is x = 3. When x = 1:

f(−3) = −(−3 + −(−2)2

52. For f(x) = 5(x − 3)2 − 1, because h = 3 and k = −1, the

1)2

+2

f(1) = 5(1 −

+2

=

When x = 2: 3)2

5(−2)2

−1

f(2) = 5(2 − 3)2 − 1

−1

= 5(−1)2 − 1

= −16 + 2

= −4 + 2

= 5(4) − 1

= 5(1) − 1

= −14

= −2

= 20 − 1

=5−1

= 19

=4

So, the points (−5, −14) and (−3, −2), and the reflected points (3, −14) and (1, −2) lie on the graph of f. The function g is of the form y = af(x), where a = —12 . So, the graph of g is a vertical shrink by a factor of —12 of the graph of f. To graph g, multiply the y-coordinates of the points on the graph of f by —12. So, the points (−5, −7), (−3, −1), (−1, 1), (1, −1), and (3, −7) lie on the graph of g. 2 −8 −6 −4

y

So, the points (1, 19) and (2, 4), and the reflected points (5, 19) and (4, 4) lie on the graph of f. The function g is of the form y = f(x) − k, where k = −6. So, the graph of g is a vertical translation 6 units down of the graph of f. To graph g, subtract 6 from the y-coordinates of the points on the graph of f. So, the points (1, 13), (2, −2), (3, −7), (4, −2), and (5, 13) lie on the graph of g. y

2 4 6 8 x −4

1

g(x) = 2f(x)

12 9

−6 −8 −10 −12 −14

6 3 −1

1

3

5 6 7 x

−6

51. For f(x) = −3(x + 5)2 − 6, because h = −5 and k = −6, the

vertex of f is (−5, −6), and the axis of symmetry is x = −5. When x = −7:

When x = −6:

f(−7) = −3(−7 + 5)2 − 6

f(−6) = −3(−6 + 5)2 − 6

= −3(−2)2 − 6

= −3(−1)2 − 6

= −3(4) − 6

= −3(1) − 6

= −12 − 6

= −3 − 6

= −18

= −9

So, the points (−7, −18) and (−6, −9), and the reflected points (−3, −18) and (−4, −9) lie on the graph of f. The function g is of the form y = af(x), where a = 2. So, the graph of g is a vertical stretch by a factor of 2 of the graph of f. To graph g, multiply the y-coordinates of the points on the graph of f by 2. So, the points (−7, −36), (−6, −18), (−5, −12), (−4, −18), and (−3, −36) lie on the graph of g.

g(x) = f(x) − 6

−9

53. For f(x) = (x + 3)2 + 5, because h = −3 and k = 5, the

vertex of f is (−3, 5), and the axis of symmetry is x = −3. When x = −5: =

(−2)2

g(x) = 2f(x)

494

x

−8 −12 −16 −20 −24 −28 −32

Algebra 1 Worked-Out Solutions

+5

f(−4) = (−4 + 3)2 + 5 = (−1)2 + 5

=4+5

=1+5

=9

=6

So, the points (−5, 9) and (−4, 6), and the reflected points (−1, 9) and (−2, 6) lie on the graph of f. The function g is of the form y = f(x − h), where h = 4. So, the graph of g is a horizontal translation 4 units right of the graph of f. To graph g, add 4 to the x-coordinates of the points on the graph of f. So, the points (−1, 9), (0, 6), (1, 5), (2, 6), and (3, 9) lie on the graph of g.

y −8 −7 −6 −5 −4 −3 −2−1

When x = −4:

f(−5) = (−5 + 3)2 + 5

16 14 12 10

y

6 4 2 −3 −2−1

g(x) = f(x − 4) 1 2 3 4 5 x

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 54. For f(x) = −2(x − 4)2 − 8, because h = 4 and k = −8, the

vertex of f is (4, −8), and the axis of symmetry is x = 4. When x = 2:

When x = 3:

f(2) = −2(2 − 4)2 − 8

f(3) = −2(3 − 4)2 − 8

−8

= −2(−1)2 − 8

= −2(4) − 8

Diving Bird 2

= −2(1) − 8

= −8 − 8

= −2 − 8

= −16

= −10

So, the points (2, −16) and (3, −10), and the reflected points (6, −16) and (5, −10) lie on the graph of f. The function g is of the form y = −f(x). So, the graph of g is a reflection in the x-axis of the graph of f. To graph g, find the opposite of the y-coordinates of the points on the graph of f. So, the points (2, 16), (3, 10), (4, 8), (5, 10), and (6, 16) lie on the graph of g. 16 14 12 10 8 6

graph of r is a vertical stretch by a factor of 2 of the graph of h. To graph r, multiply the y-coordinates of the points on the graph of h by 2. So, the points (0, 62.5), (1, 22.5), (2.5, 0), (4, 22.5), and (5, 62.5) lie on the graph of r.

Height (meters)

=

−2(−2)2

b. The function r is of the form y = ah(x), where a = 2. So, the

y 64 56 48 40 32 24 16 8 0

r(t) = 2h(t ) 0 1 2 3 4 5 6 7 8 t

Time (seconds)

y

c. The graph of r is a vertical stretch by a factor of 2 of the

graph of h. The second bird starts its dive from a greater height. Because r(t) is twice h(t), the second bird starts at a height twice as high as the first bird. g(x) = −f(x)

4 2

1

56. a. For f(x) = −—9 (x − 30)2 + 25, because h = 30 and k = 25,

the vertex is (30, 25) and the axis of symmetry is x = 30.

1 2 3 4 5 6 7 8 x

For x = 15:

1

f(15) = −—9(15 − 30)2 + 25 1

= −—9(−15)2 + 25

55. a. For h(t) = 5(t − 2.5)2, because h = 2.5, the vertex of the

graph is (2.5, 0), and the axis of symmetry is x = 2.5. When t = 0: h(0) = 5(0 −

1

= −—9(225) + 25

When t = 1: 2.5)2

= −25 + 25

h(1) = 5(1 − 2.5)2

= 5(−2.5)2

= 5(−1.5)2

= 5(6.25)

= 5(2.25)

= 31.25

= 11.25

=0 For x = 21:

1

f(21) = −—9(21 − 30)2 + 25 1

= −—9(−9)2 + 25 1

= −—9(81) + 25

So, plot (0, 31.25) and (1, 11.25), and the reflected points (5, 31.25) and (4, 11.25).

= −9 + 25 = 16

y 64 56 48 40 32 24 16 8 0

So, plot (15, 0) and (21, 16), and the reflected points (45, 0) and (39, 16). Football Punt 1

h(t) = 5(t − 2.5)2 0 1 2 3 4 5 6 7 8 t

Time (seconds)

Height (yards)

Height (meters)

Diving Bird 1

y 32 28 24 20 16 12 8 4 0

1

f(x) = − 9 (x − 30)2 + 25

0 6 12 18 24 30 36 42 48 x

Distance (yards)

A negative height for y does not make sense in this context, and the maximum value occurs at the vertex, where y = 25. So, the range is 0 ≤ y ≤ 25. The corresponding domain values that produce nonnegative values of y are 15 ≤ x ≤ 45. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

495

Chapter 8 b. The function g is of the form y = f(x − h), where h = −5.

So, the graph of g is a horizontal translation 5 units left of the graph of f. To graph g, subtract 5 from the y-coordinates of the points on the graph of f. So, the points (10, 0), (16, 16), (25, 25), (34, 16), and (40, 0) lie on the graph of g.

Height (yards)

Football Punt 2 y 32 28 24 20 16 12 8 4 0

g(x) = f(x + 5)

f(x) = a(x + 2)2 − 4 −6 = a(−1 + 2)2 − 4 −6 = a(1)2 − 4 −6 = a(1) − 4 +4

+4

−2 = a So, a quadratic function whose graph has vertex (−2, −4) and passes through (−1, −6) is f(x) = −2(x + 2)2 − 4. 60. f(x) = a(x − h)2 + k

f(x) = a(x − 1)2 + 8 0 6 12 18 24 30 36 42 48 x

Distance (yards)

To find the domain of g, subtract 5 from the domain values of f. So, the domain of g is 10 ≤ x ≤ 40. The range of g is the same as the range of f, because a horizontal translation does not change the range values. So, the range of g is 0 ≤ y ≤ 25. c. The graph of g is a horizontal translation 5 units left of

the graph of f. The kicker is closer to his goal line on his second punt. On the first possession, the ball is punted 15 yards from the kicker’s goal line. On the second possession, the graph has been translated 5 yards closer to the kicker’s goal line, and the ball is punted only 10 yards from the kicker’s goal line. 57. f(x) = a(x − h)2 + k

f(x) = a(x − 1)2 + 2 10 = a(3 − 1)2 + 2 10 = a(2)2 + 2 10 = a(4) + 2 −2 −2 8 = 4a 8 4a —=— 4 4 2=a So, a quadratic function whose graph has vertex (1, 2) and passes through (3, 10) is f(x) = 2(x − 1)2 + 2. 58. f(x) = a(x − h)2 + k

f(x) = a(x + +5 −14 = a(0 + 3)2 + 5 −14 = a(3)2 + 5 −14 = a(9) + 5 −5 −5 −19 = 9a −19 9a —=— 9 9 19 −— = a 9 So, a quadratic function whose graph has vertex (−3, 5) and 19 2 passes through (0, −14) is f(x) = −— 9 (x + 3) + 5. 3)2

496

59. f(x) = a(x − h)2 + k

Algebra 1 Worked-Out Solutions

12 = a(3 − 1)2 + 8 12 = a(2)2 + 8 12 = a(4) + 8 −8

−8

4 = 4a 4 4a —=— 4 4 1=a So, a quadratic function whose graph has vertex (1, 8) and passes through (3, 12) is f(x) = 1(x − 1)2 + 8, or f(x) = (x − 1)2 + 8. 61. f(x) = a(x − h)2 + k

f(x) = a(x − 5)2 − 2 0 = a(7 − 5)2 − 2 0 = a(2)2 − 2 0 = a(4) − 2 +2

+2

2 = 4a 2 4a —=— 4 4 1 —=a 2 So, a quadratic function whose graph has vertex (5, −2) and passes through (7, 0) is f(x) = —12 (x − 5)2 − 2. 62. f(x) = a(x − h)2 + k

f(x) = a(x + 5)2 − 1 2 = a(−2 + 5)2 − 1 2 = a(3)2 − 1 2 = a(9) − 1 +1

+1

3 = 9a 3 9a —=— 9 9 1 —=a 3 So, a quadratic function whose graph has vertex (−5, −1) and passes through (−2, 2) is f(x) = —13 (x + 5)2 − 1. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 63.

f(x) = a(x − h)2 + k

When x = 89: 1 f(89) = −—(89 − 59)2 + 300 12 1 = −—(30)2 + 300 12 1 = −—(900) + 300 12

f(x) = a(x − 25)2 + 90 0 = a(50 − 25)2 + 90 0 = a(25)2 + 90 0 = a(625) + 90 − 90

− 90

− 90 = 625a −90 625a —=— 625 625 18 −— = a 125 18 2 So, f(x) = −— 125 (x − 25) + 90 models this portion of the roller coaster.

= −75 + 300

= −243 + 300

= 225

= 57

So, plot (119, 0) and (89, 225), and the reflected point (29, 225). When the given point, (119, 0) is reflected, you get a negative x-value, which does not make sense in this context. So, use (5, 57). Path of Flare

Height (meters)

Plot the vertex (25, 90), and because h = 25, the axis of symmetry is x = 25. When x = 37.5: 18 f(37.5) = −—(37.5 − 25)2 + 90 125 18 = −—(12.5)2 + 90 125 18 = −—(156.25) + 90 125

y 320 280 240 200 160 120 80 40 0

1

f(x) = −12 (x − 59)2 + 300

0

= −22.5 + 90 = 67.5

Height (feet)

Roller Coaster Track f(x) =

18 −125(x

−

25)2

+ 90

90

120 x

(−8) b 2a 2(2) y = 2x2 − 8x + 4

8 4

65. x = −— = −— = — = 2

= 2(4) − 16 + 4 = 8 − 16 + 4 = −8 + 4 = −4 So, the vertex is (2, −4). To find a, use the vertex and the y-intercept (0, 4). y = a(x − h)2 + k

0

10

20

30

40

50 x

f(x) = a(x − h)2 + k f(x) = a(x −

59)2

+ 300

0 = a(119 − 59)2 + 300 0 = a(60)2 + 300 0 = a(3600) + 300 − 300

60

y = 2(2)2 − 8(2) + 4

Distance (feet)

64.

30

Distance (meters)

So, plot (50, 0) and (37.5, 67.5), and the reflected points (0, 0) and (12.5, 67.5).

y 100 90 80 70 60 50 40 30 20 10 0

When x = 5: 1 f(5) = −—(5 − 59)2 + 300 12 1 = −—(−54)2 + 300 12 1 = −—(2916) + 300 12

− 300

−300 = 3600a −300 3600a —=— 3600 3600 1 −— = a 12 1 2 So, f(x) = −— 12 (x − 59) + 300 models the path of the flare. Plot the vertex (59, 300), and because h = 59, the axis of symmetry is x = 59. Copyright © Big Ideas Learning, LLC All rights reserved.

y = a(x − 2)2 − 4 4 = a(0 − 2)2 − 4 4 = a(−2)2 − 4 4 = a(4) − 4 +4

+4

8 = 4a 8 4a —=— 4 4 2=a So, the rewritten quadratic function is y = 2(x − 2)2 − 4.

Algebra 1 Worked-Out Solutions

497

Chapter 8 6 b 2a 2(3) y = 3x2 + 6x − 1

−6 6

66. x = −— = −— = — = −1

= 3(−1)2 + 6(−1) − 1 = 3(1) − 6 − 1 =3−6−1 = −3 − 1 = −4 So, the vertex is (−1, −4). To find a, use the vertex and the y-intercept (0, −1). y = a(x − h)2 + k y = a(x + 1)2 − 4 −1 = a(0 + 1)2 − 4 −1 = a(1)2 − 4 −1 = a(1) − 4 +4

+4

3=a So, the rewritten quadratic function is y = 3(x + 1)2 − 4. 10 b −10 2a 2(−5) −10 f(x) = −5x2 + 10x + 3

67. x = −— = −— = — = 1

f(1) = −5(1)2 + 10(1) + 3 = −5(1) + 10 + 3 = −5 + 10 + 3 =5+3

(−4) b 4 2a 2(−1) −2 f(x) = −x2 − 4x + 2 f(−2) = −(−2)2 − 4(−2) + 2 = −4 + 8 + 2 =4+2 =6 So, the vertex is (−2, 6). To find a, use the vertex and the y-intercept (0, 2). f(x) = a(x − h)2 + k f(x) = a(x + 2)2 + 6 2 = a(0 + 2)2 + 6 2 = a(2)2 + 6 2 = a(4) + 6 −6 −6 −4 = 4a −4 4a —=— 4 4 −1 = a So, the rewritten quadratic function is f(x) = −(x + 2)2 + 6.

68. x = −— = −— = — = −2

69. A function cannot be symmetric about the x-axis because it

would not pass the vertical line test. 70. The graph is a translation 2 units left and 3 units down

of the graph of y = x2. So, h = −2 and k = −3, and y = a(x − (−2))2 + (−3) = a(x + 2)2 − 3. 71. The function h(x) = f(x) + 4 is of the form y = f(x) + k,

where k = 4. So, the graph of h is a vertical translation 4 units up of the graph of f.

=8 So, the vertex is (1, 8).

h(x) = f(x) + 4

To find a, use the vertex and the y-intercept (0, 3).

h(x) = [−(x + 1)2 − 2] + 4

f(x) = a(x −

h)2

+k

f(x) = a(x − 1)2 + 8 3 = a(0 −

1)2

+8

3 = a(−1)2 + 8 3 = a(1) + 8 −8

−8

−5 = a So, the rewritten quadratic function is f(x) = −5(x − 1)2 + 8.

= −(x + 1)2 + (−2 + 4) = −(x + 1)2 + 2 So, h(x) = −(x + 1)2 + 2. 72. The function h(x) = f(x − 5) is of the form y = f(x − h),

where h = 5. So, the graph of h(x) is a horizontal translation 5 units right of the graph of f. h(x) = f(x − 5) = 2[(x − 5) − 1]2 + 1 = 2[x + (−5 − 1)]2 + 1 = 2(x − 6)2 + 1 So, h(x) = 2(x − 6)2 + 1.

73. The function h(x) = 2f(x) is of the form y = af(x), where

a = 2. So, the graph of h is a vertical stretch by a factor of 2 of the graph of f. h(x) = 2f(x) = 2[4(x − 2)2 + 3]

⋅

⋅

= 2 4(x − 2)2 + 2 3 = 8(x − 2)2 + 6

So, h(x) = 8(x − 2)2 + 6.

