Appendix B: Practice Problems Answer Key - … · Appendix B Practice Problem Answer Key Appendix...

Appendix B Practice Problem Answer Key

Appendix B: Practice Problems Answer Key

Chapter 1 - Introduction to Functions Section 1.1: What is a Function? 1. a) time in minutes, t, distance in km, D b) {(0, 0), (20, 4003), (40, 9452), (60, 14232), (80, 18700), (100, 20200), (120, 20200)} c) Forty minutes after being launched, the satellite is 9452 km from the Earth d) Yes. Every time value has exactly one distance value. e) No. The distance value 20,200 km corresponds to two time values, 100 minutes and 120 minutes. 2. a) time in minutes, t, number of Gene copies, G b) {(0, 52), (3, 104), (5, 165), (6, 208), (8, 330), (10, 524), (12, 832)} c) After six minutes of observation, there are 208 gene copies. d) Yes. Every time value has exactly one number of Gene Copies value. e) Yes. Every number of gene copies has exactly one corresponding time value. 3. a) time in minutes, t, number of homework problems completed, H

b) {(0, 0), (10, 3), (20, 8), (30, 8), (40, 15), (50, 17), (60, 20)} c) After forty minutes, Tara completed 15 homework problems. d) Yes. Every time value has exactly one corresponding number of homework

problems completed. e) No. The number of homework problems completed value of 8 corresponds to two

time values, 20 minutes and 30 minutes. 4. a) time in minutes, t, number of hotdogs eaten, H

b) {(0, 0), (1, 8), (3, 23), (5, 37), (7, 50), (9, 63), (10, 68)} c) After seven minutes, the competitive hotdog eater had eaten 50 hotdogs. d) Yes. Every time value has exactly one corresponding number of hotdogs eaten e) Yes. Every number of hot dogs eaten has exactly one corresponding time value Section 1.2: Multiple Representations of Function 5. a) Yes b) Yes c) No d) Yes e) No 6. a) No b) Yes c) No d) Yes e) No 7. a) Yes b) Yes c) Yes d) No e) Yes f) No 8. Answers Vary 9. a) Constant b) Decreasing c) Increasing d) Decreasing e) Constant f) Increasing 10. a) Increasing b) Decreasing c) Constant

d) Constant e) Decreasing f) Increasing Section 1.3: Function Evaluation 11. a) 4 b) 7 c) 6


471

12. a) 20 b) 6 c) 14 13. a) 18 b) 13 c) 4 14. a) −18 b) −4 c) 0 15. a) −2𝑥 + 6 b) −3

1𝑥 + 6 c) −𝑥 + 9

16. a) 14 − 6𝑡 b) 14 − 31𝑡 c) 6 − 2𝑡

17. a) 8𝑐1 + 6𝑐 + 4 b) 2𝑐1 − 7𝑐 + 9 c) 2𝑥1 + 5𝑥 + 6 18. a) Given: Input Finding: Output Ordered Pair: (2, 0) b) Given: Output Finding: Input Ordered Pair: (3, 3) c) Given: Input Finding: Output Ordered Pair: (−4, −18) d) Given: Output Finding: Input Ordered Pair: (−2, −12) 19. a) Given: Input Finding: Output Ordered Pair: 4, 33

1

b) Given: Output Finding: Input Ordered Pair: 86, 3

c) Given: Input Finding: Output Ordered Pair: −8,− 1S1

d) Given: Output Finding: Input Ordered Pair: −2,− 81

20. a) −7 b) 7 c) 14 21. a) Given: Output Finding: Input Ordered Pair: 0,−12 b) Given: Input Finding: Output Ordered Pair: −4,3 c) Given: Output Finding: Input Ordered Pair: −4,3 d) Given: Input Finding: Output Ordered Pair: 2,−17 22. a) Given: Output Finding: Input Ordered Pair: 0, 5

b) Given: Input Finding: Output Ordered Pair: −2,3 c) Given: Output Finding: Input Ordered Pair: −2,3

d) Given: Input Finding: Output Ordered Pair: 3,9 23. a) Given: Output Finding: Input Ordered Pairs: 0, 5 , (6,5) b) Given: Input Finding: Output Ordered Pair: 2,−3 c) Given: Output Finding: Input Ordered Pair: 1,0 , 5,0 d) Given: Input Finding: Output Ordered Pair: 3,−4 24. y = 2x – 3

a)

x – 3 – 1 0 1 3 y – 9 – 5 – 3 – 1 3

b)


472

25. f(x) = -3x+4 a)

x – 3 – 1 0 1 3 y 13 7 4 1 – 5

b)

Section 1.4: Domain and Range 26. a) Domain: 3, 5, 7, 9, 11, 13 Range: −2,−1, 8, 4 b) Domain: −2,−1, 0, 1 Range: −5 c) Domain: −3, 1, 0, 4 Range: 2,−5,−3,−2 27. a) Domain: −10,−5, 0, 5, 10 Range: 3, 8, 12, 15, 18 b) Domain: −20,−10, 0, 10, 20, 30 Range: −4, 14, 32, 50, 68, 86 c) Domain: 1, 2, 3, 4, 8, 9, 10, 11, 12 Range: 54, 62, 66, 69, 72, 73, 74 28. a) Domain: Inequality Notation −∞ < 𝑥 < ∞, Interval Notation (−∞,∞) Range: Inequality Notation −∞ < 𝑦 < ∞, Interval Notation (−∞,∞) b) Domain: Inequality Notation−8 ≤ 𝑥 ≤ 6, Interval Notation −8, 6 Range: Inequality Notation−4 ≤ 𝑦 ≤ 4, Interval Notation −4,4 c) Domain: Inequality Notation−6 ≤ 𝑥 ≤ 7, Interval Notation −6, 7 Range: Inequality Notation−3 ≤ 𝑦 ≤ 2, Interval Notation −3,2 d) Domain: Inequality Notation−8 < 𝑥 ≤ 7, Interval Notation (−8, 7] Range: Inequality Notation−5 ≤ 𝑦 < 4, Interval Notation [−5,4) Section 1.5: Applications of Functions 29. a) C(w) = 0.5w + 20 b) 0 ≤ w ≤ 200 c) 20 ≤ C(w) ≤ 120

d) W 0 50 150 200 C(w) 20 45 95 120

e) C(50) = 45. When 50 windows are washed, the total cost is $45. f) C(50) = 45. When 50 windows are washed, the total cost is $45.

g) Solve 45 = 0.50𝑤 + 20 for w. 30. b) (0, 0) or P(0) = 0; (8, 96) or P(8) = 96

c) Input Quantity: time in hours Practical Domain: Inequality Notation 0 ≤ 𝑡 ≤ 8, Interval Notation 0, 8 d) Output Quantity: number of pizzas made Practical Domain: Inequality Notation 0 ≤ 𝑃(𝑡) ≤ 96, Interval Notation 0, 96 e) P(3) = 36. The number of pizzas made in 3 hours is 36. f) 𝑃 5 S

\= 70. Seventy pizzas are made in 5 S

\ ℎ𝑜𝑢𝑟𝑠 or 5 hrs and 50 minutes.

31. a) x is used for the input b) The number of years since 1900


473

c) L is used for the output d) The life expectancy for males in years

e)

x L(x) 0 48.3 20 53.7 40 59.1 60 64.5 80 69.9 100 75.3 120 80.7

f) Practical Domain: 0 ≤ 𝑥 ≤ 120 g) Practical Range: 48.3 ≤ 𝐿(𝑥) ≤ 80.7 h) L(43.3) ≈ 60. 1900 + 43.3 = 1943.3, so the man was born in 1943.

Chapter 2 – Functions and Function Operations

Section 2.1: Combining Functions

1. a) 2 b) 4x2 + 2x – 1 c) 3 d)

2. a) –11x + 12 b) – x + 24 c) 12x2 – 18x d)

3. a) 18x3 – 6x2 + 2x + 2 b) 8 c) 6x2 – 7x + 1 d)

4. a) b) x3 + 2x2 + x + 1 c) 4𝑥1 + 2𝑥 − 1 − UDJ31D�PDJ1

5. a) 10 b) 8 c) 4 d) −36

6. a) 8 b) 4 c) 12 d) 2 e) 22 f) 5 7. a) 7 b) 4 c) 0 d) 0 8. a) 8 b) –5 c) 12 d) 32 Section 2.2: Applications of Function Operations 9. a) $587; $370; –$40; $587; $498; $300 b) By July, she will have saved up $2302. This is enough for her trip and she will have $302 for extra spending money. 10. a) $6050 b) $1280 c) $7330 d) $4330 e) about 67 people

3x2116x2−

++

x2

34−

1x342x2+

+−

22 346

xxxx −−−


474

11. a) $1322; $2172; $3022; $3872; $4722 b) $0; $1750; $3500; $5250; $7000 c) – $1322; –$422; $478; $1378; $2278 d) C(n) = 8.50n + 1322; R(n) = 17.50n; P(n) = 9n – 1322 e) (Answers may vary) Approximately 150 bedspreads

f) bedspreads or 147 bedspreads (rounded to the nearest whole number)

Chapter 3 – Linear Equations and Functions Section 3.1: Linear Equations and Functions 1. b) 2; Positive; Increasing c) 3; Positive; Increasing d) 12/5; Positive; Increasing e) –13/8; Negative; Decreasing f) 0; Zero; Constant g) 4/7; Positive; Increasing 2. a) y = – 4x – 8; – 4; ( 0, -8); (–2, 0) b) y = – 4x + 3; – 4; (0, 3); (¾, 0) c) y = ⅓ x – 2; ⅓; (0, –2); (6, 0) d) y = – 4x + 2; – 4; (0, –2); (–½, 0) e) y = 2x + 3; 2; (0, 3); (–3/2, 0) f) y = 2x; 2; (0, 0); (0, 0) g) y = 4; 0; (0, 4); DNE h) x = – 3; undefined; DNE; (–3, 0) Section 3.2: Graphs of Linear Functions 3.

a)

x y = 3x – 2 Ordered Pair –3 y = 3(–3) – 2 = -11 (–3, –11) –2 y = 3(–2) – 2 = -8 (–2, –8) –1 y = 3(–1) – 2 = -5 (–1, –5) 0 y = 3(0) – 2 = -2 (0, –2) 1 y = 3(1) – 2 = 1 (1, 1) 2 y = 3(2) – 2 = 4 (2, 4) 3 y = 3(3) – 2 = 7 (3, 7)