498

Algebra 1 Worked-Out Solutions

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Chapter 8 1

74. The function h(x) = —3 f(x) is of the form y = af(x), where a = —13. So, the graph of h is a vertical shrink by a factor of —13

of the graph of f. h(x) = —13 f(x)

= —13[−(x + 5)2 − 6] 1

⋅

= −—3(x + 5)2 − —13 6 1

= −—3(x + 5)2 − 2 1

So, h(x) = −—3(x + 5)2 − 2. 75. Because the graph is translated 2 units right and 5 units

down, the vertex is (2, −5). So, an equation of the function in vertex form is f(x) = (x − 2)2 − 5. Simplify f(x) = (x − 2)2 − 5. f(x) = [x2 − 2(x)(2) + 22] − 5 = (x2 − 4x + 4) − 5 = x2 − 4x + (4 − 5) = x2 − 4x − 1 So, an equation of the function in standard form is f(x) = x2 − 4x − 1. Sample answer: The vertex, (2, −5), can be quickly determined from the vertex form, and the y-intercept, −1, can be quickly determined from the standard form. 76. a. true; Sample answer: If g(x) = af(x), then g(−x) = af(−x).

b. First birdbath:

1 y = —(x − 18)2 − 4 81 1 0 = —(x − 18)2 − 4 81 1 81 0 = 81 —(x − 18)2 − 81 4 81 0 = (x − 18)2 − 324 0 = [x2 − 2(x)(18) + 182] − 324 0 = (x2 − 36x + 324) − 324 0 = x2 − 36x + (324 − 324) 0 = x2 − 36x 0 = x(x − 36) x=0 or x − 36 = 0 + 36 + 36 x = 36 The ends of the first birdbath occur when x = 0 and x = 36. So, the first birdbath is 36 − 0 = 36 inches wide. Second birdbath: 2 4 y = —x2 − —x 75 5 2 4 0 = —x2 − —x 75 5 2 4 75 0 = 75 —x2 − 75 —x 75 5 0 = 2x2 − 60x

Because f is odd, g(−x) = a(−f(x)) = −af(x) = −g(x), so g is odd.

d. true; Sample answer: If h(x) = f(x) + g(x), then

h(−x) = f(−x) + g(−x). Because f and g are odd, h(−x) = −f(x) − g(x) = −(f(x) + g(x)) = −h(x), so h is odd.

e. false; Sample answer: A counterexample is illustrated by

adding f(x) = x2 + 4, an even function, to g(x) = 3x, an odd function. h(x) = f(x) + g(x) = (x2 + 4) + (3x) = x2 + 3x + 4 h(−x) = (−x)2 + 3(−x) + 4 = x2 − 3x + 4 Because h(x) = x2 + 3x + 4 and −h(x) = −x2 − 3x − 4, you can conclude that h(−x) ≠ h(x) and h(−x) ≠ −h(x). So, the sum h(x) = x2 + 3x + 4 is neither even nor odd. 1

77. a. For y = — (x − 18)2 − 4, k = −4. So, the first birdbath 81

has a depth of 0 − (−4) = 4 inches. The vertex of the graph shown has a y-value of −6. So, the second birdbath has a depth of 0 −(−6) = 6 inches. So, the second birdbath is deeper.

Copyright © Big Ideas Learning, LLC All rights reserved.

⋅

⋅

2x = 0 2x 0 —=— 2 2 x=0

b. true; Sample answer: If g(x) = af(x), then g(−x) = af(−x).

h(−x) = f(−x) + g(−x). Because f and g are even, h(−x) = f(x) + g(x) = h(x), so h is even.

⋅

⋅

⋅

0 = 2x(x − 30)

Because f is even, g(−x) = af(x) = g(x), so g is even.

c. true; Sample answer: If h(x) = f(x) + g(x), then

⋅

x − 30 =

or

0

+ 30 + 30 x = 30

The ends of the second birdbath occur when x = 0 and x = 30. So, the second birdbath is 30 − 0 = 30 inches wide. So, the first birdbath is wider. 78. Factor the right side of y = 2x2 + 8x + 8.

y = 2x2 + 8x + 8 y = 2(x2 + 4x + 4) = 2(x2 + 2(x)(2) + 22) = 2(x + 2)2 So, y = 2(x + 2)2, which is in vertex form. Now you can see that a = 2 and h = −2, which means that the graph of y = 2x2 + 8x + 8 is a vertical stretch by a factor of 2 and a horizontal translation 2 units left of the graph of y = x2. 79. Your friend is incorrect because if the absolute value

function includes a horizontal translation, it will not be symmetric about the y-axis. Maintaining Mathematical Proficiency 80. x(x − 1) = 0

x=0

or

x−1= +1 x =

0 +1 1

The roots are x = 0 and x = 1.

Algebra 1 Worked-Out Solutions

499

Chapter 8 81. (x + 3)(x − 8) = 0

x+3= −3

8.5 Monitoring Progress (pp. 451–454) x−8= 0

or

0

−3

+8

x = −3

1. Because the x-intercepts are p = −2 and q = 3, plot (−2, 0)

p + q −2 + 3 1 and (3, 0). The axis of symmetry is x = — = — = —. 2 2 2 f(x) = (x + 2)(x − 3)

+8

x=

8

The roots are x = −3 and x = 8. 82. (3x − 9)(4x + 12) = 0

3x − 9 = +9

0

or

4x + 12 =

+9

− 12

3x = 9 3x 9 —=— 3 3 x=3

0

− 12

4x = −12 4x −12 —=— 4 4 x = −3

( ) ( )( ( )( )

(

x=

The roots are x = 3 and x = −3.

and the x-intercepts are 1 and 2. b. y = −(x − 1)(x − 2); Sample answer: The graph opens

12 9

(−2, 0)

x-intercepts are 0 and −3.

e. y = −x(x + 3); Sample answer: The graph opens down

and the x-intercepts are 0 and −3.

f. y = −x(x − 3); Sample answer: The graph opens down

and the x-intercepts are 0 and 3. g. y = (x + 1)(x − 2); Sample answer: The graph opens up

and the x-intercepts are −1 and 2.

h. y = −(x + 1)(x − 2); Sample answer: The graph opens

down and the x-intercepts are −1 and 2.

2. Sample answer: It is a parabola with x-intercepts p and q. 3. a. Changing the sign of a does not change the x-intercepts,

because they are still p and q. However, changing the sign of a does change the sign of the y-intercept when the y-intercept is not 0.

2 4 6 8 x −6 −9

2. Because the x-intercepts are p = −1 and q = 4, plot (−1, 0)

and (4, 0). The axis of symmetry is p + q −1 + 4 3 x = — = — = —. 2 2 2 g(x) = −2(x − 4)(x + 1)

()

( )( ( )( )

3 3 3 g — = −2 — − 4 — + 1 2 2 2 5 5 = −2 −— — 2 2 1 25 = —, or 12— 2 2

(

)

)

1 1 So, the vertex is 1—, 12— . 2 2 x= y

3 2

(112 , 12 12 )

12

to the new value of p. = a[x(x) + x(−q) + (−p)(x) + (−p)(−q)]

( 12 , −6 14 )

1 The domain is all real numbers. The range is y ≥ −6—. 4

b. One of the x-intercepts will change from the old value of p

f(x) = a(x − p)(x − q)

(3, 0)

−8 −6 −4

c. y = x(x − 3); Sample answer: The graph opens up and the d. y = x(x + 3); Sample answer: The graph opens up and the

1 2

f(x) = (x + 2)(x − 3)

6 3

down and the x-intercepts are 1 and 2. x-intercepts are 0 and 3.

)

y

8.5 Explorations (p. 449) 1. a. y = (x − 1)(x − 2); Sample answer: The graph opens up

)

1 1 1 f — = —+2 —−3 2 2 2 5 5 = — −— 2 2 25 1 = −—, or −6— 4 4 1 1 So, the vertex is —, −6— . 2 4

8

(−3, 0) −6 −4−2

(4, 0) 2

6 8 10 x

= a(x2 − qx − px + pq) = a[x2 − (q + p)x + pq] = ax2 − a(q + p)x + apq The y-intercept is apq. So, changing the value of p will change the value of the y-intercept when the y-intercept is not 0.

500

Algebra 1 Worked-Out Solutions

g(x) = −2(x − 4)(x + 1)

1 The domain is all real numbers, the range is y ≤ 12—. 2

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 3. h(x) = 4x2 − 36

8. g(x) = x2 + x −12

h(x) = 4(x2 − 9) h(x) =

4(x2 −

Because a = 1 and 1 > 0, the parabola opens up. Because c = −12, the y-intercept is −12. So, plot (0, −12).

32)

h(x) = 4(x + 3)(x − 3)

g(x) = x2 + x −12

Because the x-intercepts are p = −3 and q = 3, plot (−3, 0) and (3, 0). The axis of symmetry is p + q −3 + 3 0 x = — = — = — = 0. 2 2 2 h(x) = 4x2 − 36

The zeros of the function are −4 and 3. So, plot (−4, 0) and (3, 0).

h(0) = 4(0)2 − 36

= (x + 4)(x − 3)

2 −8 −6

= 4(0) − 36

y

−2

2 4 6 8 x

−4 −6 −8

= 0 − 36 = −36 So, the vertex is (0, −36).

g(x) = x2 + x − 12 −14

y

(−3, 0) −4

−2−1

(3, 0) 1 2

−10 −15 −20 −25

−40

4 x

9. Sample answer:

f(x) = (x − p)(x − q) = (x + 1)(x − 1) h(x) =

4x2

= x2 − 12

− 36

= x2 − 1

(0, −36)

So, a quadratic function is f(x) = x2 − 1.

x=0

The domain is all real numbers. The range is y ≥ −36.

f(x) = (x − h)2 + k

4. f(x) = (x − 6)(x − 1)

x−6=

0

or

= (x − 8)2 + 8

x−1= 0

+6 +6

+1

x= 6

= [x2 − 2(x)(8) + 82] + 8

+1

= (x2 − 16x + 64) + 8

x= 1

= x2 − 16x + (64 + 8)

So, the zeros of the function are 1 and 6. 5. g(x) =

=

3x2

− 12x + 12

3(x2

6. h(x) =

− 1)

=

x(x2

−

− 4x + 4)

= 3(x − 2)2 So, the zero of the function is a repeated root of 2.

= x2 −16x + 72

x(x2

= 3(x2 − 2(x)(2) + 22)

12)

= x(x + 1)(x − 1) So, the zeros of the function are 0, −1, and 1.

So, a quadratic function is f(x) = x2 −16x + 72. 11. f(x) = a(x − p)(x − q)

f(x) = a(x − 0)(x − 10) 12 = a(4 − 0)(4 − 10) 12 = a(4)(−6) 12 = −24a

7. f(x) = (x − 1)(x − 4)

The zeros of the function are 1 and 4. So, plot (1, 0) and (4, 0). f(x) = (x − 1)(x − 4) = x(x) + x(−4) + (−1)(x) + (−1)(−4) = x2 − 4x − x + 4

y

= x2 − 5x + 4 Because a = 1 and 1 > 0, the parabola opens up. Because c = 4, the y-intercept is 4. So, plot (0, 4).

10. Sample answer:

4 2 −6 −4−2

Copyright © Big Ideas Learning, LLC All rights reserved.

−4

6 8 10 x

f(x) = (x − 1)(x − 4)

−24a 12 —=— −24 −24 1 −— = a 2 1 f(x) = −—(x − 0)(x − 10) 2 1 = −—x(x − 10) 2 1 1 = −—x(x) − —x(−10) 2 2 1 = −—x2 + 5x 2 1 So, a quadratic function is f(x) = −—x2 + 5x. 2

Algebra 1 Worked-Out Solutions

501

Chapter 8 12. f(x) = a(x − p)(x − q)

15. f(x) = a(x − p)(x − q)(x − r)

f(x) = a(x + 5)(x − 4)

f(x) = a(x + 3)(x + 1)(x − 1)

−16 = a(3 + 5)(3 − 4)

−3 = a(0 + 3)(0 + 1)(0 − 1)

−16 = a(8)(−1)

−3 = a(3)(1)(−1)

−16 = −8a −16 −8a —=— −8 −8 2=a

−3 = −3a −3 −3a —=— −3 −3 1=a

f(x) = 2(x + 5)(x − 4)

f(x) = 1(x + 3)(x + 1)(x − 1)

= 2[x(x) + x(−4) + 5(x) + 5(−4)]

= (x + 3)[x(x) + x(−1) + 1(x) + 1(−1)]

= 2(x2 − 4x + 5x − 20)

= (x + 3)(x2 − x + x −1)

=

= (x + 3)(x2 − 1)

+ x − 20)

2(x2

= 2(x2) + 2(x) − 2(20)

= x(x2) + x(−1) + 3(x2) + 3(−1)

=

= x3 − x + 3x2 − 3

2x2

+ 2x − 40

So, a quadratic function is f(x) = 2x2 + 2x − 40. 13. g(x) = (x − 1)(x − 3)(x + 3)

The zeros of the function are −3, 1, and 3. So, plot (−3, 0), (1, 3), and (3, 0). −1

2

g(x)

16

−5

function f(x) = a(x − p)(x − q).

Monitoring Progress and Modeling with Mathematics

6 3

3. Because p = 1 and q = −3, the x-intercepts are −3 and 1.

−2−1

4 x

1 2

−6

g(x) = (x − 1)(x − 3)(x + 3)

− 6x + 5)

= x(x − 1)(x − 5) The zeros of the function are x = 0, x = 1, and x = 5. So, plot (0, 0), (1, 0), and (5, 0). x

0.5

3.5

h(x)

1.125

−13.125

Plot (0.5, 1.125) and (3.5, −13.125). 2 −2−1

y 2 3 4

p + q −3 + 1 −2 x = — = — = — = −1 2 2 2 So, the axis of symmetry is x = −1. 4. Because p = 2 and q = 5, the x-intercepts are 2 and 5.

14. h(x) = x3 − 6x2 + 5x

=

2. Sample answer: Find the x-coordinate of the vertex using

p+q x = —, then evaluate the function to determine the 2 maximum value or minimum value.

y

9

x(x2

8.5 Exercises (pp. 455–458) 1. The values of p and q are x-intercepts of the graph of the

Plot (−1, 16) and (2, −5).

−4

So, a cubic function is f(x) = x3 + 3x2 − x − 3.

Vocabulary and Core Concept Check

x

18

= x3 + 3x2 − x − 3

6 x

p+q 2+5 7 x=—=—=— 2 2 2 7 So, the axis of symmetry is x = —. 2 5. Because p = −7 and q = 5, the x-intercepts are −7 and 5.

p + q −7 + 5 −2 x = — = — = — = −1 2 2 2 So, the axis of symmetry is x = −1. 6. Because p = 0 and q = −8, the x-intercepts are −8 and 0.

p + q 0 − 8 −8 x = — = — = — = −4 2 2 2 So, the axis of symmetry is x = −4.

h(x) = x3 − 6x2 + 5x

502

Algebra 1 Worked-Out Solutions

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Chapter 8 7. Because the x-intercepts are p = −4 and q = −1, plot

(−4, 0) and (−1, 0). The axis of symmetry is 5 p + q −4 − 1 x = — = — = −—. 2 2 2 f(x) = (x + 4)(x + 1)

( ) ( ) ( −52 + 1 ) ( )( )

5 5 f −— = −— + 4 2 2 3 3 = — −— 2 2 9 = −— 4

—

(

9. Because the x-intercepts are p = −6 and q = 4, plot (−6, 0)

and (4, 0). The axis of symmetry is p + q −6 + 4 −2 x = — = — = — = −1. 2 2 2 y = −(x + 6)(x − 4) = −(−1 + 6)(−1 − 4) = −(5)(−5) = 25 So, the vertex is (−1, 25). y

)

5 9 So, the vertex is −—, −— . 2 4

(−1, 25) y = −(x + 6)(x − 4)

5

x =− 2

y

12 10 8 6

(−6, 0) f(x) = (x + 4)(x + 1)

−8

(4, 0) −4−2

2

6 8 x

4

(−4, 0)

(−1, 0)

−6 −5

1 2 x

(−52 , −94 )

−4

9 The domain is all real numbers. The range is y ≥ −—. 4 8. Because the x-intercepts are p = 2 and q = −2, plot (2, 0)

and (−2, 0). The axis of symmetry is p+q 2−2 0 x = — = — = — = 0. 2 2 2 y = (x − 2)(x + 2)

x = −1

The domain is all real numbers. The range is y ≤ 25. 10. Because the x-intercepts are p = 7 and q = 3, plot (7, 0)

and (3, 0). The axis of symmetry is p + q 7 + 3 10 x = — = — = — = 5. 2 2 2 h(x) = −4(x − 7)(x − 3) h(5) = −4(5 − 7)(5 − 3) = −4(−2)(2) = 16

y = (0 − 2)(0 + 2)

So, the vertex is (5, 16).

= (−2)(2) = −4 So, the vertex is (0, −4). 4 3

x=0

2 1

(−2, 0) −4 −3

y

−1 −2

(2, 0) 1

3 4 x

y = (x − 2)(x + 2) (0, −4)

16 14 12 10 8 6 4 2

y

(5, 16) h(x) = −4(x − 7)(x − 3)

(3, 0) 1 2

(7, 0) 4

8 x

x=5

The domain is all real numbers. The range is y ≤ 16.

The domain is all real numbers. The range is y ≥ −4.

Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

503

Chapter 8 11. Because the x-intercepts are p = −1 and q = −2, plot (−1, 0)

and (−2, 0). The axis of symmetry is 3 p + q −1 − 2 x = — = — = −—. 2 2 2 g(x) = 5(x + 1)(x + 2)

( ) ( )( 1 1 = 5( − )( ) 2 2

3 3 3 g −— = 5 −— + 1 −— + 2 2 2 2

)

— —

5 = −— 4

(

13. y = x2 − 9

= x2 − 32 = (x + 3)(x − 3) Because the x-intercepts are p = −3 and q = 3, plot (−3, 0) and (3, 0). The axis of symmetry is p + q −3 + 3 __ x = _____ = _______ = 0 = 0. 2 2 2 y = x2 − 9 = 02 − 9 =0−9

)

3 5 So, the vertex is −—, −— . 2 4

= −9 So, the vertex is (0, −9).

3

x =− 2 y

6

12 10 8

g(x) = 5(x + 1)(x + 2)

(−2, 0)

(

5 −4

4 2

(−3, 0) −8 −6 −4

(3, 0) 2 4 6 8 x

−2

y = x2 − 9

(−1, 0)

−6 −5 −4 −3 3 −2,

y

−10

1 2 x

)

(0, −9)

x=0

5 The domain is all real numbers. The range is y ≥ −—. 4 12. Because the x-intercepts are p = 3 and q = −4, plot (3, 0)

and (−4, 0). The axis of symmetry is 1 p+q 3−4 x = — = — = −—. 2 2 2 y = −2(x − 3)(x + 4)

( )( 7 7 = −2( − )( ) 2 2

1 1 = −2 −— − 3 −— + 4 2 2

)

— —

1 49 = —, or 24— 2 2

(

The domain is all real numbers. The range is y ≥ −9. 14. f(x) = x2 − 8x

= x(x − 8) Because the x-intercepts are p = 0 and q = 8, plot (0, 0) and (8, 0). The axis of symmetry is p + q 0 + 8 __ x = _____ = _____ = 8 = 4. 2 2 2 f(x) = x2 − 8x f(4) = 42 − 8(4) = 16 − 32

)

1 1 So, the vertex is −—, 24— . 2 2

= −16 So, the vertex is (4, −16). y

y

(−12 , 24 12 )

(−4, 0) −8 −6

x=

y = −2(x − 3)(x + 4)

(3, 0) −2

2 4 6 8 x

1 −2

(0, 0)

3

−4−2 −6 −9 −12 −15 −18

(8, 0) 2

6

10 12 x

f(x) = x2 − 8x (4, −16) x=4

The domain is all real numbers. The range is y ≥ −16.

1 The domain is all real numbers. The range is y ≤ 24—. 2

504

Algebra 1 Worked-Out Solutions

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Chapter 8 15. h(x) = −5x2 + 5x

17. q(x) = x2 + 9x + 14

= −5x(x − 1)

= (x + 7)(x + 2)

Because the x-intercepts are p = 0 and q = 1, plot (0, 0) and (1, 0). The axis of symmetry is p + q 0 + 1 __ x = _____ = _____ = 1. 2 2 2 h(x) = −5x2 + 5x

Because the x-intercepts are p = −7 and q = −2, plot (−7, 0) and (−2, 0). The axis of symmetry is

1 = −5 __ 1 2 + 5 __ 1 h __ 2 2 2 5 1 +— = −5 __ 4 2 5 5 = −— + — 4 2 5 =— 4 5 . 1 , __ So, the vertex is __ 2 4

9 9 2 9 q −__ = −__ + 9 −__ + 14 2 2 2 81 − ___ 81 + 14 = ___ 4 2 81 = −— + 14 4 25 ___ =− 4

()

() ()

( ) ( ) ( )

()

( ) 2

(0, 0) −2

y

p + q −7 − 2 9 x = — = — = −__. 2 2 2 q(x) = x2 + 9x + 14

y = x2 + 9x + 14 y

( 12 , 54 )

1

6

(1, 0)

−1

2 x

−8

h(x) = −5x2 + 5x

−6 −5

−3

−1

(−7, 0) −5 −6

(−92 , −254 )

9

1 2

5. The domain is all real numbers. The range is y ≤ __ 4 16. y = 3x2 − 48

= 3(x2 − 16) = 3(x2 − 42) = 3(x + 4)(x − 4) Because the x-intercepts are p = −4 and q = 4, plot (−4, 0) and (4, 0). The axis of symmetry is p + q −4 + 4 __ x = _____ = _______ = 0 = 0. 2 2 2 y = 3x2 − 48 = 3(0)2 − 48

x

−4 −6

x = −2 x=

4 2

(−2, 0)

25 The domain is all real numbers. The range is y ≥ −___. 4 18. p(x) = x2 + 6x −27

= (x + 9)(x − 3) Because the x-intercepts are p = −9 and q = 3, plot (−9, 0) and (3, 0). The axis of symmetry is p + q −9 + 3 ___ x = _____ = _______ = −6 = −3. 2 2 2 p(x) = x2 + 6x −27 p(−3) = (−3)2 + 6(−3) −27 = 9 − 18 −27 = −9 − 27 = −36 So, the vertex is (−3, −36).

= 3(0) − 48 = 0 − 48

y

(−9, 0) −12

= −48

−6

(3, 0) 6 9 x

So, the vertex is (0, −48). (−4, 0) −3 −2−1 −12 −18 −24 −30 −36

y

(4, 0) 1 2 3

x

x=0

p(x) = x2 + 6x − 27 (−3, −36) x = −3

The domain is all real numbers. The range is y ≥ −36. y = 3x2 − 48 (0, −48)

The domain is all real numbers. The range is y ≥ −48. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

505

Chapter 8 19. y = 4x2 − 36x + 32

21. y = −2(x − 2)(x − 10)

= 4(x2 − 9x + 8) = 4(x − 1)(x − 8) Because the x-intercepts are p = 1 and q = 8, plot (1, 0) and (8, 0). The axis of symmetry is p + q 1 + 8 __ x = _____ = _____ = 9. 2 2 2 y = 4x2 − 36x + 32

()

9 9 2 − 36 — = 4 __ + 32 2 2 81 − 162 + 32 = 4 ___ 4 = 81 − 162 + 32

() ( )

(

24. y = x2 − 17x + 52

)

(1, 0) x=

= (x − 4)(x − 13) The zeros of the function are 4 and 13.

(8, 0)

2 3 4 5 6 7

−5 −10 −15 −20 −25 −30 −35 −40 −45 −50

9 10 x

25. y = 3x2 − 15x − 42

9 2

= 3(x − 5x − 14) = 3(x + 2)(x − 7) The zeros of the function are −2 and 7. 26. g(x) = −4x2 − 8x − 4

y = 4x2 − 36x + 32

( 92 , −49)

The domain is all real numbers. The range is y ≥ −49. 20. y = −2x2 − 4x + 30

= −2(x2 + 2x − 15) Because the x-intercepts are p = −5 and q = 3, plot (−5, 0) and (3, 0). The axis of symmetry is p + q −5 + 3 −2 x = _____ = _______ = — = −1. 2 2 2 y = −2x2 − 4x + 30 =

− 4(−1) + 30

= −2(1) + 4 + 30 = −2 + 4 + 30 = 32 y

(−1, 32)

28. h(x) = (x2 − 36)(x − 11)

= (x2 − 62)(x − 11) = (x + 6)(x − 6)(x − 11) The zeros of the function are −6, 6, and 11.

y = −2x2 − 4x + 30

= x(x2 − 49) = x(x2 − 72) = x(x + 7)(x − 7) The zeros of the function are −7, 0, and 7. 30. y = x3 − x2 − 9x + 9

(−5, 0) −10 −8 −6

(3, 0) −2

x = −1

506

= (x + 5)(x2 − 22) = (x + 5)(x + 2)(x − 2) The zeros of the function are −5, −2, and 2.

29. y = x3 − 49x

= 2 + 30 So, the vertex is (−1, 32). The domain is all real numbers. The range is y ≤ 32.

= −4(x2 + 2x + 1) = −4(x2 + 2(x)(1) + 12) = −4(x + 1)2 = −4(x + 1)(x + 1) The zero of the function is the repeated root −1. 27. f(x) = (x + 5)(x2 − 4)

= −2(x + 5)(x − 3)

−2(−1)2

x+5= 0 or x−1= 0 −5 −5 +1 +1 x = −5 x= 1 The zeros of the function are −5 and 1. = (x + 8)(x − 3) The zeros of the function are −8 and 3.

9 , −49 . So, the vertex is __ 2 y

1

22. f(x) = —3 (x + 5)(x − 1)

23. g(x) = x2 + 5x − 24

= −81 + 32 = −49

x−2= 0 or x − 10 = 0 +2 +2 + 10 + 10 x= 2 x = 10 The zeros of the function are 2 and 10.

Algebra 1 Worked-Out Solutions

2 4 6 x

= (x3 − x2) + (−9x + 9) = x2(x − 1) − 9(x − 1) = (x − 1)(x2 − 9) = (x − 1)(x2 − 32) = (x − 1)(x + 3)(x − 3) The zeros of the function are −3, 1, and 3. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 31. D; The zeros of the function are −5 and −3. So, it matches

the graph whose x-intercepts are also −5 and −3.

39. y = x2 − 11x + 18

Because a = 1 and 1 > 0, the graph opens up.

32. F; The zeros of the function are −5 and 3. So, it matches the

graph whose x-intercepts are also −5 and 3.

Because c = 18, the y-intercept is 18. So, plot (0, 18). y = x2 − 11x + 18

33. C; The zeros of the function are 5 and −3. So, it matches the

graph whose x-intercepts are also 5 and −3.

= (x − 2)(x − 9) The zeros of the function are 2 and 9. So, plot (2, 0) and (9, 0).

34. B; The zeros of the function are 3 and 5. So, it matches the

graph whose x-intercepts are also 3 and 5.

20 16

35. A; The zeros of the function are −5 and 5. So, it matches the

12

graph whose x-intercepts are also −5 and 5.

36. E; The zeros of the function are −3 and 3. So, it matches the

graph whose x-intercepts are also −3 and 3.

y

y = x2 − 11x + 18

8 4 −2

4 6 8 10 12 14 x

−8 −12

37. f(x) = (x + 2)(x − 6)

The zeros of the function are −2 and 6. So, plot (−2, 0) and (6, 0). f(x) = (x + 2)(x − 6)

40. y = x2 − x − 30

Because a = 1 and 1 > 0, the graph opens up. Because c = −30, the y-intercept is −30. So, plot (0, −30).

= x(x) + x(−6) + 2(x) + 2(−6)

y = x2 − x − 30

= x2 − 6x + 2x − 12

= (x + 5)(x −6)

= x2 − 4x − 12

The zeros of the function are −5 and 6. So, plot (−5, 0) and (6, 0).

Because a = 1 and 1 > 0, the graph opens up. Because c = −12, the y-intercept is −12. So, plot (0, −12). 6 3 −6 −4

2 4

−12 −15 −18

y −8 −6

y

−2

2 4

8 x

−8 −12 −16

8 10 x

y = x2 − x − 30

−32

f(x) = (x + 2)(x − 6)

41. y = −5x2 − 10x + 40 38. g(x) = −3(x + 1)(x + 7)

Because a = −5 and −5 < 0, the graph opens down.

The zeros of the function are −1 and −7. So, plot (−1, 0) and (−7, 0). g(x) = −3(x + 1)(x + 7)

Because c = 40, the y-intercept is 40. So, plot (0, 40). y = −5x2 − 10x + 40 = −5(x2 + 2x − 8)

= −3[x(x) + x(7) + 1(x) + 1(7)]

= −5(x + 4)(x − 2)

= −3(x2 + 7x + x + 7)

The zeros of the function are −4 and 2. So, plot (−4, 0) and (2, 0).

= −3(x2 + 8x + 7) = −3(x2) − 3(8x) − 3(7)

48

= −3x2 − 24x − 21 Because a = −3 and −3 < 0, the graph opens down.

y = −5x2 − 10x + 40

g(x) = −3(x + 1)(x + 7)

Because c = −21, the y-intercept is −21. So, plot (0, −21).

Copyright © Big Ideas Learning, LLC All rights reserved.

−10 −8

32 24 16 8 −4−2 −16 −24 −32

y

2 4 x −10

−6

y

36 30 24 18 12 6

−2

4 6 x

Algebra 1 Worked-Out Solutions

507

Chapter 8 47. Sample answer:

42. h(x) = 8x2 − 8

f(x) = (x − p)(x − q)

Because a = 8 and 8 > 0, the graph opens up.

= (x − 1)(x − 9)

Because c = −8, the y-intercept is −8. So, plot (0, −8).

= x(x) + x(−9) + (−1)(x) + (−1)(−9)

h(x) = 8x2 − 8 =

8(x2

= x2 − 9x − x + 9

− 1)

= x2 − 10x + 9

= 8(x2 − 12)

So, a quadratic function is f(x) = x2 − 10x + 9.

= 8(x + 1)(x − 1) The zeros of the function are −1 and 1. So, plot (−1, 0) and (1, 0). 6

48. Sample answer:

f(x) = (x − p)(x − q)

y

= (x + 2)(x + 5)

4 2

= x(x) + x(5) + 2(x) + 2(5)

−4 −3 −2

= x2 + 5x + 2x + 10

2 3 4 x

= x2 + 7x + 10 h(x) =

8x2

So, a quadratic function is f(x) = x2 + 7x + 10.

−8

−8 −10

49.

43. To find the zeros, each factor should be set equal too. Then

each equation should be solved.

−3

or

0 −3

x−2= +2

x = −3

0 +2

x=

2

The zeros of the function are −3 and 2. 44. The binomial x2 − 9 needs to be factored.

y = (x +

4)(x2

f(x) = a(x + 4)(x − 3) −18 = a(2 + 4)(2 − 3) −18 = a(6)(−1)

y = 5(x + 3)(x − 2) x+3=

f(x) = a(x − p)(x − q)

−18 = −6a

−6a −18 = ____ ____ −6 −6 3=a

f(x) = 3(x + 4)(x − 3) = 3[x(x) + x(−3) + 4(x) + 4(−3)] = 3(x2 − 3x + 4x − 12)

− 9)

= (x + 4)(x2 − 32)

= 3(x2 + x − 12)

= (x + 4)(x + 3)(x − 3)

= 3(x2) + 3(x) + 3(−12)

The zeros of the function are −4, −3, and 3.

So, a quadratic function is f(x) = 3x2 + 3x − 36.

45. Sample answer:

50. f(x) = a(x − p)(x − q)

f(x) = (x − h)2 + k

f(x) = a(x + 5)(x + 1)

f(x) = (x − 7)2 − 3 =

(x2

− 2(x)(7) +

72)

−3

= (x2 − 14x + 49) − 3 =

x2

= 3x2 + 3x − 36

− 14x + (49 − 3)

= x2 − 14x + 46 So, a quadratic function is f(x) = x2 − 14x + 46. 46. Sample answer:

3 = a(−4 + 5)(−4 + 1) 3 = a(1)(−3) 3 = −3a −3a 3 —=— −3 −3 −1 = a f(x) = −(x + 5)(x + 1)

f(x) = (x − h)2 + k

= −[x(x) + x(1) + 5(x) + 5(1)]

f(x) = (x −

= −(x2 + x + 5x + 5)

4)2

+8

= (x2 − 2(x)(4) + 42) + 8

= −(x2 + 6x + 5)

= (x2 − 8x + 16) + 8

= −x2 − 6x − 5

= x2 − 8x + (16 + 8)

So, a quadratic function is f(x) = −x2 − 6x − 5.

= x2 − 8x + 24 So, a quadratic function is f(x) = x2 − 8x + 24.

508

Algebra 1 Worked-Out Solutions

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Chapter 8 51. Sample answer:

57. f(x) = a(x − p)(x − q)

f(x) = (x − p)(x − q)

f(x) = a(x + 3)(x − 2)

f(x) = (x − 7)(x − 0)

−8 = a(1 + 3)(1 − 2)

= (x − 7)(x)

−8 = a(4)(−1)

= x(x) − 7(x) = x2 − 7x So, a quadratic function is f(x) = x2 − 7x. 52. Sample answer:

f(x) = 2(x + 3)(x − 2)

f(x) = (x − p)(x − q)

= 2[x(x) + x(−2) + 3(x) + 3(−2)]

f(x) = (x − 0)(x − 6)

= 2(x2 − 2x + 3x − 6)

= x(x − 6)

= 2(x2 + x − 6)

= x(x) − x(6) =

x2

= 2(x2) + 2(x) + 2(−6)

− 6x

So, a quadratic function is f(x) = x2 − 6x. 53. Sample answer:

5 = a(6 − 1)(6 − 7)

f(x) = (x + 5)2 + 0

5 = a(5)(−1)

= (x + 5)2 = x2 + 2(x)(5) + 52 = x2 + 10x + 25 So, a quadratic function is f(x) = x2 + 10x + 25. 54. Sample answer:

Use the vertex (4, 0) and a = −1. +k

=

− 2(x)(4) +

= −x2 + 8x − 16 So, a quadratic function is f(x) = −x2 + 8x − 16. 55. Sample answer:

Use the vertex (0, − 3). f(x) = (x − h)2 + k f(x) = (x − 0)2 − 3 = x2 − 3 So, a quadratic function is f(x) = x2 − 3. 56. Sample answer:

Use the vertex (0, 10) and a = −1. f(x) = a(x −

+k

f(x) = −(x − 0)2 + 10 =

−x2

f(x) = −(x − 1)(x − 7) = −[x(x) + x(−7) + (−1)(x) + (−1)(−7)] = −(x2 − 7x − x + 7) = −x2 + 8x − 7

42]

= −(x2 − 8x + 16)

h)2

5 = −5a −5a 5 —=— −5 −5 −1 = a

= −(x2 − 8x + 7)

f(x) = −(x − 4)2 + k −[x2

So, the quadratic function is f(x) = 2x2 + 2x − 12. f(x) = a(x − 1)(x − 7)

f(x) = (x − h)2 + k

f(x) = a(x −

= 2x2 + 2x − 12 58. f(x) = a(x − p)(x − q)

Use the vertex (−5, 0).

h)2

−8 = −4a −8 −4a —=— −4 −4 2=a

+ 10

So, a quadratic function is f(x) = −x2 + 10.