98146


475

b) b)

x y = – 2x + 4 Ordered Pair

–3 y = –2(–3) + 4 = 10 (–3, 10) –2 y = –2(–2) + 4 = 8 (–2, 8) –1 y = –2(–1) + 4 = 6 (–1, 6) 0 y = –2(0) + 4 = 4 (0, 4) 1 y = –2(1) + 4 = 2 (1, 2) 2 y = –2(2) + 4 = 0 (2, 0) 3 y = –2(3) + 4 = -2 (3, –2)

c)

x 𝑦 = −32𝑥 + 1 Ordered

Pair

–4 𝑦 = −32 −4 + 1 = 7 (–4, 7)

–2 𝑦 = −32 −2 + 1 = 4 (–2, 4)

0 𝑦 = −32 0 + 1 = 1 (0, 1)

2 𝑦 = −32 2 + 1 = −2 (2, –2)

4 𝑦 = −32 4 + 1 = −5 (4, -5)

d) b)

x 𝑦 =

2𝑥5 − 3 Ordered

Pair –10

𝑦 =2(−10)

5 − 3 = −7 (–10,-7)

–5 𝑦 =

2(−5)5 − 3 = −5

(–5,-5)

0 𝑦 =

2(0)5 − 3 = −3

(0, -3)

5 𝑦 =

2(5)5 − 3 = −1

(5, -1)

10 𝑦 =

2(10)5 − 3 = 1

(10,1)


476

4. y-intercept: (0, -1); x-intercept: (⅓, 0) 5. y-intercept: (0, 2); x-intercept: (2, 0) 6. y-intercept: (0, 3); x-intercept: (9, 0) 7. y-intercept: (0, -4); x-intercept: (6, 0) 8.

a)

Many possible answers such as (–2, 4), (–4, –4)

x y –3 0 –5 –8 –1 8

b)

Many possible answers such as (–2, 7), (1, 1)

x y –3 9 0 3 6 –9

c)

Many possible answers such as (5, –5), (–1, –3)

x y 2 –4 8 –6

–10 0

d)

Many possible answers such as (–1, 3), (1, 4)

x y –3 2 –7 0 7 7


477

Section 3.3: Horizontal and Vertical Lines 9. b) 0; DNE; (0, 3) c) undefined; (3, 0); DNE d) 0; DNE; (0, –2) e) undefined; (-4, 0); DNE f) 0; (all real numbers, 0); (0, 0) g) undefined; (0, 0); (0, all real numbers) 10.

a)

b)

c)

d)

e)

f)

11.

a) 𝑦 = 3 b) 𝑥 = 5


478

c) y = 3 d) 𝑥 = – 4

Section 3.4: Writing the Equation of a Line

12. a) y = 2x – 3; b) y = – 4x + ⅔; c) ; d) y = –

2.37x+6.35 13. a) y = 2x – 7; f(x) = 2x – 7 b) y = – 4x + 16; f(x) = – 4x + 16

c)

d) y = –1.4x + 5.14; f(x) = –1.4x + 5.14 14. a) m = 5; y = 5x – 13; f(x) = 5x – 13 b) m = –3; y = –3x – 3; f(x) = –3x – 3

c) m =

d) m = 1.3; y = 1.3x + 1.6; f(x) = 1.3x + 1.6

15. a) m =

b) m = –3; y = –3x + 13; f(x) = –3x + 13 c) m = 0; y = –5; f(x) = –5 16. a) m = –½; y = –½x – 2; f(x) = –½x – 2 b) m = 3; y = 3x – 5; f(x) = 3x – 5 c) m = 0; y = –5; f(x) = –5

17. a) y = x – 3 b) c)

18. a) Increasing b) (0, 2) c) (– 4, 0) d) m = ½

e)

19. y = –2x + 200

20. a) y = 1.8x + 0.5 b) cups c) Bucket weighs 0.5 lb

Chapter 4 – Linear Functions and Applications Section 4.1: Review of Linear Functions

5x83y −=

25

165)(;

25

165

−=−= xxfxy

35

34)(;

35

34;

34

+−=+−=− xxfxy

6x23xf6x

23y

23

−=−= )(;;

4x34y +−= 5x

35y +=

221

+= xy

91111


479

1. a) (0, 875) – Edward is 875 miles away from Bella when he begins his trip. b) (12.5, 0) – After 12.5 hours, Edward has traveled his full trip and is zero miles from Bella c) D(4) = 595 – After 4 hours, Edward is 595 miles from Bella. d) t = 5.3 or D(5.3) = 504 – After 5.3 hours, Edward is 504 miles away from Bella. e) Decreasing because the slope is negative. f) −70 miles per hour – Every hour, Edward’s distance from Bella decreases by 70 miles. g) Practical Domain: 0 ≤ t ≤12.5; Practical Range: 0 ≤ D(t) ≤ 875

Section 4.2: Average Rate of Change 2. a) not linear b) linear, m = 2 c) linear, m = ─3 d) linear, m = ½ e) not linear 3. a) Vertical Intercept: (0, 0); Average Rate of Change: 6200 views per hour b) V(t) = 6200t c) V(8) = 49,600 – After 8 hours, the video had been viewed 49,600 times. d) After approximately 16 hours, the video had been viewed 100,000 times. 4. a) Vertical Intercept: (0, 6), Average Rate of Change: 1.25 pounds per year b) W(t) = 1.25t + 6 c) After 8 years, the cat weighs 16 pounds. d) After approximately 11 years, the cat will weigh 20 pounds. 5. a) 4 pushups per minute

b) 4 pushups per minute c) 3.6 pushups per minute. d) On average, over the 5-week interval, the number of pushups Tim could do increased

by 3.6 pushups per week. e) No. The increase per week is a constant 4 pushups per minute for weeks 0 – 4.