Copyright © Big Ideas Learning, LLC All rights reserved.

So, the quadratic function is f(x) = −x2 + 8x − 7. 59. f(x) = a(x − p)(x − q)

f(x) = a(x + 2)(x − 4) 32 = a(2 + 2)(2 − 4) 32 = a(4)(−2) 32 = −8a −8a 32 —=— −8 −8 −4 = a f(x) = −4(x + 2)(x − 4) = −4[x(x) + x(−4) + 2(x) + 2(−4)] = −4(x2 − 4x + 2x − 8) = −4(x2 − 2x − 8) = −4(x2) − 4(−2x) − 4(−8) = −4x2 + 8x + 32 So, a quadratic function is f(x) = −4x2 + 8x + 32.

Algebra 1 Worked-Out Solutions

509

Chapter 8 60. f(x) = a(x − p)(x − q)

63. h(x) = (x − 2)(x + 2)(x + 7)

f(x) = a(x − 6)(x − 10)

The zeros of the function are −7, −2, and 2. So, plot (−7, 0), (−2, 0), and (2, 0).

−2 = a(8 − 6)(8 − 10) −2 = a(2)(−2) −2 = −4a −2 −4a —=— −4 −4 1=a __ 2 1 (x − 6)(x − 10) f(x) = __ 2 1 [x(x) + x(−10) + (−6)(x) + (−6)(−10)] = __ 2 1 (x2 − 10x − 6x + 60) = __ 2 1 (x2 − 16x + 60) = __ 2 1 (x2) − __ 1 (16x) + __ 1 (60) = __ 2 2 2 1 x2 − 8x + 30 = __ 2 1 x2 − 8x + 30. So, a quadratic function is f(x) = __ 2 61. y = 5x(x + 2)(x − 6)

The zeros of the function are −2, 0, and 6. So, plot (−2, 0), (0, 0), and (6, 0). x

−1

3

y

35

−225

x

−5

0

h(x)

42

−28

So, plot (−5, 42) and (0, −28). y

50 40 30 20 10 −10 −8

−4

4 6 x

h(x) = (x − 2)(x + 2)(x + 7)

−30

64. y = (x + 1)(x − 5)(x − 4)

The zeros of the function are −1, 4, and 5. So, plot (−1, 0), (4, 0), and (5, 0). x

1

y

24

4.5 −1.375

So, plot (1, 4.5) and (24, −1.375). 28 24 20

y

So, plot (−1, 35) and (3, −225). y −8 −6 −4

4 2 4

−2

8 x

1 2 3 4 5 6 x

y = (x + 1)(x − 5)(x − 4)

65. f(x) = 3x3 − 48x

= 3x(x2 − 16)

y = 5x(x + 2)(x − 6)

= 3x(x + 4)(x − 4)

62. f(x) = −x(x + 9)(x + 3)

The zeros of the function are −9, −3, and 0. So, plot (−9, 0), (−3, 0), and (0, 0). x

−7

−1

f(x)

−56

16

= 3x(x2 − 42) The zeros of the function are −4, 0, and 4. So, plot (−4, 0), (0, 0), and (4, 0). x

−2

2

f(x)

72

72

So, plot (−2, 72) and (2, 72).

So, plot (−7, −56) and (−1, 16). y

−14

−10

f(x) = −x(x + 9)(x + 3)

510

−6 −4

80

y

2 x −20 −30 −40 −50 −60

Algebra 1 Worked-Out Solutions

−8 −6

−2 −40 −60 −80

2

6 8 x

f(x) = 3x3 − 48x

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 66. y = −2x3 + 20x2 − 50x

69. f(x) = a(x − p)(x − q)(x − r)

f(x) = a(x + 1)(x − 0)(x − 4) f(x) = a(x)(x + 1)(x − 4) −24 = a(1)(1 + 1)(1 − 4) −24 = a(1)(2)(−3) −24 = −6a −24 −6a —=— −6 −6 4=a f(x) = 4(x)(x + 1)(x − 4) = 4x[x(x) + x(−4) + 1(x) + 1(−4)] = 4x(x2 − 4x + x − 4) = 4x(x2 − 3x − 4) = 4x(x2) + 4x(−3x) + 4x(−4) = 4x3 − 12x2 − 16x So, the cubic function is f(x) = 4x3 − 12x2 − 16x.

= −2x(x2 − 10x + 25) =

−2x(x2

− 2(x)(5) +

52)

= −2x(x − 5)2 The zeros of the function are 0 and a repeated root of 5. So, plot (0, 0) and (5, 0). y

x

2

6

y

−36

−12

−1

1 2 3 4 5 6 7 x

−10 −15 −20 −25 −30 −35 −40

So, plot (2, −36) and (6, −12).

y = −2x3 + 20x2 − 50x

67. y = −x3 − 16x2 − 28x

y = −x(x2 + 16x + 28)

70.

f(x) = a(x + 3)(x − 0)(x − 2)

= −x(x + 14)(x + 2)

f(x) = a(x)(x + 3)(x − 2)

The zeros of the function are −14, −2, and 0. So, plot (−14, 0), (−2, 0), and (0, 0). x

−10

−1

y

−320

13

−36 = a(−1)(−1 + 3)(−1 − 2) −36 = a(−1)(2)(−3) −36 = 6a 6a −36 = ___ ____ 6 6 −6 = a

So, plot (−10, −320) and (−1, 13). 50 −18

f(x) = a(x − p)(x − q)(x − r)

−9 −6

y

f(x) = −6(x)(x + 3)(x − 2) = 6x[x(x) + x(−2) + 3(x) + 3(−2)]

3 x

= −6x(x2 − 2x + 3x − 6)

f(x) = −x3 − 16x2 − 28x

68. g(x) =

6x3

+

30x2

= −6x(x2 + x − 6)

−200 −250 −300 −350

= −6x(x2) − 6x(x) − 6x(−6) = −6x3 − 6x2 + 36x So, the cubic function is f(x) = −6x3 − 6x2 + 36x.

− 36x

71.

= 6x(x2 + 5x − 6)

f(x) = a(x − p)(x − q)(x − r) f(x) = a(x + 7)(x + 4)(x − 0)

= 6x(x + 6)(x − 1) The zeros of the function are −6, 0, and 1. So, plot (−6, 0), (0, 0), and (1, 0). x

−4

0.5

g(x)

240

−9.75

So, plot (−4, 240) and (0.5, −9.75). g(x) = 6x3 + 30x2 − 36x 280

f(x) = a(x)(x + 7)(x + 4) 40 = a(−2)(−2 + 7)(−2 + 4) 40 = a(−2)(5)(2) 40 = −20a −20a 40 —=— −20 −20 −2 = a f(x) = −2(x)(x + 7)(x + 4)

y

= −2x[x(x) + x(4) + 7(x) + 7(4)]

240

= −2x(x2 + 4x + 7x + 28) = −2x(x2 + 11x + 28) = −2x(x2) − 2x(11x) − 2x(28) −10 −8

−4−2

2 4 x

Copyright © Big Ideas Learning, LLC All rights reserved.

= −2x3 − 22x2 − 56x So, the cubic function is f(x) = −2x3 − 22x2 − 56x.

Algebra 1 Worked-Out Solutions

511

Chapter 8 72. f(x) = a(x − p)(x − q)(x − r)

f(x) = a(x − 1)(x − 3)(x − 6) −40 = a(5 − 1)(5 − 3)(5 − 6) −40 = a(4)(2)(−1) −40 = −8a −40 −8a —=— −8 −8 5=a f(x) = 5(x − 1)(x − 3)(x − 6) = [5(x) − 5(1)][x(x) + x(−6) + (−3)(x) + (−3)(−6)] = (5x − 5)(x2 − 6x − 3x + 18) = (5x − 5)(x2 − 9x + 18) = 5x(x2) + 5x(−9x) + 5x(18) + (−5)(x2) + (−5)(−9x) + (−5)(18) = 5x3 − 45x2 + 90x − 5x2 + 45x − 90 = 5x3 + (−45x2 − 5x2) + (90x + 45x) − 90 = 5x3 − 50x2 + 135x − 90 So, the cubic function is f(x) = 5x3 − 50x2 + 135x − 90. 73. Sample answer:

f(x) = (x − p)(x − q)(x − r) = (x + 2)(x − 3)(x − 8) = (x + 2)[x(x) + x(−8) + (−3)(x) + (−3)(−8)] = (x + 2)(x2 − 8x − 3x + 24) = (x + 2)(x2 − 11x + 24) = x(x2) − x(11x) + x(24) + 2(x2) − 2(11x) + 2(24) = x3 − 11x2 + 24x + 2x2 − 22x + 48 = x3 + (−11x2 + 2x2) + (24x − 22x) + 48 = x3 − 9x2 + 2x + 48 So, a cubic function is f(x) = x3 − 9x2 + 2x + 48.

76. The graph must have a y-intercept of 6 in order to pass

through the point (0, 6). Sample answer: So, a cubic function is f(x) = x3 + 6. 77.

f(x) = a(x − 0)(x − 7) f(x) = a(x)(x − 7) 30 = a(2)(2 − 7) 30 = a(2)(−5) 30 = −10a −10a 30 —=— −10 −10 −3 = a f(x) = −3(x)(x − 7) = −3x(x − 7) = −3x(x) − 3x(−7) = −3x2 + 21x So, a quadratic function is f(x) = −3x2 + 21x. 78. f(x) = a(x − p)(x − q)

f(x) = a(x + 3)(x − 4) −72 = a(1 + 3)(1 − 4) −72 = a(4)(−3) −72 = −12a −12a −72 = _____ ____ −12 −12 6=a f(x) = 6(x + 3)(x − 4) = 6[x(x) + x(−4) + 3(x) + 3(−4)] = 6(x2 − 4x + 3x − 12) = 6(x2 − x − 12)

74. Sample answer:

f(x) = (x − p)(x − q)(x − r) = (x + 7)(x + 5)(x − 0) = (x)(x + 7)(x + 5) = x[x(x) + x(5) + 7(x) + 7(5)] = x(x2 + 5x + 7x + 35) = x(x2 + 12x + 35) = x(x2) + x(12x) + x(35) = x3 + 12x2 + 35x So, a cubic function is f(x) = x3 + 12x2 + 35x. 75. Sample answer:

f(x) = (x − p)(x − q)(x − r) = (x − 1)(x − 7)(x − 0) = x(x − 1)(x − 7) = x[x(x) + x(−7) + (−1)(x) + (−1)(−7)] = x(x2 − 7x − x + 7) = x(x2 − 8x + 7) = x(x2) − x(8x) + x(7) = x3 − 8x2 + 7x So, a cubic function is f(x) = x3 − 8x2 + 7x.

512

Algebra 1 Worked-Out Solutions

f(x) = a(x − p)(x − q)

= 6(x2) −6(x) − 6(12) = 6x2 − 6x − 72 So, a quadratic function is f(x) = 6x2 − 6x − 72. 79.

f(x) = a(x − p)(x − q)(x − r) f(x) = a(x + 4)(x + 3)(x − 3) −180 = a(0 + 4)(0 + 3)(0 − 3) −180 = a(4)(3)(−3) −180 = −36a −180 −36a —=— −36 −36 5=a f(x) = 5(x + 4)(x + 3)(x − 3) = [5(x) + 5(4)](x2 − 32) = (5x + 20)(x2 − 9) = 5x(x2) + 5x(−9) + 20(x2) + 20(−9) = 5x3 − 45x + 20x2 − 180 = 5x3 + 20x2 − 45x − 180 So, a cubic function is f(x) = 5x3 + 20x2 − 45x − 180. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 80. f(x) = a(x − p)(x − q)(x − r)

84. Sample answer:

f(x) = a(x + 8)(x + 3)(x − 0)

y

f(x) = a(x)(x + 8)(x + 3) −36 = a(−6)(−6 + 8)(−6 + 3) −36 = a(−6)(2)(−3)

= −x[x(x) + x(3) + 8(x) + 8(3)] + 3x + 8x + 24)

= −x(x2 + 11x + 24) = −x(x2) − x(11x) − x(24) So, a cubic function is f(x) = −x3 − 11x2 − 24x. 81. Because the x-intercepts are p = −4 and q = 2, plot (−4, 0)

and (2, 0). p + q −4 + 2 ___ x = _____ = _______ = −2 = −1 2 2 2 So, the vertex is (−1, −3). y

6 4 2 −8 −6

4 6 8 x −4

82. Sample answer: 21 18

( 21 )

1 foot deep. b. The satellite dish is 0 − −__ = __ 2

c. Use the x-intercepts p = −3 and q = 3 and the

y-intercept −1.5.

y = a(x − p)(x − q) y = a(x + 3)(x − 3) −1.5 = a(0 + 3)(0 − 3) −1.5 = a(3)(−3) −1.5 = −9a −9a −1.5 = — _____ −9 −9 1=a __ 6 1 (x + 3)(x − 3) y = __ 6 1 (x2 − 32) y = __ 6 1 (x2 − 9) y = __ 6 1 (x2) − __ 1 (9) y = __ 6 6 3 1 x2 − __ y = __ 2 6 A quadratic function that models the cross section of a 3 satellite dish is y = —16x2 − —2.

y

1 20 1 −—[32 − 19(3) + 48] 20 1 −—(9 − 57 + 48) 20 1 −—(−48 + 48) 20 1 −—(0) 20

86. a. y = −—(x2 − 19x + 48)

15 12 9

? 0=

6 3 −2

7 x

85. a. The satellite dish is 2 − (−2) = 4 feet wide.

= −x3 − 11x2 − 24x

12 10 8

1 2 3 4 5

−3 −4 −5

f(x) = −(x)(x + 8)(x + 3) =

−1 −2

−36 = 36a −36 36a — = ____ 36 36 −1 = a

−x(x2

1

? 0= 2 4 6 8 10

14 x

83. Sample answer: 6

? 0=

0=0✓

y

5 4 3

Because the location of the hoop is a solution of the equation, the ball is on path to go through the hoop. So, the player will make the shot.

2 1 −4 −3 −2−1

? 0=

1 2 3 4 x

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Algebra 1 Worked-Out Solutions

513

Chapter 8 b. The x-intercepts of the graph that models the shot are

92. a. Sample answer: The graph appears to intersect the x-axis

p = 3 and q = 13. y = a(x − p)(x − q) y = a(x − 3)(x − 13) 1.4 = a(10 − 3)(10 − 13) 1.4 = a(7)(−3) 1.4 = −21a −21a 1.4 —=— −21 −21 1 =a −___ 15 1 (x − 3)(x − 13) y = −___ 15 1 [x(x) + x(−13) + (−3)(x) + (−3)(−13)] y =−___ 15 1 (x2 − 13x − 3x + 39) =−___ 15 1 (x2 − 16x + 39) =−___ 15 1 (x2) − ___ 1 (−16x) − ___ 1 (39) =−___ 15 15 15 13 1 x2 + ___ 16 x − ___ =−___ 15 5 15 A quadratic function that models the path of the shot is 1 2 16 13 y = −— x−— . 15 x + — 15 5

at about p = 75 and q = 425. b. Sample answer: Using the estimates of p and q from part

(a), the width of the arch is about 425 − 75 = 350 feet. From the graph, the arch appears to have a maximum height of about 61 feet. To calculate a, first use the values of p and q to calculate the x-coordinate of the vertex. Then substitute this value for x in the function from part (a). Also substitute the estimate of 61 for f(x), and the estimates, of 75 and 425 for p and q. Solve the function for a. 93. a. f(x) = −3(x + 1)2 + 27

= −3[x2 + 2(x)(1) + 12] + 27 = −3(x2 + 2x + 1) + 27 = −3(x2) − 3(2x) − 3(1) + 27 = −3x2 − 6x − 3 + 27 = −3x2 − 6x + 24 = −3(x2 + 2x − 8) = −3(x + 4)(x − 2) So, the rewritten quadratic function is f(x) = −3(x + 4)(x − 2). b. 28

f(x) = −3(x + 1)2 + 27

87. D; y = −x2 + 5x

20 16

= −x(x − 5) Because the function is quadratic, the graph is a parabola. Also, a = −1 and −1 < 0. So, the parabola opens down. Because p = 0 and q = 5, the x-intercepts are 0 and 5.

12 8 4 −5

88. A; y = x2 − x − 12

= (x + 3)(x − 4) Because the function is quadratic, the graph is a parabola. Also, a = 1 and 1 > 0. So, the parabola opens up. Because p = −3 and q = 4, the parabola has one negative x-intercept and one positive x-intercept.