However, the increase is only 2 pushups per minute between weeks 4 and 5. A perfectly linear function would have a constant rate of change for any pair of weeks.

6. a) $62 per inch b) $62 per inch c) For every increase in depth of 1 inch of coins, the value of the coins in the coffee can

increases by $62. d) V(d) = 62d

e) Ordered pair: ; Function notation: = 1000. The value of the

coins in the coffee can will be $1000 when the coins are inches deep.

Section 4.4: Scatterplot and Linear Modeling 9. a) All answers are $48 per day

b) Yes – average rates of change of intervals are constant so data could be exactly linear

⎟⎠

⎞⎜⎝

⎛ 1000,31416 ⎟

⎠

⎞⎜⎝

⎛31416V

41316


480

d) The data appear to be exactly linear. They look like they are all falling on a straight line.

e) y = 48x; E(t) = 48t f) Every day, Sara earns $48 more. This is the same amount as the computed average

rate of change.

10. a) i.) 2 degrees per day; ii.) 2 degrees per day; iii.) degrees per day

b) No – not all of the average rates of change are the same c) No – the temperature seem to increase and then decrease

11. a) i.) 10 pounds per day; ii.) 10.2 pounds per day; iii.) pounds per day

b) Yes – although the average rates of change are not the same, they are quite close to each other.

c) Data appear to be approximately linear. Except for the point (12, 147) the rest of the data points appear approximately linear.

12. b) Approximately linear – All of the data do not fall exactly on the line, but they are fairly close to the line, if drawn. d) One possible answer: y = –51.65x + 448.2; S(t) = –51.65t + 448.2 e) m = –51.65; TType equation here.he number of newspaper subscriptions in

Middletown, USA is decreasing by 51,650 per year since 2002. f) From our linear model: Vertical Intercept: (0, 448). In 2002, there were

approximately 448,000 newspaper subscriptions in Middletown, USA. g) 189, 950 newspaper subscriptions h) According to our model, the number of newspaper subscriptions in 2004 was

approximately 344.9 thousand or 344,900. Using our model, our answer does not match the data value but comes close to the data value.

i) Year 2009 13. b) Approximately linear – data do not fall exactly on the line, but they are fairly close to

the line, if drawn. d) y = 18.01x + 9.22; D(t) = 18.01t + 9.22

e) slope is 18.01 – Scott hiked approximately 18.01 miles per day. f) 909.72 miles g) Approximately 55 days

Chapter 5 - Absolute Value Equations and Inequalities Section 5.1: Absolute Value Equations 1. a) x = – 8, 4 b) x = 4, 8 c) x = –8, 0 d) x = –½

e) No Solution f) No Solution g) x = –1, 5 h) x = 4,

Section 5.2: Absolute Value Inequality 2. a) Inequality Notation: –2 ≤ x ≤ 2; Interval Notation: [–2, 2]

b) Inequality Notation: x ≤ –10 or x ≥ 10 Interval Notation: (–∞, –10] U [10, ∞) c) Inequality Notation: –14 < x < 6; Interval Notation: (–14, 6) d) No Solution

78

−

6110

34

−


481

e) Inequality Notation: x = ; Interval Notation:

f) Inequality Notation: x < –7 or x > 9; Interval Notation: (–∞, –7) U (9, ∞) g) Inequality Notation: x ≤ –1 or x ≥ 1; Interval Notation: (–∞, –1] U [1, ∞) h) Inequality Notation: –∞ < x < ∞; Interval Notation: (–∞, ∞) i) Inequality Notation: –2 < x < 9; Interval Notation: (–2, 9)

j) Inequality Notation: ; Interval Notation:

3. a) x = –8, 6 b) Inequality Notation: x < –8 or x > 6; Interval Notation: (–∞, –8) U (6, ∞) c) Inequality Notation: –8 ≤ x ≤ 6; Interval Notation: [–8, 6] Section 5.3: Absolute Value Applications 4. | x – 18 | > 0.36 where x is the weight of the oatmeal container; Reject pile: x < 17.64 oz or x > 18.36 oz. 5. | x – 21.7 | ≤ 3.2 where x is the body mass; Healthy body mass range: 18.5 ≤ x ≤ 24.9 6. a) 7.125 seconds < t < 8.875 seconds b) t ≤ 6.75 seconds or t ≥ 9.25 seconds c) t = 8 seconds 7. a) x = –6, 4/3 b) x = –½ 8. One possible answer: |x – 6| ≤ 3

Chapter 6 – Solving Quadratic Equations Section 6.1: Factoring Quadratic Expressions 1. a) (x + 6)(x + 1) b) 3𝑥(𝑥1 + 4) c) (𝑥1 + 9)(𝑥 + 3)(𝑥 − 3) d) (2x – 11) (x – 1) e) (5x – 1)(x – 2) f) 3(x + 1)(x + 3) g) (6 − 𝑥)(6 + 𝑥) h) (3x – 1)(2x + 1) i) (𝑛 + 1)(𝑛 − 1)(𝑛 − 4) j) 3(x2 – 2x + 8) Section 6.2: Solving Quadratic Equations by Factoring 2. a) x = 0, 2 b) x = ⅔, 0 c) x = 0, 2 3. a) x = –6, –2 b) x = –6, 7 c) x = –1, 5 d) x = –6, 6 4. a) x = -4, 4, -1, 1 b) x = –2, ⅓ c) x = –3, 5

5. a) b) c) d)

Section 6.3: Completing the Square

6. a) x = ± 4 b) ≈ – 0.44; 3.44

7. a) ≈ –1.32, 5.32

b) c) ≈ –2.37, –0.63

56

⎭⎬⎫

⎩⎨⎧56

817

823

≤≤− x ⎥⎦

⎤⎢⎣

⎡−817,

823

HVB 3

=BAxCy −

=A

ABC+

=1 lw

lwAh222+

−=

2153±

=x

112±=x

25,

21

−=x233±−

=x


482

Section 6.4: The Quadratic Formula

8. a) x = –1, 2 b) ≈ –0.64, 3.14 c) x = –1,

d) ≈ –0.19, 0.86 e) ≈ –3.31, 1.81

Section 6.5: The Quadratic Formula 9. a) 9i b) c) –2 – 10i d) 12 + 6i

e) 7 + i f) –8 – 20i g) 3 + 4i h)

i) ⅓ + 2i j)

Section 6.6: Complex Solutions to Quadratic Equations

10. a) x = –5 + 3i or –5 – 3i b)

c) d)

11. a) x2 + 80x + 1500 = 40,000 b) length: 190.25 feet; width: 210.25 feet c) Corn: 25,680.0625 ft2; Cucumber: 4807.5 ft2; Kale: 8,012.5 ft2; Collard: 1,500 ft2 12. a) 180 sq. ft. b) 1.93 ft. 13. a) One possible answer: f(x) = x2 – x – 6 b) One possible answer: g(x) = 2x2 – 11x + 12 c) One possible answer: h(x) = x3 – 2x2 – x + 2 14. a) Incorrect factor; (x – 6)(x + 1) b) Factor only - No solving; (x – 6)(x – 2)

15. Solution has 2 answers and ; x = –1 or x = 7 b) Cannot divide by a variable; x = 0 or x = ⅓

c) Variable being solved needs to be isolate;

d) Answer not simplified;

Chapter 7 – Quadratic Graphs Section 7.1: Solutions to Quadratic Equations 1. a) 1; x = 3 b) No Real Number Solution

4575 ±

=x23

6102 ±

=x41053±−

=x

i11

i32

32−

i23

21−

ixorix261

261 −−=+−=

ixorix23

23

−== ixorix317

32

317

32

−=+=

( ) |3|3 2 −=− xx

bgcda−

=

hAb 2

=


483

c) 2; x = –3, 0.5 d) 2; x = –2, 0

Section 7.2: Characteristics of Quadratic Functions 2. a) a = 2, b = – 4, c = – 4; Opens upward since a > 0; (0, – 4) b) a = –1, b = 6, c = – 4; Opens downward since a < 0; (0, – 5) c) a = –2, b = –6, c = 1; Opens downward since a < 0; (0, 1) d) a = 1, b = –3, c = 0; Opens upward since a > 0; (0, 0) e) a = ½, b = 0, c = –3, Opens upward since a > 0; (0, –3) 3. a) 0.5, –2.5, (0.5, –2.5), x = 0.5 b) 3, –2.5, (3, –2.5), x = 3 c) 0, 3, (0, 3), x = 0 d) –0.25, –0.25, (–0.25, –0.25), x = –0.25 e) 0, 0, (0, 0), x = 0 4. a) Domain: (–∞, ∞), Range: (–∞, –2.5), x-intercepts: None b) Domain: (–∞, ∞), Range: (–2.5, ∞), x-intercepts: (0.76, 0), (5.24, 0) c) Domain: (–∞, ∞), Range: (–∞, 3), x-intercepts: (–1.73, 0), (1.73, 0) d) Domain: (–∞, ∞), Range: (–0.25, ∞), x-intercepts: (–0.5, 0), (0, 0) e) Domain: (–∞, ∞), Range: (0, ∞), x-intercept: (0, 0) 5. a) Vertex: (1.5, 7.5); VI: (0, 3); Axis of Symmetry: x = 1.5; Point Symmetric to VI: (3,

3); HI: (–0.436, 0), (3.436, 0); Domain: (–∞, ∞); Range: (–∞, 7.5) b) Vertex: (1⅓, –1⅓); VI: (0, 0); Axis of Symmetry: x = 1⅓; Pt Symmetric to VI: (2⅔, 0); HI: (0, 0), (2.667, 0); Domain: (–∞, ∞); Range: (–1.⅓, ∞) c) Vertex: (0, 4); VI: (0, 4); Axis of Symmetry: x = 0; Point Symmetric to VI: (0, 4); HI: None; Domain: (–∞, ∞), Range: (4, ∞) Section 7.3: Solving Quadratic Equations Algebraically and Graphically 6. a) x = – 4, 1 b) x = –½, 0 c) x = 1, 4 7. a) x = 0, 2 b) c) x = –½, –2 d) x = 0, 2

e) ≈ – 0.58, 4.58 f) ≈ –1.66, 1.56

Section 7.4: Applications of Quadratic Functions 8. a) 15 feet b) 3.25 seconds c) 17.1125 feet d) 12.5 seconds e) [0, 12.5] f) [0, 17.1125] g) Vertex: (3.25, 17.1125); HI: (12.5, 0); VI: (0, 15) 9. a) P(x) = –2x2 + 680x – 10,000 b) (0, –10000) - If the company sells 0 items, it loses $10,000 c) 170 items d) $47,800 e) 16 items f) Practical Domain: [0, ∞); Practical Range: (–∞, 47800] 10. The arrow reaches its maximum height at approximately 2.81 seconds. b) The maximum height of the arrow is 132.56 feet. c) The arrow hits the ground at approximately 5.69 seconds.