−3 −2−1

= 2[x2 − 2(x)(1) + 12] − 2 = 2(x2 − 2x + 1) − 2 = 2(x2) − 2(2x) + 2(1) − 2 = 2x2 − 4x + 2 − 2 = 2x2 − 4x = 2x(x − 2) So, the rewritten quadratic function is g(x) = 2x(x − 2). b.

6 5 4

Because the function is cubic, the graph has two curves. The polynomial that defines the function is not factorable. So, you know that the graph does not have an x-intercept of 0. So, rule out graph C. The only remaining cubic graph is graph B.

514

Algebra 1 Worked-Out Solutions

y

g(x) = 2(x − 1)2 − 2 −3 −2−1

by the table. Sample answer: Because −5 and 1 are the x-intercepts, the axis of symmetry is x = −2. The points (−3, 12) and (−1, 4) are the same horizontal distance from the axis of symmetry, so for both of them to lie on the parabola they would have to have the same y-coordinate.

3 x

94. a. g(x) = 2(x − 1)2 − 2

= x(x2 − 2x − 8) = x(x + 2)(x − 4) Because the function is cubic, the graph has two curves. Because p = 0, q = −2, and r = 4, the graph has one negative x-intercept, one positive x-intercept, and it passes through the origin.

91. It is not possible to write a quadratic function represented

1

Sample answer: Plot the vertex (−1, 27), which can be determined from the vertex form. Then plot the x-intercepts (−4, 0) and (2, 0), which can be determined from the intercept form. Draw a smooth curve through these points.

89. C; y = x3 − 2x2 − 8x

90. B; y = x3 − 4x2 − 11x + 30

y

1

3 4 5 x

−2

Sample answer: Plot the vertex (1, −2), which can be determined from the vertex form. Then plot the x-intercepts (0, 0) and (2, 0), which can be determined from the intercept form. Draw a smooth curve through these points. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 b. First, factor the function so it is in intercept form.

95. A function with exactly one real zero can be written in

g(x) = x(x2 − 1)(x2 − 4)

intercept form. Sample answer: Let p and q be real zeros of a quadratic function. When p = q, the function has exactly one real zero and can be written as y = a(x − p)(x − p).

= x(x2 − 12)(x2 − 22) = x(x + 1)(x − 1)(x + 2)(x − 2) Because the zeros of the function are −2, −1, 0, 1, and 2, plot (−2, 0), (−1, 0), (0, 0), (1, 0), and (2, 0).

96. Your friend is correct. Sample answer: In standard form,

f(x) = ax2 + bx + c, missing terms can be included by using 0 for b or c. In vertex form, f(x) = a(x − h)2 + k, missing values can be included by using 0 for h or k. 97.

Next, calculate the values for several additional points. x

f(x) = a(x + 5)(x + 2)(x − 1)(x − 4)(x − 8)

g(x)

−32 = a(0 + 5)(0 + 2)(0 − 1)(0 − 4)(0 − 8) −32 = a(5)(2)(−1)(−4)(−8)

−—2

—2

1

1—12

9 3— 32

13 −1— 32

13 1— 32

9 −3— 32

Plot these points to get an idea of the general shape of the curve. Draw a smooth curve through the plotted points.

−32 = −320a −320a −32 —=— −320 −320 1 —=a 10

8 6

y

4 2

So, the function represented by the graph is 1 f(x) = —(x + 5)(x + 2)(x − 1)(x − 4)(x − 8). 10

−4 −3

−1

f(x) = (x2 − 1)(x2 − 4)

99.

= (x2 − 12)(x2 − 22)

3 4 x

−4 −6 −8

98. a. First, factor the function so it is in intercept form.

g(x) = x(x2 − 1)(x2 − 4)

f(x) = kx2 − k2x − 2k3 = k(x2 − kx − 2k2)

= (x + 1)(x − 1)(x + 2)(x − 2)

= k(x + k)(x − 2k)

Because the zeros of the function are −1, 1, −2, and 2, plot (−1, 0), (1, 0), (−2, 0), and (2, 0).

x+k=

Next, calculate the values for several additional points.

−k

x

−1.5

0

1.5

f(x)

3 −2— 16

4

3 −2— 16

Plot these points to get a general idea of the general shape of the curve. Draw a smooth curve through the plotted points. 5 4

1

−1—12

y

0

or

x − 2k =

−k

+ 2k

x=−k

x=

0 + 2k 2k

So, the zeros of the function are −k and 2k. 100. Sample answer: Use p = −4 and q = 2.

When a = 1: y = a(x − p)(x − q) y = (x + 4)(x − 2) When a = 2: y = a(x − p)(x − q) y = 2(x + 4)(x − 2)

−4 −3

3 4 x −2 −3

f(x) = (x2 − 1)(x2 − 4)

Copyright © Big Ideas Learning, LLC All rights reserved.

So, two quadratic equations are y = (x + 4)(x − 2) and y = 2(x + 4)(x − 2). The two given points are x-intercepts, so the graphs of any two quadratic equations having these two intercepts would intersect at these points.

Algebra 1 Worked-Out Solutions

515

Chapter 8 101. Sample answer:

3 + 7 10 The axis of symmetry is x = — = — = 5. So, h = 5. 2 2 When k = 2:

106. −2 −6 −2

−54 −54 −18

—=3

—=3

The sequence has a common ratio of 3. So, it is geometric.

6 = a(3 − 5)2 + 2

107. 26

6 = a(−2)2 + 2

18 18 − 26 = −8

6 = 4a + 2

10 10 − 18 = −8

2 2 − 10 = −8

−6 −6 − 2 = −8

The sequence has a common difference of −8. So, it is arithmetic.

−2

4 = 4a 4 4a —=— 4 4 1=a

108. 4

5

9

5−4=1

9−5=4

4

5

When k = 10:

14 − 9 = 5

23 23 − 14 = 9

14

9 5

—

6 = a(3 − 5)2 + 10

14

9

5 4

y = a(x − h)2 + k

23

14 9

—

23 14

—

—

The sequence has neither a common difference nor a common ratio. So, it is neither arithmetic nor geometric.

6 = a(−2)2 + 10 6 = 4a + 10 − 10

−18 −18 −6

—=3

y = a(x − h)2 + k

−2

−6

8.6 Explorations (p. 459)

− 10

1.

−4 = 4a −4 4a —=— 4 4 −1 = a

t

y=t

t

y = 2t − 1

t

y = t2

0

0

0

0

0

0

0.2

0.2

0.2

0.15

0.2

0.04

So, two quadratic equations are y = (x − 5)2 + 2 and y = −(x − 5)2 + 10.

0.4

0.4

0.4

0.32

0.4

0.16

0.6

0.6

0.6

0.52

0.6

0.36

The two given points have the same y-coordinate, so the axis of symmetry of any parabola passing through them would be halfway between, which is x = 5. The vertex form can be used with h = 5 and any selected value of k, where k ≠ 6, to find quadratic equations that would pass through these points. Any two of these equations would form a system with these two points as solutions.

0.8

0.8

0.8

0.74

0.8

0.64

1.0

1.0

1.0

1.00

1.0

1.00

102. Find the point that has an x-coordinate of 12. It has a

y-coordinate of 300. So, the burger that contains 12 grams of fat has 300 calories. 103. Find the point that has a y-coordinate of 600. It has an

of grams of fat increases.

11 − 3 = 8

3

21 21 − 11 = 10

11 11 3

—

33 33 − 21 = 12

21 21 11

—

47 47 − 33 = 14

33 33 21

11 7

—=—

0.6

y=t 0.4 0.2

y = t2 0

0.2

0.4

0.6

0.8

1.0 t

Time (minutes)

104. The number of calories tends to increase as the number

11

y = 2t − 1

0.8

0

x-coordinate of about 29. So, the burger that contains 600 calories has about 29 grams of fat.

105. 3

y 1.0

Distance (miles)

Maintaining Mathematical Proficiency

Motion of cars

47

Sample answer: All three cars have the same average speed over the interval. You can tell the first car has a constant speed because it is modeled by a linear function and is traveling the same distance in each time interval. The third car, which is modeled by a quadratic function, is accelerating the most because it has the greatest change in speed over each time interval.

47 33

—

The sequence has neither a common difference nor a common ratio. So, it is neither arithmetic nor geometric.

516

Algebra 1 Worked-Out Solutions

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 2.

3.

t

y=t

t

y = 2t − 1

t

y = t2

1

1

1

1

1

1

8

1.5

1.5

1.5

1.83

1.5

2.25

4 2

2

2

2

3

2

4

2.5

2.5

2.5

4.66

2.5

6.25

3

3

3

7

3

9

3.5

3.5

3.5

10.31

3.5

12.25

4

4

4

15

4

16

4.5

4.5

4.5

21.66

4.5

20.25

5

5

5

31

5

25

Sample answer: The first car has a constant speed. The second and third cars are accelerating. The second car, which is modeled by an exponential equation, is accelerating at a greater rate than the third car. So, this car eventually overtakes the others because its speed is significantly greater than the other two by the last few time intervals. 3. Sample answer: Graph functions of each type in the same

coordinate plane and compare the shapes of the graphs.

10 6

−5 −4 −3 −2−1

1 2 3 x

−4 −6

The points appear to model a linear function.

⤻ +1 ⤻⤻ +1 +1 ⤻ +1

4.

x

−1

0

1

2

3

y

1

3

9

27

81

⤻⤻⤻⤻ ×3

×3

×3

×3

Consecutive y-values have a common ratio. So, the table represents an exponential function. +1 ⤻⤻ +1 +1 ⤻ +1 ⤻

5.

x

−1

0

1

2

3

y

16

8

4

2

1

⤻ ⤻ ⤻1 ⤻1 1 1

4. The growth rate of an exponential function is eventually

much greater than the growth rates of the other two functions. Sample answer: An exponential graph begins with small changes in each interval, but the change increases more rapidly as the value of the exponent increases.

y

×—2

×—2

×—2

×—2

Consecutive y-values have a common ratio. So, the table represents an exponential function. When x = 0, y = 8. So, a = 8.

8.6 Monitoring Progress (pp. 460–464) 1.

5

y

y = abx x

()

4 3

y = 8 —12

2 1

So, the exponential function is y = 8 —12 .

−3 −2−1

()

x

1 2 3 4 5 x

−2 −3

The points appear to represent a quadratic function. 2.

The common ratio is b = —12 .

32 28 24 20 16

y

f(b) − f(a) 1400 − 650 average rate of change = — = — b−a 10 − 0 750 = — = 75 10 Website B: Use (0, 550) and (10, 850). f(b) − f(a) 850 − 550 average rate of change = — = — b−a 10 − 0 300 = — = 30 10

12 8 4 −5 −4 −3 −2−1

6. Website A: Use (0, 650) and (10, 1400).

1 2 3 x

From Day 0 to Day 10, Website A membership increases at an average rate of 75 people per day and Website B membership increases at an average rate of about 30 people per day. So, Website A membership is growing faster.

The points appear to model an exponential function.

Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

517

Chapter 8 7. Let x represent the number of years since 1900.

10. 16

Littleton: L(x) = 50x + 1000 Tinyville: T(x) = 500(1.08)x Use a graphing calculator to graph each function in the same viewing window. Use the intersect feature to find the value of x for which L(x) = T(x). The graphs intersect when x ≈ 17.

6 4 2

2500

T

y

14 12 10 8

−4 −3 −2−1

L

1 2 3 4 x

The points appear to lie on a continually increasing curve. So, they appear to represent an exponential function.

Intersection

0 X=16.999902 Y=1849.9951 25 0

11. 12 10 8 6

So, the populations were about equal in 1917. 8.6 Exercises (pp. 465–468)

4 2

Vocabulary and Core Concept Check −4 −3 −2−1

1. Three types of functions that you can use to model data are

linear, exponential, and quadratic. Linear functions are of the form y = mx + b, and their graphs are straight lines. Exponential functions are of the form y = abx, and their graphs are continually increasing or decreasing curves. Quadratic functions are of the form y = ax2 + bx + c, and their graphs are parabolas.

The points appear to lie on a parabola. So, they appear to represent a quadratic function. 12.

18 15 12 9

determine what shape they appear to take.

y

6 3

3. Find the slope of the line through (a, f(a)) and (b, f(b)).

−4 −3 −2−1

4. The graph that does not belong with the other three is the

1 2 3 4 x

−6

graph of function n, because it is an exponential function. The rest are quadratic functions.

5. The points appear to lie on a parabola. So, they appear to

1 2 3 4 x

−4

2. Sample answer: Plot the points from the data set and

Monitoring Progress and Modeling with Mathematics

y

The points appear to lie on a parabola. So, they appear to represent a quadratic function. 13.

y −4 −3 −2−1

represent a quadratic function.

1 2 3 4 x

−1

6. The points appear to lie on a continually decreasing curve.

−2

So, they appear to represent an exponential function.

−3

7. The points appear to lie on a continually increasing curve. −4

So, they appear to represent an exponential function.

The points appear to lie on a straight line. So, they appear to represent a linear function.

8. The points appear to lie on a straight line. So, they appear to

represent a linear function. 9.

4

14.

y

8 7 6 5 4 3

3 2 1 −4 −3 −2−1

1 2 3 4 x

−2 −3 −4

The points appear to lie on a straight line. So, they appear to represent a linear function.

518

Algebra 1 Worked-Out Solutions

y

2 1 −4

−3

−2

−1

x

The points appear to lie on a continually increasing curve. So, they appear to represent an exponential function. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 +1 ⤻⤻ +1 +1 ⤻ +1 ⤻

15.

⤻ +1 ⤻ +1 ⤻ +1 ⤻ +1 ⤻ +1

21.

x

−2

−1

0

1

2

x

−2

−1

0

1

2

3

y

0

0.5

1

1.5

2

y

8

0

−4

−4

0

8

⤻⤻⤻⤻

⤻ ⤻⤻⤻⤻ +(−8) +(−4) +0 +4 +8

+0.5 +0.5 +0.5 +0.5

⤻ ⤻⤻⤻

+4 +4 +4 +4 The second differences are constant. So, the table represents a quadratic function.

The first differences are constant. So, the table represents a linear function.

⤻ +1 ⤻⤻ +1 +1 ⤻ +1

16.

y = a(x − p)(x − q)

x

−1

0

1

2

3

y

0.2

1

5

25

125

y = a(x + 1)(x − 2) −4 = a(0 + 1)(0 − 2)

⤻⤻⤻⤻ ×5

×5

×5

−4 = a(1)(−2)

×5

−4 = −2a −4 −2a —=— −2 −2 2=a

Consecutive y-values have a common ratio. So, the table represents an exponential function.

⤻ +1 ⤻ +1 ⤻ +1 ⤻ +1

17.

y = 2(x + 1)(x − 2)

x

2

3

4

5

6

y

2

6

18

54

162

= 2[x(x) + x(−2) + 1(x) + 1(−2)]

⤻⤻⤻⤻ ×3

×3

×3

= 2(x2 − 2x + x − 2)

×3

= 2(x2 − x − 2)

Consecutive y-values have a common ratio. So, the table represents an exponential function.

= 2(x2) − 2(x) − 2(2) = 2x2 − 2x − 4

⤻ +1 ⤻ +1 ⤻ +1 ⤻ +1

18.

So, the quadratic function is y = 2x2 − 2x − 4.

x

−3

−2

−1

0

1

y

2

4.5

8

12.5

18

⤻ ⤻⤻⤻ +2.5 +3.5 +4.5 +5.5 ⤻ ⤻ ⤻ +1 +1 +1

The second differences are constant. So, the table represents a quadratic function.

Time, t Distance, d

0.5

1

3

5

0.335

0.67

2.01

3.35

Cost, c

3

4

5

6

63.90

113.60

177.50

255.60

⤻ +63.9 ⤻ +78.1 ⤻ +49.7 ⤻ ⤻ +14.2

−2

−1

0

1

8

4

2

1

0.5

⤻ ⤻ ⤻ ⤻1 1 1 1

+14.2

The data can be modeled by a quadratic function because the second differences are constant.

×—2

×—2

×—2

Consecutive y-values have a common ratio of —12. So, the table represents an exponential function with b = —12 . When x = 0, y = 1. So, a = 1.

⤻ ⤻ ⤻

+1 +1 +1 ⤻ ⤻ ⤻ Diameter, d

−3

y

y = abx

+0.335 +1.34 +1.34 1.34 1.34 0.335 — = 0.67 — = 0.67 — = 0.67 2 2 0.5 The data have a constant rate of change. So, they can be modeled by a linear function. 20.

x

×—2

⤻ ⤻ +0.5 ⤻ +2 +2

19.

+1 ⤻ +1 ⤻ +1 ⤻ +1 ⤻

22.

x

() y = (—) y = 1 —12

1 x 2

()

1 x So, the exponential function is y = —2 .

⤻ +1 ⤻ +1 ⤻⤻ +1 +1

23.

x y

−2

−1

0

1

2

4

1

−2

−5

−8

⤻ ⤻ ⤻⤻ –3 –3 –3 –3

The first differences are constant. So, the table represents −3 a linear function whose rate of change is m = — = −3. 1 When x = 0, y = −2. So, the y-intercept is b = −2. y = mx + b y = −3x − 2 So, the linear function is y = −3x − 2.

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Algebra 1 Worked-Out Solutions

519

Chapter 8 ⤻ +1 ⤻⤻ +1 +1 ⤻ +1

24.

28. The factor for the first intercept should be (x + 2), not (x − 2).

x

−1

0

1

2

3

y

2.5

5

10

20

40

⤻ +1 ⤻ +1 ⤻ +1 ⤻ +1

⤻ ⤻⤻⤻ ×2 ×2 ×2 ×2

x

−3

−2

−1

0

1

y

4

0

−2

−2

0

⤻ ⤻ ⤻⤻ −4 −2 +0 +2 ⤻ ⤻ ⤻ +2 +2 +2

Consecutive y-values have a common ratio of 2. So, the table represents an exponential function with b = 2. When x = 0, y = 5. So, a = 5.

The second differences are constant. So, the table represents a quadratic function.

y = abx y = 5(2)x So, the exponential function is y = 5(2)x.

f(x) = a(x + 2)(x − 1) 4 = a(−3 + 2)(−3 − 1)

⤻ +1 ⤻ +1 ⤻ +1 ⤻ +1

25.

4 = a(−1)(−4)

x

−2

−1

0

1

2

y

−12

−3

0

−3

−12

⤻ ⤻ ⤻ ⤻ +9 +3 −3 −9 ⤻ ⤻ ⤻ −6 −6 −6

The second differences are constant and the points appear to lie on a parabola. So, the data represent a quadratic function whose vertex and only x-intercept is (0, 0).

4 = 4a 4 4a —=— 4 4 1=a f(x) = 1(x + 2)(x − 1) = (x + 2)(x − 1) = x(x) + x(−1) + 2(x) + 2(−1) = x2 − x + 2x − 2

y = a(x − h)2 + k

= x2 + x − 2

y = a(x − 0)2 + 0

So, the quadratic function is f(x) = x2 + x − 2.

y = a(x)2 −3 = a(1)2

29. a.

Football Game Attendance

−3 = a(1) −3 = a People

So, the quadratic function is y = −3x2.

⤻⤻⤻⤻ +1 +1 +1 +1

26.

p 400 350 300 250 200

x

−3

−2

−1

0

1

150 100

y

4

2

0

−2

−4

50 0

⤻ ⤻⤻ ⤻ −2 −2 −2 −2

The first differences are constant and the points appear to lie on a straight line. So, the data represent a linear function −2 whose rate of change is m = — = −2. 1 When x = 0, y = −2. So, the y-intercept is b = −2.

Game

b. This situation cannot be represented by a linear,

exponential, or quadratic function because the points do not appear to follow the shape of any of these types of functions. 30. a.

y = mx + b

Breathing Rate of Cyclist

27. Consecutive y-values have a constant ratio. They do not

change by a constant amount.

⤻⤻⤻⤻ +1 +1 +1 +1

x

1

2

3

4

5

y

3

9

27

81

243

⤻ ⤻⤻ ⤻ ×3 ×3 ×3 ×3

Consecutive y-values have a common ratio. So, the table represents an exponential function.

520

Algebra 1 Worked-Out Solutions

Breathing rate (liters per minute)

y = −2x − 2 So, the linear function is y = −2x − 2.

0 1 2 3 4 5 6 7 8 g

y 80 70 60 50 40 30 20 10 0

0 19 20 21 22 23 24 25 26 x

Speed (miles per hour)

The points appear to lie on an exponential curve. So, an exponential function best represents this situation. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 57.1 63.3 70.3 51.4 57.1 63.3 So, let b = 1.11

78 70.3

b. — ≈ — ≈ — ≈ — ≈ 1.11

Town A:

Because the first y-value in the table is 51.4, let a = 51.4. This initial value occurs when x = 20. Therefore, the graph of the data must be a horizontal translation 20 units right of the graph of y = abx = 51.4(1.11)x. So, the exponential function that models the data is y = 51.4(1.11)x−20. c. y = 51.4(1.11)x−20

= 51.4(1.11)18−20 = 51.4(1.11)−2 ≈ 51.4(0.8116) ≈ 41.7 So, the breathing rate is about 41.7 liters of air per minute. 31. a.

t

0

0.5

1

1.5

2

2.5

3

f(t)

3

23

35

39

35

23

3

b.

Height (feet)

Volleyball Motion y 40 35 30 25 20 15 10 5 0

f(t) = −16t2 + 48t + 3

f(b) − f(a) 6220.8 − 4320 average rate of change = — = —— b−a 4−2 1900.8 = — = 950.4 2 Town B: Use the graph to estimate points when x = 2 and x = 4. Use (2, 4900) and (4, 6250). f(b) − f(a) 6250 − 4900 average rate of change = — = —— b−a 4−2 1350 = — = 675 2 From 1990 to 2010, the population of Town A increased at an average rate of about 950 people per decade, and the population of Town B increased at an average rate of about 675 people per decade. So, the population of Town A is growing faster. b. In 2010, the populations of both towns are about equal

and the average rate of change of Town A, which shows exponential growth, exceeds the average rate of change of Town B, which appears to show exponential or quadratic growth. So, Town A will have a greater population after 2030. 33. a. For Organization A, the number of donations collected is

multiplied by a common ratio of 4. So, their number of donations can be represented by an exponential function. +1 +1 +1 +1 +1 +1 ⤻⤻⤻⤻⤻⤻

0

1

2

3

4 t

Time (seconds)

Time (hours), t

0

1

2

3

4

5

6

Number of donations, y

0

4

8

12

16

20

24

⤻⤻⤻⤻⤻⤻ +4 +4 +4 +4 +4 +4

c. The function is increasing between 0 and 1.5 seconds, and the

function is decreasing between 1.5 seconds and about 3 seconds. 20 35 − 23 12 23 − 3 d. — = — = 40 — = — = 24 1 − 0.5 0.5 0.5 − 0 0.5 4 35 − 39 −4 39 − 35 —=—=8 — = — = −8 0.5 0.5 1.5 − 1 2 − 1.5 3 − 23 −20 23 − 35 −12 — = — = −40 — = — = −24 3 − 2.5 2.5 − 2 0.5 0.5 Over the interval 0 to 1.5, as the function increases, the average rate of change decreases. Over the interval 1.5 to 3, as the function decreases, the average rate of change increases in the negative direction. 32. a. Write an exponential function to model the population of

Town A, whose initial population is a = 3000 and whose rate of change is r = 20%, or 0.2. f(x) = a(1 + r)x f(x) = 3000(1 + 0.2)x f(x) = 3000(1.2)x To find the populations in 1990 and 2010 of Town A, find f(2) and f(4), respectively. f(2) = 3000(1.2)2

f(4) = 3000(1.2)4

= 4320

= 6220.8

Copyright © Big Ideas Learning, LLC All rights reserved.

For Organization B, the first differences are constant. So, their number of donations can be represented by a linear function. For Organization C, the points appear to lie on a parabola. So, their number of donations can be represented by a quadratic function. b. Organization A: First, write an exponential function to

model the number of donations, whose initial number of donations is a = 1 and whose common ratio is b = 4. y = abt y = 1(4)t y = 4t Time (hours), t

0

1

2

Number of donations, y = 4t

1

4

16 64 256 1024 4096

4−1 3 1−0 1 64 − 16 48 — = — = 48 1 3−2 1024 − 256 768 — = — = 768 1 5−4 —=—=3

3

4

5

6

16 − 4 12 2−1 1 256 − 64 192 — = — = 192 4−3 1 4096 − 1024 3072 —— = — = 3072 6−5 1 — = — = 12

Algebra 1 Worked-Out Solutions

521

Chapter 8 Organization B: 8−4 4 4−0 4 —=—=4 —= —=4 1−0 1 2−1 1 4 16 − 12 4 12 − 8 = = 4 = — — — —=4 3−2 1 4−3 1 24 − 20 4 20 − 16 4 —=—=4 —=—=4 5−4 1 6−5 1 Organization C: 16 − 4 12 4−0 4 —=—=4 — = — = 12 1−0 1 2−1 1 36 − 16 20 — = — = 20 3−2 1 100 − 64 36 64 − 36 28 — = — = 28 — = — = 36 4−3 1 5−4 1 144 − 100 44 — = — = 44 6−5 1 c. The average rate of change will increase most quickly for Organization A. So, Organization A will have the most donations, followed by Organization C, then Organization B.

35. The average rate of change of a linear function is constant

because the dependent variable of a linear function increases by the same amount for each constant change in the independent variable. The average rate of change of a quadratic or exponential function is not constant because the dependent variable of a quadratic or exponential function changes by a different amount for each constant change in the independent variable. 36. a. D; A line is the graph of a linear function. b. C; The graph appears to represent an exponential decay

function, which has a base between 0 and 1. Choice C is a translation of this type of function. c. B; The graph appears to represent an exponential growth

function, which has a base greater than 1. Choice B is a translation of this type of function. d. A; A parabola is the graph of a quadratic function. 37.

34. a. Let x represent how many additional nights a vacation

lasts beyond the first 3. Write a function to model the cost of a vacation at each resort. For Blue Water Resort, the cost per night is increasing at a constant rate of $112 per additional night. So, it can be modeled by a linear equation using m = 112 and initial cost b = 1100. B(x) = mx + b B(x) = 112x + 1100 For Sea Breeze Resort, the cost can be represented by an exponential growth function with an initial cost of a = 1000 and a percent increase of r = 10%, or 0.1. S(x) = a(1 + r)x S(x) = 1000(1.1)x Use a graphing calculator to graph each function in the same viewing window. Use the intersect feature to find the value of x for which B(x) = S(x). 3000

S

2

3

4

5

y

3n − 1

10n + 2

26n

51n − 7

85n − 19

+(7n + 3) + (16n − 2) + (25n − 7) + (34n − 12) + (9n − 5) + (9n − 5) + (9n − 5) The second differences have a constant value of 9n − 5. So, the ordered pairs represent a quadratic function. 38. Sample answer: For simplicity, let the origin, (0, 0), be the

vertex. In order for the second differences to be 3 and for the origin to be the vertex, the first differences on either side of the origin must have a difference of 3, and the points must be an equal distance from the origin. So, the first differences 3 must be —32 = 1.5 to the right of the origin and −—2 = −1.5 to the left of the origin. Use the second difference of 3 to find more first differences and then more y-values. x

−2

−1

0

1

2

y

6

1.5

0

1.5

6

−1.5 +3

Intersection

0 X=6.007714 Y=1772.864 0

12

The graphs intersect when x ≈ 6. So, each resort costs about the same for a vacation that lasts 3 + 6 = 9 nights. b. Sea Breeze Resort will eventually be more expensive than

Blue Water Resort because the cost at Sea Breeze Resort shows exponential growth, and Blue Water Resort shows linear growth. c. The cost of the first three nights at Sea Breeze Resort

($1200) is greater than the cost of the first three nights at Blue Water Resort ($1100). Also, the cost of each additional night at Sea Breeze Resort (fourth night: $1200 0.1 = $120, fifth night: $1320 0.1 = $132, sixth night: $1452 0.1 = $145.20, and so on) is greater than the cost of each additional night at Blue Water Resort ($112). So, the total cost is always greater at Sea Breeze Resort.

522

1

−4.5

B

⋅

x

⋅

Algebra 1 Worked-Out Solutions

⋅

+3

+1.5

+4.5

+3

Then, use h = 0, k = 0, and one of the points, such as (2, 6), to solve for a and write an equation of the quadratic function. y = a(x − h)2 + k y = a(x − 0)2 + 0 y = ax2 6 = a(2)2 6 = 4a 6 4a —=— 4 4 1.5 = a So, a quadratic function that has constant second differences of 3 is y = 1.5x2.

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 39. The graph of a set of points is not always enough to

determine what kind of function the points represent. Sample answer: There may not be enough points to clearly determine the shape of the graph.

Figure number, n

1

2

3

4

Number of new branches, y

1

2

4

8

y = abx =

2

3

1

3

7

4 15

Notice that 1 is 1 less than = 2, 3 is 1 less than = 4, 7 is 1 less than 23 = 8, and 15 is 1 less than 24 = 16. In other words, each y-value is one less than the value of 2n. So, an exponential function that represents the total number y of branches on the nth figure is y = 2n − 1.

Because consecutive y-values have a common ratio of 2, this pattern can be represented by an exponential function with b = 2. The initial value of 1 occurs when n = 1, not 0. So, use n − 1 as the exponent.

2n−1

1 21

Pattern 1: A pattern is formed by the number of new branches on each figure.

=

Figure number, n Total number of branches, y

40. Sample answer:

1(2)n−1

Pattern 4: A pattern is formed by the total number of branches.

22

41. Your friend is incorrect because it is always the case that

the graph of an exponential function will eventually have a greater average rate of change and therefore have greater y-values than the graph of a quadratic function. 42. Sample answer: Enter the data into a graphing calculator

So, an exponential function that represents the number y of new branches on the nth figure is y = 2n−1. Pattern 2: A pattern is formed by the length of the new branches on each figure.

and use the linear regression, quadratic regression, and exponential regression features of the graphing calculator to find three different functions that could model the data. Then graph the points and the three functions in the same viewing window. The quadratic function appears to be the best fit for the data points. So, a quadratic function that models the data is y = −0.414x2 + 6.17x + 47.0. Use this function to make a prediction for Year 7.

Figure number, n Length (in units) of new branches, y

1

2

3

4

1

1 —2

1 —4

1 — 16

Let x = 7. y = −0.414x2 + 6.17x + 47.0 y = −0.414(7)2 + 6.17(7) + 47.0

1 —2 ,

Because consecutive y-values have a common ratio of this pattern can be represented by an exponential function with 1 b = —2. The initial value of 1 occurs when n = 1, not 0. So, use n − 1 as the exponent. y = abx

n−1 1 —12 1 n−1

() = (—) =

y = −20.286 + 43.19 + 47.0 y = 69.904 So, residents will spend about $69.9 billion in Year 7. Maintaining Mathematical Proficiency —

So, an exponential function that represents the length y of the n−1 new branches on the nth figure is y = —12 .

()

Pattern 3: A pattern is formed by the total length of all branches.

= 11 3—

1

2

3

4

Total length (in units) of all branches, y

1

2

3

4

Because the first differences are a common 1, this pattern can be represented by a linear function with m = 1. If there were no figure, then n = 0 and the total length is y = 0. So, the y-intercept is b = 0. y = mx + b = 1n + 0 =n So, a linear function that represents the total length y of all branches on the nth figure is y = n.

44. √ 125 = √ 5

⋅8 ⋅8

46. √ 243 = √ 3

3—

5—

⋅5 ⋅5

3—

=5

3—

45. √ 512 = √ 8

=8

Figure number, n

⋅ 11

—

43. √ 121 = √ 11

2

Copyright © Big Ideas Learning, LLC All rights reserved.

y = −0.414(49) + 6.17(7) − 6.3 + 47.0

⋅3 ⋅3 ⋅3 ⋅3

5 ——

=3

47. (x + 8)(x − 8) = x2 − 82

= x2 − 64 48. (4y + 2)(4y − 2) = (4y)2 − 22

= 16y2 − 4 49. (3a − 5b)(3a + 5b) = (3a)2 − (5b)2

= 9a2 − 25b2 50. (−2r + 6s)(−2r − 6s) = (−2r)2 − (6s)2

= 4r2 − 36s2

Algebra 1 Worked-Out Solutions

523

Chapter 8 8.4–8.6 What Did You Learn? (p. 469)

3.

1. Sample answer: Graph the function on a graphing calculator

and compare it to your graph.

x

−4

−2

0

2

4

g(x)

−12

−3

0

−3

−12

2. Sample answer: The equation can be factored into intercept

y −4 −3 −2

form, and the width can be determined from the intercepts.

2 3 4 x 3

g(x) = −4 x2

−2 −3 −4 −5 −6 −7 −8

3. Sample answer: The area of a circular rug, πr2, is a quadratic

function, so it makes sense that the cost, based on the amount of material needed to make the rug, would also be a quadratic function. Chapter 8 Review (pp. 470–472) 1.

x

–2

–1

0

1

2

p(x)

28

7

0

7

28

32 28 24 20 16 12

y

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of g opens down and is wider than the graph of f(x) = x2. So, the graph of g is a vertical shrink by a factor of —34 and a reflection in the x-axis of the graph of f. 4.

x

−2

−1

0

1

2

h(x)

−24

−6

0

−6

−24

y

p(x) = 7x2

8 4 −4 −3 −2−1

−4 −3 −2−1

x q(x)

8 7 6 5 4 3 2 1 −4 −3 −2−1

h(x) = −6x2

−9

1 2 3 4 x

−15 −18 −21 −24

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of p is narrower than the graph of f(x) = x2. So, the graph of p is a vertical stretch by a factor of 7 of the graph of f. 2.

1 2 3 4 x

−4

−2

0

2

4

8

2

0

2

8

y

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of h opens down and is narrower than the graph of f(x) = x2. So, the graph of h is a vertical stretch by a factor of 6 and a reflection in the x-axis of the graph of f. 5. The vertex is (1, −3). The axis of symmetry is x = 1. The

domain is all real numbers. The range is y ≥ −3. When x < 1, y increases as x decreases. When x > 1, y increases as x increases.

6. q(x) =

1 2 2x

x

−2

−1

0

1

2

9

6

5

6

9

g(x)

1 2 3 4 x

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of q is wider than the graph of f(x) = x2. So, the graph of q is a vertical shrink by a factor of —12 of the graph of f.