46.332x ±≈±=

3306 ±

=x2010411±−

=x


484

d) (0, 6) – The arrow is initially shot from a height of 6 feet. e) H(3) = 132 – After 3 seconds, the height of the arrow is 132 feet. f) t = 1, 4.625 – At 1 second and 4.625 seconds, the height of the arrow is 80 feet. g) Practical Domain: [0, 5.69] h) Practical Range: [0, 132.56] 11. a) 28.8 feet b) 16 feet c) 10.8 feet d) Vertex: (0, 28.8); HI: (8, 0); VI: (0, 28.8)


485

Section 7.5: Quadratic Modeling 12. a) F(t) = –15.75t2 + 68.85t + 40.75

b) Based on the model, the height of the launching tower is 40.75 feet because this is the output when the time value is 0.

c) The model predicts the height of the fireworks to be 105.55 feet at 3 seconds. The data table shows a height of 103 feet. The model’s prediction is 2.55 feet more than the actual height. This is because the model approximates data and doesn’t fit the data exactly.

d) t = 0.572 seconds and t = 3.8 seconds e) t = 4.9 seconds f) Practical Domain: [0, 4.9]; Practical Range: [0, 116]

13. a) J(t) = –38.57t2 + 303.429t + 23 b) The hill is 23 feet high because this is the output when the input is 0. c) The model predicts the height of the rocket to be 586.148 feet at 3 seconds. The data

table shows a height of 590 feet. The model’s prediction is approximately 4 feet less than the actual height. This is because models approximate data and don’t fit the data exactly.

d) t = 1.836 seconds and t = 6.031 seconds e) t = 7.942 seconds f) Practical Domain: [0, 7.942]; Practical Range: [0, 619.751]

14. a) P(x) = –2x2 + 679.24x – 9970.78 b) VI: (0, –9970.78) – If the company sells 0 items, it will lose $9970.78. c) Vertex: (169.81, 47700.09) – The maximum profit of $47,700.09 will be earned

when approximately 170 items are sold. d) The company needs to sell 16 items to break even.

Chapter 7: Extension 15. a) Answers may vary b) Answers may vary c) Answers may vary 16. x = 0 17. (–5, 7)


486

Chapter 8 – Introduction to Exponential & Logarithmic Functions

Section 8.1: Linear Functions vs. Exponential Functions 1. For both functions: VI (0, $1000) - $1000 was invested in a simple interest account. b)

t S(t) C(t) 1 $1050 $1040 5 $1250 $1216.65 10 $1500 $1480.24 12 $1600 $1601.03 15 $1750 $1800.94 18 $1900 $2025.82 20 $2000 $2191.12 25 $2250 $2665.84

c) Answers may vary 2. a) VI: (0, 1); Slope: -3; f(x) = –3x + 1 b) VI: (0, -1); Slope = 2; f(x) = 2x-1 3. a) f(x) = 4(2)x = 21 2D =2x + 2 b) f(x) = 2(2.1)x

4. a) Exponential – there is a common ratio of 5 between successive outputs

b) Linear – there is a common difference of 0.875 between successive outputs c) Linear – there is a common difference of -2.5 between successive outputs d) Linear – there is a common difference of -1.25 between successive outputs e) Exponential – there is a common ratio of ⅓ between successive outputs Section 8.2: Introduction to Logarithms 5. a) (5, 1.6094) b) (3, 4.3944) c) (2, 1.2041) d) (6, 2.5850)

6. a) b) c)

d) 45 = 1024 e) 253/2 = 125 f) e2x = 200 Section 8.3: Computing Logarithms 7. a) 1 < n < 2 b) –2 < n < –1 c) –3 < n < –2 d) –1 < n < 0 e) 4 < n < 5 f) 3 < n < 4 g) –2< n < –1 8. a) 0 b) 1 c) Does not exist d) 1/3

e) 1 f) –3 g) 5 i) –2 j) 999 Section 8.4: Solving Logarithmic Equations

9. a) b) c) 2401 d) 49

10. a) b) 30 c) 731 d) 1014

29log3 =214log16 = 3

81log2 −=

169

2161

71


487

11. a) 100.1 or 1.26

b) Alaska earthquake is 100.8 ≈ 6 times more intense than San Francisco earthquake 12. a) 10–1 = 0.1 b) 1.5 c) 101.8 ≈ 63 d) pH = 6

Section 8.5: Solving Exponential Equations

13. a) b)

c) x = ln 14 ≈ 2.6391 d)

14. a) x = log 6 ≈ 0.7782 b) x = ln 18 – 2 ≈ 0.8904 c) x = log 4 + 2 ≈ 2.6021 d) x = 0 15. a) 1,500,000 b) 0.015 c) 195.8554 d) –3.6466

16. a) b) ≈ 0.2218 c) ln 3 ≈ 1.0986 d) ln 5.5 ≈ 1.7047

Section 8.6: Applications of Exponential Functions 17. a) 3.80% b) P(t) = 210,679e0.038t c) 264,63

18. a) P(t) = 382,872e–0.026t b) 273,063

c) approximately 9 years d) approximately 27 years

19. ≈ 8.66 years

20. ≈ 4.95 hours

21. Lead-211 Thorium-234 Cesium-137 Decay Rate 19.25% 2.89% 2.31% Exponential Decay Model

Time to reach 12 grams 2.65 minutes 17.68 days 22.11 years

Time to reach 1% of original 23.92 minutes 159.35 days 199.36 years

22. United States Thorium-234 Cesium-137 Exponential Growth Model

Population in 10 years 324,527,388 7,542,469 20,042,456

Time it takes to double 138.63 years 38.51 years 25.67 years

23. a) P(t) = 17,770t + 634,635 b) 705,715 c) P(t) = 634,635e0.0276t d) 708,713

2162.097ln≈=x 1461.0

75log −≈⎟⎠

⎞⎜⎝

⎛=x

4515.028log≈=x

4771.031log −≈⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛35log

08.02ln

=t

14.021ln ⎟⎠

⎞⎜⎝

⎛

−=t

tetA 1925.020)( −= tetA 0289.020)( −= tetA 0231.020)( −=

tetP 005.07.308)( = tetP 018.03.6)( = tetA 027.03.15)( =


488

Chapter 9 – Radical Functions Section 9.1: Roots, Radicals, and Rational Exponents 1. b) 201/5; 20^(1/5) ≈ 1.82 c) 24/3; 2^(4/3) ≈ 2.52 d) –72/4 = –71/2; –7^(1/2) ≈ –2.65 e) (–8)1/3; (–8)^(1/3) = –2

f) ; 3^(1/4) ≈ 1.32 g) ; 11^(1/7) ≈ 1.41 h) ; – 4^(1/2) = –2 i) ; (–2)^(2/3) ≈ 1.59 2. a) –2 b) DNE c) 6 d) –24 e) –2/5 f) –1 3. a) –6.7823 b) DNE c) 4 d) –1 4. a) 0, 1, 2 b) 0, 1, 2 c) 0, 6, 9 d) , 1, 0 e) 2, 0, 1 f) 2, 0, 4 5. a) w1/7 b) y9/2 c) m6/5 d) n3/4

e) f) g) h) Section 9.2: Operations with Radical Expressions 6. a) b) c) d)

e) f) g) h)

i) j) 7. a) b) –110 c) d) 4 e) f) –14 Section 9.3: Solving Radical Equations 8. a) –331 b) –13/2 c) 15 d) No Solution e) –76 f) –8/5 9. a) 1 or 7 b) 9 c) 9 d) 3 e) 9 f) No Solution

10. a) b)

c) 53 in; 259.6 lbs; 60.3 in; 180.3 lbs; 76.5 in; 143.6 lbs 11. a) 8 ohms; b) 63.25 volts c) 83.33 watts 13. ≈ 0.59 or 3.14 but ≈ 0.59 does not check

43

711

4−3 2)2(−

3

3c 5 3g xx4 32 pp

5215 + 3 42 9109 + 3 472 +3553 − 21583− 22240+ 43 +

5614 − 612212632 −−+

77218 − 6615 +61230 −

MW703

703

2MH

22±=x 22−


489

Chapter 10 – Rational Functions Section 10.1: Rational Function Operation

1. a) b)

2. a) b) 4y

c) d)

3. a) b) x c) d)

Section 10.2: Solving Rational Equations 4. a) x = 4 b) x = 7 c) x = ≈ ±2.45 d) x = 1, 2 e) x = –8 f) x = 0, 9 5. a) No Solution b) x = – 4, 8 c) x = –24/7

d) w = –3, 0 e) x = 2, f) ≈ –1.39, 2.89

6. a) b)

7. a) –1 b) x = c) d)

8. a) –2 b) x = 4 c) d)

9. a) b) c) d) –3(x + 2)

10. a) c = S – PS b) A = QC + B c)

d) e) f)

g) h)

Section 10.3: Applications of Rational Functions 11. a) $258 – Average cost of a cap when 1 cap is produced b) $33 – Average cost per cap when 10 caps are produced c) 36 caps d) $10.50; $8.25; $8.03; $8.00; $8.00 e) As more caps are produced, the average cost to produce each cap approaches $8.

3222

−

+−

xxx

yxyx

−

+

xxxor

xx

6)12)(12(

614 2 −+−

152

)2(6+

+

xxx

xxx )1)(43( +−

1212

+

−

xx

)(1hxx +

−

6±

25

4733±

=y

35 4,

72

=x

25

)1(23xx− x

x25)2(3

−

−

2

2

3143xx

+

+

2314

−

−

xx

121

23

±=x3

2x

AArErorR

AEr −

=−=

AEARr−

=tb

bta−

=BC

BCA−

=

2

24TLg π

=TmvL2

=


490

12. a) $7 – Average cost per refill with the Looney Tunes Cup when 1 refill is purchased b) 25 refills c) $0.81; $0.76; $0.76; $0.75; $0.75 d) As more soft drink refills are purchased, the average cost per refill approaches $0.75.