16 14 12 10 8 6 4 2 −4 −3 −2−1

y

g(x) = x2 + 5 1 2 3 4 x

Both graphs open up and have the same axis of symmetry, x = 0. The vertex, (0, 5), of the graph of g is above the vertex, (0, 0), of the graph of f(x) = x2. So, the graph of g is a vertical translation 5 units up of the graph of f.

524

Algebra 1 Worked-Out Solutions

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Chapter 8 7.

x

−2

−1

0

1

2

h(x)

−8

−5

−4

−5

−8

y = x2 − 2x + 7 = (1)2 − 2(1) + 7

=6 h(x) = −x2 − 4

So, the vertex is (1, 6). The y-intercept is 7. So, plot (0, 7) and the reflected point (2, 7).

−6 −7 −8

16 14 12

The graph of h opens down, but the graph of f(x) = x2 opens up. Both graphs have the same axis of symmetry, x = 0. The vertex, (0, −4), of the graph of h is below the vertex, (0, 0), of the graph of f. So, the graph of h is a reflection in the x-axis, followed by a vertical translation 4 units down of the graph of f. x

−2

−1

0

1

2

m(x)

−2

4

6

4

−2

6

−4 −3 −2−1

The domain is all real numbers. The range is y ≥ 6.

()

2 1 1

3 4 x

The graph of m opens down and is narrower than the graph of f(x) = x2, which opens up. Both graphs have the same axis of symmetry, x = 0. The vertex, (0, 6), of the graph of m is above the vertex, (0, 0), of the graph of f. So, the graph of m is a vertical stretch by a factor of 2, a reflection in the x-axis, and a vertical translation 6 units up of the graph of f. x

−4.5

−3

−1.5

0

1.5

3

4.5

n(x)

1.75

−2

−4.25

−5

−4.25

−2

1.75

−2

y

2

6 8 x

−2 −3 1

−6

b 2a

3 2(−3)

−3 −6

1 2

() () 3 1 = −3( ) + − 4 2 4 2

—

—

−1 −2

−8 −6

1 2 3 4 x

1 1 1 f — = −3 — + 3 — − 4 2 2 2

4 3

2 1

y = x2 − 2x + 7

4 2

f(x) = −3x2 + 3x − 4

m(x) = −2x2 + 6

9.

y

11. The axis of symmetry is x = −— = −— = — = —.

y

−4 −3

2 2

= −1 + 7

1 2 3 4 x

−2 −3

8.

(−2) 2(1)

=1−2+7

y −4 −3 −2−1

b 2a

10. The axis of symmetry is x = −— = −— = — = 1.

3 3 = −— + — − 4 4 2 3 =—−4 4 13 1 = −—, or −3— 4 4 1 13 So, the vertex is —, −— . 4 2 The y-intercept is −4. So, plot (0, −4) and the reflected point (1, −4).

(

)

y −2

−1

1 −2 −3 −4

2 x

f(x) = −3x2 + 3x − 4

−7 −8

13 The domain is all real numbers. The range is y ≤ −—. 4

n(x) = 3 x2 − 5

Both graphs open up and have the same axis of symmetry, x = 0. The graph of n is wider, and its vertex, (0, −5), is below the vertex, (0, 0), of the graph of f(x) = x2. So, the graph of n is a vertical shrink by a factor of —13 and a vertical translation 5 units down of the graph of f. Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

525

Chapter 8 (−6)

b

6

12. The axis of symmetry is x = −— = −— = — = 6. 2a 1 2 —12 1 y = —x2 − 6x + 10

()

2 1 y = —(6)2 − 6(6) + 10 2 1 = —(36) − 36 + 10 2 = 18 − 36 + 10

16.

h(x) = 3x2 − 2x h(−x) = 3(−x)2 − 2(−x) = 3x2 + 2x Because h(x) = 3x2 − 2x and −h(x) = −3x2 + 2x, you can conclude that h(−x) ≠ h(x) and h(−x) ≠ −h(x). So, the function is neither even nor odd.

17. h(x) = 2(x − 4)2

= −18 + 10

Because h = 4 and k = 0, graph the axis of symmetry x = 4 and plot the vertex (4, 0).

= −8 So, the vertex is (6, −8). The y-intercept is 10. So, plot (0, 10) and the reflected point (12, 10).

When x = 2: h(2) = 2(2 − 4)2 = 2(−2)2

y

= 2(4)

9

=8

6 3

When x = 3:

−2

4 6 8

h(3) = 2(3 − 4)2

12 14 x

= 2(−1)2

−6 −9

= 2(1)

1

y = 2 x2 − 6x + 10

=2

The domain is all real numbers. The range is y ≥ −8. b 2a

88 2(−16)

−88 −32

11 4

13. t = −— = −— = — = —, or 2.75

f(t) = −16t2 + 88t + 12

( )

—

8 7 6 5 4 3

= −121 + 242 + 12

2 1

( ) + 88( 114 ) + 12 121 = −16( + 242 + 12 16 )

11 11 f — = −16 — 4 4

2

—

= 121 + 12

(

)

11 The vertex is — , 133 . So, the pumpkin reaches a maximum 4 height of 133 feet 2.75 seconds after it is launched.

w(x) = 5x w(−x) = 5−x 1 = —x 5 Because w(x) = 5x and − w(x) = −5x, you can conclude that w(−x) ≠ w(x) and w(−x) ≠ −w(x). So, the function is neither even nor odd.

15.

r(x) = −8x r(−x) = −8(−x)

h(x) = 2(x − 4)2

Both graphs open up. The graph of h is narrower than the graph of f(x) = x2. The axis of symmetry, x = 4, and the vertex, (4, 0), of the graph of h are 4 units right of the axis of symmetry, x = 0, and the vertex, (0, 0), of the graph of f. So, the graph of h is a vertical stretch by a factor of 2 and a horizontal translation 4 units right of the graph of f. 1

18. g(x) = —2 (x − 1)2 + 1

Because h = 1 and k = 1, graph the axis of symmetry x = 1 and the vertex (1, 1). When x = −3: 1

g(−3) = —2 (−3 − 1)2 + 1 1

= —2(−4)2 + 1 1

= 8x

= —2(16) + 1

= −r(x)

=8+1

Because r(−x) = −r(x), the function is odd.

526

y

1 2 3 4 5 6 7 8 x

= 133

14.

So, plot (2, 8) and (3, 2) and the reflected points (6, 8) and (5, 2).

Algebra 1 Worked-Out Solutions

=9

Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 20. g(x) = −3(x + 2)2 − 4

When x = −1:

Because h = −2 and k = −4, the axis of symmetry is x = −2 and the vertex is (−2, −4).

1

g(−1) = —2 (−1 − 1)2 + 1 = =

1 —2 (−2)2 + 1 —2 (4) + 1

1

When x = −4: g(−4) = −3(−4 + 2)2 − 4

=2+1

= −3(−2)2 − 4

=3

= −3(4) − 4

So, plot (−3, 9) and (−1, 3), and the reflected points (5, 9) and (3, 3). 8 7 6 5

y

= −12 − 4 = −16 When x = −3: g(−3) = −3(−3 + 2)2 − 4 = −3(−1)2 − 4

4

= −3(1) − 4

2 1 −8 −6 −4−2

= −3 − 4

1

g(x) = 2(x − 1)2 + 1

= −7

2 4 6 8 x

Both graphs open up. The graph of g is wider than the graph of f(x) = x2. The axis of symmetry, x = 1, of the graph of g is 1 unit right of the axis of symmetry, x = 0, of the graph of f. Also, the vertex, (1, 1), of the graph of g is 1 unit right and 1 unit up of the vertex, (0, 0), of the graph of f. So, the graph of g is a vertical shrink by a factor of —12 and a translation 1 unit right and 1 unit up of the graph of f.

So, the points (−4, −16) and (−3, −7), and the reflected points (0, −16) and (−1, −7) lie on the graph of g. The function h(x) = g(x − 1) is of the form y = f(x − h), where h = 1. So, the graph of h is a horizontal translation 1 unit right of the graph of g. So, add 1 to the x-coordinates of the points that lie on the graph of g. So, the points (−3, −16), (−2, −7), (−1, −4), (0, −7), and (1, −16) lie on the graph of h. y

19. q(x) = −(x + 4)2 + 7

−4 −3 −2−1

Because h = −4 and k = 7, graph the axis of symmetry x = −4 and plot the vertex (−4, 7).

1 2 3 4 x

h(x) = g(x − 1) −8 −10 −12 −14 −16

When x = −8: q(−8) = −(−8 + 4)2 + 7 = −(−4)2 + 7 = −16 + 7

21. f(x) = a(x − h)2 + k

= −9

f(x) = a(x − 3)2 + 2

When x = −6:

7 = a(4 − 3)2 + 2

q(− 6) = −(−6 + 4)2 + 7

7 = a(1)2 + 2

= −(−2)2 + 7

7 = a(1) + 2

= −4 + 7

7=a+2

=3 So, plot (−8, −9) and (−6, 3), and the reflected points (0, −9) and (−2, 3).

−2 9

The graph of q opens down, 6 3 while the graph of f(x) = x2 opens up. The axis of symmetry, −8 −6 −5 −4 −3 −2 x = −4, of the graph of q is 4 units left of the axis of −9 symmetry, x = 0, of the graph −12 −15 of f. Also, the vertex, (−4, 7), q(x) = −(x + 4)2 + 7 of the graph of q is 4 units left and 7 units up of the vertex, (0, 0), of the graph of f. So, the graph of q is a reflection in the x-axis and a translation 4 units left and 7 units up of the graph of f. Copyright © Big Ideas Learning, LLC All rights reserved.

y

−2

5=a f(x) = 5(x − 3)2 + 2

x

= 5[x2 − 2(x)(3) + 32] + 2 = 5(x2 − 6x + 9) + 2 = 5(x2) − 5(6x) + 5(9) + 2 = 5x2 − 30x + 45 + 2 = 5x2 − 30x + 47 So, a quadratic function is f(x) = 5x2 − 30x + 47.

Algebra 1 Worked-Out Solutions

527

Chapter 8 22. y = (x − 4)(x + 2)

24. y = x2 − 8x + 15

Because the x-intercepts are p = 4 and q = −2, plot (4, 0) and (−2, 0). The axis of symmetry is p+q 4−2 2 x = — = — = — = 1. 2 2 2 y = (1 − 4)(1 + 2) = (−3)(3) = −9 So, the vertex is (1, −9).

(−2, 0) −3

(4, 0) 2 3

5 x

−4 −8 −10

= (x − 3)(x − 5) The zeros are 3 and 5. So, plot (3, 0) and (5, 0). p+q 3+5 8 The axis of symmetry is x = — = — = — = 4. 2 2 2 y = (4 − 3)(4 − 5) = −1

x=1

−1

y = x2 − 8x + 15

= (1)(−1)

y 4 2

Because a = 1 and 1 > 0, the parabola opens up. Because c = 15, the y-intercept is 15. So, plot (0, 15).

y = (x − 4)(x + 2) (1, −9)

So, the vertex is (4, −1). 6 5

Because the x-intercepts are p = −3 and q = −1, plot (−3, 0) and (−1, 0). The axis of symmetry is p + q −3 − 1 −4 x = — = — = — = −2. 2 2 2 f(−2) = −3(−2 + 3)(−2 + 1) = −3(1)(−1)

x=4

4 3 2 1

The domain is all real numbers. The range is y ≥ −9. 23. f(x) = −3(x + 3)(x + 1)

y

y = x2 − 8x + 15 (3, 0)

(5, 0)

1 2 3

5 6 7 8 x

(4, −1)

−2

The domain is all real numbers. The range is y ≥ −1. 25. y = −2x2 + 6x + 8

Because a = −2 and −2 < 0, the parabola opens down. Because c = 8, the y-intercept is (0, 8). y = −2x2 + 6x + 8 = −2(x2 − 3x − 4)

=3 So, the vertex is (−2, 3). (−2, 3)

4 3 2 1

(−3, 0) −6 −5 −4

= −2(x + 1)(x − 4) The zeros of the function are p = −1 and q = 4. So, plot (−1, 0) and (4, 0).

y

14 12 10

(−1, 0) 1 2 x

y = −2x2 + 6x + 8

8

x = −2

f(x) = −3(x + 3)(x + 1)

y

−2

1 2 3

5 6 x

The domain is all real numbers. The range is y ≤ 3.

528

Algebra 1 Worked-Out Solutions

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Chapter 8 30. a. Let t represent how long (in years) the money has been in

26. f(x) = x2 + x − 2

Because a = 1 and 1 > 0, the parabola opens up. Because c = −2, the y-intercept is (0, −2). f(x) = x2 + x − 2 5 4 3

= (x + 2)(x − 1) The zeros of the function are p = −2 and q = 1. So, plot (−2, 0) and (1, 0).

y

Your account: Calculate the values of y for x = 2 and x = 7. y = 200(1.1)t = 200(1.1)2

2 1 −4 −3

the account.

f(x) = x2 + x − 2

−1

2 3 4 x

= 200(1.21) = 242 y = 200(1.1)t = 200(1.1)7

−3

≈ 200(1.9487) 27. f(x) =

2x3

− 18x

≈ 389.74

= 2x(x2 − 9) = 2x(x2 − 32) = 2x(x + 3)(x − 3) The zeros are −3, 0, and 3. So, plot (−3, 0), (0, 0), and (3, 0). x

−2

2

f(x)

20

−20

Write a function to model the balance of your friend’s account. The balance is increasing at a constant rate of 20 dollars per year. So, it can be modeled by a linear equation using m = 20 and initial balance b = 250.

So, plot (−2, 20) and (2, −20) also. 24 18

Use (2, 242) and (7, 389.74). f(b) − f(a) 389.74 − 242 average rate of change = — = —— b−a 7−2 147.74 = — ≈ 29.55 5 Friend’s account:

g(x) = mx + b

y

g(x) = 20x + 250 Calculate g(2) and g(7).

−4

−2−1

g(x) = 20x + 250

4 x

1 2

g(2) = 20(2) + 250

−12 −18 −24

= 40 + 250 f(x) =

2x3

= 290

− 18x

g(x) = 20x + 250

28. Sample answer:

g(7) = 20(7) + 250

f(x) = (x − p)(x − q)

= 140 + 250

= (x − 4)(x − 6)

= 390

= x(x) + x(−6) + (−4)(x) + (−4)(−6)

Use (2, 290) and (7, 390).

= x2 − 6x − 4x + 24 = x2 − 10x + 24 So, a quadratic function is f(x) = x2 − 10x + 24. +1 ⤻ +1 ⤻ +1 ⤻ +1 ⤻

29.

x

−1

0

1

2

3

y

512

128

32

8

2

⤻ ⤻ ⤻ ⤻ 1

×—4

1

×—4

1

×—4

1

×—4

Consecutive y-values have a common ratio of —14 . So, the table 1 represents an exponential function with b = —4. When x = 0, y = 128. So, a = 128.

b. After 7 years, the balances are about equal and the average rate of change of your balance, which shows exponential growth, exceeds the average rate of change of your friend's balance, which shows linear growth. So, your account will have a greater balance after 10 years.

x

()

y = abx = 128 —14

g(b) − g(a) 390 − 290 average rate of change = — = — b−a 7−2 100 = — = 20 5 From year 2 to year 7, your account increased at an average rate of about $29.55 per year and your friend’s account increased at an average rate of $20 per year. So, your account is growing faster.

()

x

So, the exponential function is y = 128 —14 . Copyright © Big Ideas Learning, LLC All rights reserved.

Algebra 1 Worked-Out Solutions

529

Chapter 8 Chapter 8 Test (p. 473) 1.

x

When x = −5: 1

−2

−1

0

1

2

5

−1

−3

−1

5

h(x)

5

p(−5) = —2(−5 + 1)2 − 1 1

= —2(−4)2 − 1 1

= —2(16) − 1 =8−1

y

=7

4 3

So, the point (−5, 7) and its reflection (3, 7) lie on the graph of p.

2 1 −4 −3 −2

2 3 4 x

7 6 5 4 3

h(x) = 2x2 − 3

Both graphs open up and have the same axis of symmetry, x = 0. The graph of h is narrower than the graph of f(x) = x2, and the vertex, (0, −3), of the graph of h is below the vertex, (0, 0), of the graph of f. So, the graph of h is a vertical stretch by a factor of 2 and a vertical translation 3 units down of the graph of f. 2.

x

−4

−2

0

2

4

g(x)

−8

−2

0

−2

−8

y −4 −3 −2

2 3 4 x −2 −3 −4 −5 −6 −7 −8

2 1 −8 −6 −4

1

p(x) = 2(x + 1)2 − 1 2 4 6 8 x

Both graphs open up. The graph of p is wider than the graph of f(x) = x2. The axis of symmetry, x = −1, of the graph of p is 1 unit left of the axis of symmetry, x = 0, of the graph of f. Also, the vertex, (−1, −1), of the graph of p is 1 unit left and 1 unit down from the vertex, (0, 0), of the graph of f. So, the graph of p is a vertical shrink by a factor of —12, and a translation 1 unit left and 1 unit down of the graph of f. 4. a. The domain is all real numbers. The range is y ≤ 8. The

zeros of the function are p = 3 and q = 7.

1

g(x) = − 2 x2

b. f(x) = a(x − p)(x − q)

f(x) = a(x − 3)(x − 7) 8 = a(5 − 3)(5 − 7) 8 = a(2)(−2)

The graphs have the same vertex, (0, 0), and the same axis of symmetry, x = 0, but the graph of g opens down and is wider than the graph of f(x) = x2. So, the graph of g is a vertical shrink by a factor of —12 and a reflection in the x-axis of the graph of f. 3. Because h = −1 and k = −1, the vertex of the graph is

(−1, −1) and the axis of symmetry is x = −1.

8 = −4a 8 = −4a −4 = −4 −2 = a f(x) = −2(x − 3)(x − 7) = −2[x(x) + x(−7) + (−3)(x) + (−3)(−7)]

1

= −2(x2 − 7x − 3x + 21)

1

= −2(x2 − 10x + 21)

p(x) = —2(x + 1)2 − 1 = —2[x2 + 2(x)(1) + 12] − 1

= −2(x2) − 2(−10x) −2(21)

1

= —2(x2 + 2x + 1) − 1 1 1 1 = —2(x2) + —2(2x) + —2(1) − 1 1 2 1 = —2x + x + —2 − 1 1 1 = —2x2 + x − —2 1 Because c = −—2, the graph passes

y-axis, and the reflection of this

530

y

= −2x2 + 20x − 42 So, the function is f(x) = −2x2 + 20x − 42. c. The graph of f opens down, while the graph of g(x) = x2

(

) on the

1 through 0, −—2 1 point is −2, −—2 .

Algebra 1 Worked-Out Solutions

(

)

opens up, and the graph of f is narrower than the graph of g. The axis of symmetry, x = 5, of the graph of f is 5 units right of the axis of symmetry, x = 0, of the graph of g. Also, the vertex, (5, 8), of the graph of f is 5 units right and 8 units up from the vertex, (0, 0), of the graph of g. So, the graph of f is a vertical stretch by a factor of 2, a reflection in the x-axis, and a translation 5 units right and 8 units up of the graph of g. Copyright © Big Ideas Learning, LLC All rights reserved.

Chapter 8 d. The function h(x) = f(x − 6) is of the form y = f(x − h), where

h = 6. So, the graph of h is a horizontal translation 6 units right of the graph of f. To graph h, add 6 to the x-coordinates of the points on the graph of f. So, plot (9, 0), (11, 8), and (13, 0), and draw a smooth curve through the points. 8 6

7. h(x) = 16x2 − 4

= 4(4x2 − 1) = 4[(2x)2 − 12] = 4(2x + 1)(2x − 1) 2x + 1 = 0 or 2x − 1 = 0 +1 +1 −1 −1 2x = −1 2x = 1 −1 2x 1 2x — —=— —= 2 2 2 2 1 1 x = −— x=— 2 2 1 1 1 1 The zeros are −— and —. So, plot −—, 0 and —, 0 . 2 2 2 2 1 1 p + q −—2 + —2 0 = — = 0. The axis of symmetry is x = — = — 2 2 2 2 h(0) = 16(0) − 4 y

h(x) = f(x − 6)

y

4 2 2 4 6 8 10 12 14 16 x

−2 −4 −6 −8

( ) ( )

5. f(x) = 2x2 − 8x + 8

Because a = 2 and 2 > 0, the graph opens up. Because c = 8, the y-intercept is 8. So, plot (0, 8). f (x) = 2x2 − 8x + 8

4 3

=0−4

2 1

= −4

= 2(x2 − 4x + 4) = 2(x2 − 2(x)(2) + 22)

So, the vertex is (0, −4).

= 2(x − 2)2

The domain is all real numbers. The range is y ≥ −4.

So, the function has one zero, 2, which means (2, 0) is the vertex. So, plot (2, 0). Then, reflect (0, 8) in the axis of symmetry x = 2, and plot (4, 8). 8 7 6 5 4 3

= 16(0) − 4

2 1

1 2 3 4 5 6 7 8 x

6. y = −(x + 5)(x − 1)

Because p = −5 and q = 1, plot (−5, 0) and (1, 0). p + q −5 + 1 −4 The axis of symmetry is x = — = — = — = −2. 2 2 2 y = −(−2 + 5)(−2 − 1) = −(3)(−3) =9 So, the vertex is (−2, 9). y = −(x + 5)(x − 1)

10 8

y

4 2 −6

−4 −3 −2−1

2 x

−4 −6

The domain is all real numbers. The range is y ≤ 9. Copyright © Big Ideas Learning, LLC All rights reserved.

1

2 x

h(x) = 16x2 − 4

−4

x

−1

0

1

2

3

y

4

8

16

32

64

⤻ ⤻ ⤻ ⤻ ×2 ×2 ×2 ×2

Consecutive y-values have a common ratio of 2. So, the table represents an exponential function with b = 2. When x = 0, y = 8. So, a = 8. An exponential function is y = 8(2)x.

f(x) = 2x2 − 8x + 8

The domain is all real numbers. The range is y ≥ 0.

−1

⤻ +1 ⤻ +1 ⤻ +1 ⤻ +1

8.

y

−2

+1 ⤻ +1 ⤻ +1 ⤻ +1 ⤻

9.

x

−2

−1

0

1

2

y

−8

−2

0

−2

−8

⤻ ⤻ ⤻ ⤻ +6 +2 −2 −6 ⤻ ⤻ ⤻ −4 −4 −4

The second differences are constant. So, the table represents a quadratic function. The vertex is (0, 0). y = a(x − h)2 + k y = a(x − 0)2 + 0 y = ax2 −8 = a(2)2 −8 = a(4) −8 = 4a −8 4a —= — 4 4 −2 = a So, a quadratic function is y = −2x2.

Algebra 1 Worked-Out Solutions

531

Chapter 8 10. The given points indicate that the x-intercepts are −8 and

12. Sample answer: To be even, the vertex needs to be on the

y-axis, so it must be in the form y = ax2 + c. Because the range is y ≥ 3, a must be positive and c = 3. Any positive value of a is acceptable. So, a quadratic function is f(x) = x2 + 3.

−2. So, use the intercept form to write a function.

f(x) = a(x − p)(x − q) f(x) = a(x + 8)(x + 2) Use the other point, (−6, 4), to find the value of a. f(x) = a(x + 8)(x + 2)

13. Sample answer: Use the x-intercept as the vertex. So, use

h = 4 and k = 0 to write an equation in intercept form.

4 = a(−6 + 8)(−6 + 2)

f(x) = a(x − h)2 + k

4 = a(2)(−4)

f(x) = a(x − 4)2 + 0

4 = −8a −8a 4 —=— −8 −8 1 −— = a 2 Use the value of a to write a function.

f(x) = a(x − 4)2 Substitute the second point into the equation to find a. f(x) = a(x − 4)2 9 = a(1 − 4)2 9 = a(−3)2

f(x) = a(x + 8)(x + 2) 1 f(x) = −—(x + 8)(x + 2) 2 Simplify the equation to write it in standard form. 1 f(x) = −—[x(x) + x(2) + 8(x) + 8(2)] 2 1 = −—(x2 + 2x + 8x + 16) 2 1 = −—(x2 + 10x + 16) 2 1 1 1 = −—(x2) − — (10x) − —(16) 2 2 2 1 = −—x2 − 5x − 8 2 1 So, a quadratic function is f(x) = −—x2 − 5x − 8. 2

9 = a(9) 9 = 9a 9 9a —=— 9 9 1=a Write the equation in intercept form with the value of a. f(x) = a(x − 4)2 f(x) = 1(x − 4)2 f(x) = (x − 4)2 Simplify the equation to write it in standard form. f(x) = (x − 4)2 = x2 − 2(x)(4) + 42 = x2 − 8x + 16

11. The given points indicate that the x-intercepts are 0 and 10.

So, a quadratic function is f(x) = x2 − 8x + 16.

So, use the intercept form to write a function. f(x) = a(x − p)(x − q) f(x) = a(x + 0)(x − 10)

+1 ⤻ +1 ⤻ +1 ⤻ +1 ⤻

14.

f(x) = ax(x − 10)

Time, t

1

Use the other point, (9, −27), to find the value of a.

Distance, d

19

f(x) = ax(x − 10) −27 = a(9)(9 − 10) −27 = a(9)(−1) −27 = −9a −9a −27 — —= −9 −9 3=a Use the value of a to write a function. f(x) = ax(x − 10) f(x) = 3x(x − 10) Simplify the equation to write it in standard form.

2

3

4

5

38

57

76

95

⤻ ⤻ ⤻ ⤻ +19 +19 +19 +19

The first differences are a constant 19. So, the data can be modeled by a linear function with m = 19. y = mx + b y = 19x + b 19 = 19(1) + b 19 = 19 + b − 19 − 19 0=b So, a linear function that models the data is y = 19x.

f(x) = 3x(x) − 3x(10) = 3x2 − 30x So, a quadratic function is f(x) = 3x2 − 30x.

532

Algebra 1 Worked-Out Solutions

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Chapter 8 0.17 −0.17 b 2a 2(−0.005) −0.01 y = −0.005x2 + 0.17x + 3

2.

15. a. x = −—= −— = — = 17

= −0.005(17)2 + 0.17(17) + 3 = −0.005(289) + 2.89 + 3 = 1.445 + 3

3

4

5

6

7

1

2

3

5

8

13 21 34 55 89

8

9

10 11

b x = −— 2a v0 7 — = −— 2(−16) 8 7 v0 —=— 8 32 v 7 32 — = 32 ___0 32 8 28 = v0

The maximum height is 4.445 feet. b. y = −0.005x2 + 0.17x + 3

= −0.005(30)2 + 0.17(30) + 3 = −0.005(900) + 5.1 + 3 = −4.5 + 5.1 + 3

⋅

= 0.6 + 3 = 3.6 After traveling a horizontal distance of 30 feet, the height is 3.6 feet. So, the ball will clear the net by 3.6 − 3 = 0.6 foot. 16. a. Sample answer: For a function to be even, its vertex must

be on the y-axis. So, the value of b must be 0. Any value of a and c would work. So, the function f(x) = ax2 + bx + c is even if a = 1, b = 0, and c = 0. b. Sample answer: For a function to be odd, it must be linear

and pass through the origin. So, both a and c must be zero. Any value of b would work. So, the function f(x) = ax2 + bx + c is even if a = 0, b = 1, and c = 0. c. Sample answer: A quadratic function whose vertex is not

on the y-axis will be neither even nor odd. So, the function f(x) = ax2 + bx + c is neither even nor odd if a = 1, b = 2, and c = 3. 17. f(x) = x2 + 4

x

0

1

2

3

f(x)

4

5

8

13

1 1

2

1

3. f(t) = −16t2 + v0t + s0

= 4.445

5−4 1−0

1

Term, f(n)

So, the range values between 0 and 100 are 1, 2, 3, 5, 8, 13, 21, 34, 55, and 89.

= −1.445 + 2.89 + 3

—=—=1

Position, n

8−5 2−1

3 1

—=—=3

13 − 8 3−2

5 1

—=—=5

The average rates of change are increasing by a constant amount of 2. Chapter 8 Standards Assessment (pp. 474–475) 1. C; The parabola opens down and is wider than the graph

⋅

The initial velocity is 28 feet per second. 1

4. In the systems y = 6x + 9 and y = −—6 x + 9 and 2x + 4y = −2

and 10x + 4y = −2, the coefficients of the y-terms are the same, but the coefficients of the x-terms are not the same. So, only one variable will be eliminated if the equations are subtracted. These systems each have one solution.

In the system 3x + y = 5 and −15 + 3y + 9x = 0, the equation −15 + 3y + 9x = 0 can be rewritten as 9x + 3y = 15. Then, you can multiply the other equation 3x + y = 5 by 3, and the result is also 9x + 3y = 15. So, the equations in the system are equivalent, and the system has infinitely many solutions. 3

In the system y − 2x = —2 and −3 + 2y = 4x, the equation −3 + 2y = 4x can be rewritten as 2y − 4x = 3. Then, you 3 can multiply the other equation y − 2x = —2 by 2, and the result is also 2y − 4x = 3. So, the equations in the system are equivalent, and the system has infinitely many solutions. In the system 7x + 4y = 12 and 8y − 12 = −14x, the equation 8y − 12 = −14x can be rewritten as 14x + 8y = 12. If you multiply the other equation 7x + 4y = 12 by 2, you get 14x + 8y = 24, not 14x + 8y = 12. So, the equations are not equivalent. However, both variables would be eliminated if the equations were subtracted, resulting in the false equation 0 = −12. So, the system has no solution. In the system y = −3x + 5 and y = −3x + 9, both variables would be eliminated if the equations were subtracted, resulting in the false equation 0 = −4. So, the system has no solution.

of y = x2. So, the equation that matches is the one with a negative value for a, whose absolute value is between 0 and 1.

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Algebra 1 Worked-Out Solutions

533

Chapter 8 ⤻ +2 ⤻ +2 ⤻ +2

5. Your friend is correct, because a parabola can intersect the

x-axis in one, two, or no points as shown below. Sample answer: two real zeros 6 5

one real zero y

4 3

8 7 6 5

2 1

4 3

−4 −3 −2

−4 −3 −2−1

1 2 3 4 x

y

1 2 3 4 x

6. C; The side of each shaded region can be represented by the

binomial x − a. So, the area is (x − a)2 . Using the square of a binomial pattern, this product in simplest form is (x − a)2 = x2 − 2ax + a2.

⤻ +1 ⤻ +1 ⤻ +1

7. a.

x p(x)

0 −4

1

2

3

−16

−28

−40

⤻ −12 ⤻ −12 ⤻ −12

The first differences are constant. So, the table represents a linear function.

⤻ +1 ⤻ +1 ⤻ +1

x

1

2

3

4

r(x)

0

15

40

75

⤻ ⤻ ⤻ +15 +25 +35 ⤻ ⤻ +10 +10

The second differences are constant. So, the table represents a quadratic function.

⤻ ⤻ ⤻ +1 +1 +1

x

1

2

3

4

s(x)

72

36

18

9

⤻ ⤻ ⤻ — — — 1

×2

1

×2

3

5

7

−5

−21

−45

⤻ ⤻ ⤻ −8 −16 −24 ⤻ ⤻ −8 −8

b. p(x):

1 −4 −3 −2−1

t(x)

3

The second differences are constant. So, the table represents a quadratic function.

no real zeros 8 7 6 5 4 3

1

y

2 1

2 3 4 x

x

f(b) − f(a) average rate of change = — b−a −40 − (−16) = —— 3−1 −24 = — = −12 2 r(x): f(b) − f(a) average rate of change = — b−a 40 − 0 20 = — = — = 10 3−1 2 s(x): f(b) − f(a) average rate of change = — b−a 18 − 72 −54 = — = — = −27 3−1 2 t(x): f(b) − f(a) average rate of change = — b−a −5 − 3 −8 = — = — = −4 3−1 2

So, from least to greatest according to the average rates of change between x = 1 and x = 3, the functions are s(x), p(x), t(x), and r(x). 8. a. The vertex form of a quadratic function is y = a(x − h)2 + k.

So, when the vertex of a function is (−3, 4), the equation is f(x) = 5(x + 3)2 + 4. b. The intercept form of a quadratic function is

y = a(x − p )(x − q). So, when the x-intercepts are −8 and 2, the equation is g(x) = −(x − 2)(x + 8). c. The standard form of a quadratic function is y = ax2 + bx + c.

If a is positive, the parabola opens up, and c is the value of the y-intercept. So, when the range is y ≥ −6, the equation is h(x) = 3x2 − 6. d. The binomials have to be (x + 1) and (x − 1) in either

order so that their product is x2 − 1. Then, the equation will be of the form y = ax2 + c in standard form, and the vertex will be on the y-axis. Because the range is y ≤ −4, the parabola must open down. So, the sign of a is negative. The equation could be j(x) = −4(x + 1)(x − 1) or j(x) = −4(x − 1)(x + 1).

1

×2

Consecutive y-values have a common ratio. So, the table represents an exponential function.

9. a. The graph represents a linear function. Sample answer:

The points appear to lie on a straight line, and the first difference is constant. b. The domain is the set of whole numbers from 1 to 6. It is

discrete, because it is not a range of values, but rather a set of individual values.

534

Algebra 1 Worked-Out Solutions

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Chapter 8 c. Use (2, 90) and (4, 180) to calculate the rate of change of

the function. y2 − y1 180 − 90 90 = — = 45 m = _______ x2 − x1 = — 4−2 2 Use point-slope form of an equation. y − y1 = m(x − x1) y − 90 = 45(x − 2) y − 90 = 45(x) − 45(2) y − 90 = 45x − 90 + 90

+ 90

y = 45x So, a linear function that models the data is y = 45x. d. The referee makes $45 per hour, and he gets paid for

whole numbers of hours only. If you divide $500 by the 1 500 100 hourly wage, you get — = — = 11—, which is not a 9 45 9 whole number. So, the referee cannot make exactly $500. 10. (b−5)−4 = b(−5)(−4) = b20

(b−4)−5 = b(−4)(−5) = b20 (b10)2 = b10⋅2 = b20

b12b8 = b12+8 = b20 So, the expressions that are equivalent to (b−5)−4 are (b−4)−5, (b10)2, b20, and b12b8.

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Algebra 1 Worked-Out Solutions

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