Chapter 11 – Course Review Functions 2. a) yes b) no c) yes 3. a) yes b) yes c) no 4. a) yes b) no c) yes 5. a) true b) false c) true d) true e) false 6. a) 2 b) –9 c) 13 d) 5 e) {3, 5, 6, 9, 13} f) {–9, 3, 2, 1} 7. a) 6 b) 8 c) –3 d) 2 e) {2, 0, 5, –3} f) {5, 6, 8, 7} 8. a) {4, 3, 0} b) 8, 11, 6} c) (4, 8), (3, 11), (0, 6) 9. a) [–3, 6] b) [–2, 4] c) 4 d) 2 e) –3 f) 3 10. a) (–∞, ∞) b) (–∞, ∞) c) 2 d) – 4 e) 1 f) – 4 11. a) [–3, 2) b) [–5, 4] c) 3 d) 4 e) –3 f) –2 12. a) W(–10) = 210; (–10, 210) b) p = 4 or p = 5; (4, 0) or (5, 0) 13. a) h(5) = 4; (5, 4) b) h(81) = 4; (81, 4)

14. a) b) c) d)

15. a) 5 b) 11 c) 4 d) 8 e) 80 f) 2 16. a) x2 + x + 8 b) –15 c) – 4 d) –64

e) x3 – 3x2 – 7x – 1 f)

17. a) After 12 months Darby has paid $10,462.08 to lease the car. b) After 24 months Darby has paid $17,432.28 to lease the car. c) Darby has paid $30,000.00 after 45.64 months. d) m = 580.85; Each month $580.85 is paid e) (0, 3491.88) At the beginning of the contract Darby pays $3491.88. f) 0 ≤ n ≤ 48 months g) $3,491.88 ≤ T(n) ≤ $31,372.68

38

3160

=x940a

3)1(20 −x

5387−

++x

x


491

18. a) 48 candy canes b) The machine makes 48 candy canes in 12 minutes. c) 4 candy canes are produced each minute d) Yes; C(t) = 4t e) 240 candy canes in one hour 19. c) y = 5.1x d) D(t) = 5.1t e) 117.3 (100,000s of miles) f) 249 days g) 102 (100,000s of miles) from earth 20. a) (0, 24.50); When the tow truck arrives, Bill pays $24.50. b) m = 5.5; For every mile towed, Bill pays $5.50.

21.

22. y = 2 23. x = –3 24. a) After 10 hours of driving, Donna is 242.5 miles from Phoenix.

b) After 13.83 hours of driving, Donna will be in Phoenix. c) m = –63.24; Donna travels 63.24 mph. d) (0, 874.9); When Donna begins her trip, she has 874.9 miles to drive before arriving

Phoenix. e) If Donna wants to make the trip home in 10 hours, she would need to travel 24.25

miles more each hour at an average speed of 87.49 mph.

25. a) Upward; (0,4); (1,0) and (4, 0); ; 𝑥 = S1; (5, 4); All Real Numbers; 𝑦 ≥ − T

4

b) Downward; (0, 16); (– 4, 0) and (4, 0); (0, 16); x = 0; (0, 16); all Real Numbers; y ≤ 16 c) Upward; (0, 5); None; (1, 4); x = 1; (2, 5); All Real Numbers; y ≥ 4 26. a) 3x (x – 3) b) (x – 1) (x – 3) c) x (x + 6)(x – 5) d) 2x (2x + 3)(2x – 3) 27. x = 0 or x = 6

28. a) x = 1 or x = 3 b) x = or x = 1 c) x = 2 d) x = 1 ± 2i

e) x = –2 or x = 5 f) g) h)

29. a) 6 + 15i b) 1 + 4i c) 9 – 7i 30. Numerous answers – parabola must be opening up with a vertical intercept on the

negative y-axis. 31. a) t = 3.5 seconds – After 3.5 seconds, ball is 24 feet from the ground. b) –16t2 + 40t + 80 = 0; t = 3.81 seconds c) 105 feet; maximum height is the H(t) value of the vertex d) Practical domain: [0, 3.81] and practical range is [0, 105] 32. a) Growth; 32%; (0, 24); 118.87; 9.16 b) Decay; 8%; (0, 332); 222.55; –3.80 c) Growth; 4%; (0, – 4); –3.78; 153.01 33. a) Exponential; f(x) = 5x b) Linear; f(x) = 0.875x + 0.375 c) Linear; f(x) = –2.5x – 8 d) Exponential; f(x) = 2(2x)

831

85

+= xy

⎟⎠

⎞⎜⎝

⎛ −49,

25

23

−

271 ix ±

=2351 ix ±

= 266 ix ±−=


492

34. a) 𝑙𝑜𝑔\216 = 3 b) 𝑙𝑜𝑔S

31S= −2 c) 𝑙𝑜𝑔U1 = 0

d) 7S = 16,807 e) 10S = 𝑥 35. a) 0 b) 1 c) DNE d) n

e) DNE f) ½ 36. a) 0 b) –1 c) 1 d) 2 e) –2 f) ⅓ g) 3/2 h) 15 i) 15 j) 600 37. a) x = 1/7 b) x = ln 2 ≈ 0.693

c) x = 20 d) 2.498

e) f)

g) No solution h) x = 10 38. a) 6.02 Million b) Year 2053 39. a) (0, 0); (0, 0); f (5) = 1.72; x = 81 b) (0, 3); (–9, 0); f (5) = 3.74; x = 0 c) (0, 2,29); (12, 0); f (5) = 1.91; x = 15 40. a) x = –331 b) x = 1 or x = 7 c) x = 15 d) No solution

e) x = 3 f)

41. a) None; None; x = 0; 𝑓 5 = 1S; 𝑥 = 1

S b) (0, 3); (–6, 0); x = 4; f (5) = 22; x = U

8

42. a) 𝑥 = \81

b)

≈+

=295log3x

209.023401250ln

31

≈⎟⎠

⎞⎜⎝

⎛−=x 642.623/210 ≈+=x

5216

−=x

175.2,575.084111

≈±

=x

Appendix B: Practice Problems Answer Key - … · Appendix B Practice Problem Answer Key Appendix...

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