Answers to Algebra 1 Unit 2 Practice 10. a. Answers will vary.

1. {(1, 0), (3, 0), (25, 2)} 2. A

21

3. The input 0 is paired with 2 outputs, 3 and 29.

6 12

4. Change one of the 0s in the table to a number other than 18 or 12. Then each input will be paired with exactly one output.

8

b. No; if there are fewer elements in the domain than in the range, then at least one domain value must be paired with two range values, which cannot happen in a function.

5. {(B, 2), (C, 4), (A, 5), (B, 6)} 6. a. Domain: {15, 3, 212, 21, 9}; Range: {2, 0, 21, 2}

11. D

b. Domain: {25, 1, 3}; Range: {22, 1}

12. x represents the number of movies Manny rents; f(x) represents the total cost for the movies.

7. The domain is whole numbers because you can purchase neither a negative number of bagels nor partial bagels; the range includes $0 and positive multiples of $0.90.

1 2

13. f(23) 5 21; f(0) 5 5; f   5 6;  f(10) 5 25 14. Hudson evaluated the function at x 5 17, instead of finding the value of x for which f(x) 5 17. The correct answer is 6.

8. B 9. a. Answers may vary. Sample answer: y

15. Answers will vary. Students should write a 1 sequence of numbers whose third term is . 4

5 4 3 2 1 0 2524232221 21

23

16. There are no breaks or gaps in the line. 1 2 3 4 5

x

17. B

22 23 24 25

18. a. Domain: {x | 0 # x # 60}; range: {y | 0 # y # 400} b. The domain for the second balloon is {x | 0 # x # 65} because the ride lasted 65 minutes instead of 60; the ranges are the same because neither balloon ever went higher than the other.

b. The domain and range values can be paired in 6 different ways, so there are 6 possible functions: {(21, 25), (0, 2), (3, 4)}  {(21, 25), (0, 4), (3, 2)}

19. a. The y-intercept is (0, 0); it represents the height at the start of the ride (when time 5 0).

{(21, 2), (0, 25), (3, 4)}  {(21, 2), (0, 4), (3, 25)}

b. No; all balloon rides must start from the ground, where height 5 0.

{(21, 4), (0, 2), (3, 25)} {(21, 4), (0, 25), (3, 2)}

20. The point where the graph intersects the y-axis is the y-intercept. 21. D

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SpringBoard Algebra 1, Unit 2 Practice

22. There are no relative or absolute minima.

33. Kathleen is incorrect. To check using the equation, substitute 25 for w: d 5 2(25) 5 52 inches. To check using the graph, find the point on the graph whose x-coordinate is 25. The y-coordinate of this point, 50, is the correct length of the stretch.

23. Answers may vary. All values of x correspond to a point on the graph. 24. No; the domain must be restricted so that the denominator is not 0. The correct domain is {x | x 5 4}.

34.

Time (seconds) 0 1 2 3 4 5 6 7 8 9

25. a. The independent variable is x, the number of hours. The dependent variable is the total cost. b. The reasonable domain is {x | x $ 0} because a negative number of hours does not make sense; the reasonable range is {y | y $ 0} because a negative cost does not make sense. 26. C 27. The function is discrete, and the least value for x is 0 and the greatest is 5; therefore, all real numbers greater than 0 cannot be the domain.

The object reaches the ground when the height is 0; the table shows that this happens between 8 and 9 seconds.

28. Answers will vary. Jason had a box containing 5 granola bars. Each day he ate one bar.

35. B

29. C

36.



Height (ft)

31. The reasonable domain is {w | w $ 0} because a weight cannot be negative; the reasonable range is {d | d $ 0} because a distance cannot be negative.

Spring Stretch (cm)

y 60 55 50 45 40 35 30 25 20 15 10 5

y 1200 1100 1000 900 800 700 600 500 400 300 200 100

30. d 5 2w

32.

y 5 1200216x2

0 y 5 2x

0

5 10 15 20 25 30 35 40 45 50 55 60 Weight (oz)

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Height (ft) 1200 1184 1136 1056 944 800 624 416 176 –96

1 2 3 4 5 6 7 8 9 10 Time (seconds)

x

Explanations may vary. The object falls for a length of time between 8 and 9 seconds, so viewing the x-axis from 0 to 10 is appropriate. The object starts at 1200 feet and falls to the ground (0 feet), so view the y-axis from 0 to 1200.

x

37. (0, 1200); the height of the object at time 5 0 (before it is dropped)

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SpringBoard Algebra 1, Unit 2 Practice

50. C

38. When the object reaches the ground, its height will be 0. Find the point on the graph that has y-coordinate 0; use the trace or intersect features of the calculator to find that the x-coordinate of this point is about 8.66. The object reaches the ground in about 8.66 seconds. To check using the equation, substitute 8.66 for t: 1200 2 16(8.66)2 5 0.704 ≈ 0.

51. Answers may vary. a. g(x) 5 x2 2 5 b. g(x) 5 (x 2 3)2 c. g(x) 5 (x 2 6)2 2 9 d. g(x) 5 (x 1 2)2 1 1

39. Years Since Purchase 0 1 2 3

Value of Computer

To check by graphing, graph the functions above and check that each vertex matches the vertex given in the problem.

$540 $360 $240 $160

52. (5, 4); the graph of g(x) will be translated 5 units to the right from the graph of f(x). The graph of f(x) has vertex (0, 4) so the graph of g(x) will have vertex (5, 4). 53. a. f(x) 5 5x 1 8.50

40. B

b. Substitute x 2 3 for x in the function from part a: f(x 2 3) 5 5(x 2 3) 1 8.50

41. Explanations may vary. One-half of any amount greater than 0 will always be greater than 0.

c. The graph of the function in part b is translated 3 units to the right from the graph of the function in part a.

42. Graph A; explanations may vary. Graph B shows an amount that is increasing, not decaying, over time.

54. B

43. 10 grams; the amount is the y-coordinate when time 5 2s; the line x 5 2 intersects the graph at (10, 2).

55. 3 56. Answers will vary. (0, 0) and (1, 2); 220 slope 5 5 2. 12 0 57. a. –2

44. translation 4 units down 45. a. the y-coordinate of the y-intercept of g(x) b. g(2)

b. Answers will vary; y 5 4. If y 5 4, then the 420 5 1 . 22. slope of line b is 3 2 (21) 58. Yes; the value of y increases by 4 as x increases by 1.

c. f(4) 46. B 47. No; the graph of g(x) 5 x2 1 3 would be 3 units above the graph of f(x) 5 x2.

59. a. f(x) 5 75x 1 250 b. Tables may vary.

48. a. b( x )55.25 x 1 4

Months 0 1 2 3

b. a( x )5 5.25 x 1 5 c. The graph of a(x) is one unit above the graph of b(x), because a(x) 5 b(x) 1 1. d. The graph will be translated down by the amount of the reduction.

Total Amount $250 $325 $400 $475

49. The graph of g(x) is translated 1 unit to the left and 12 units up from the graph of f(x). © 2014 College Board. All rights reserved.

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SpringBoard Algebra 1, Unit 2 Practice

Total Amount ($)

y 1000 900 800 700 600 500 400 300 200 100

66. 6; use any two ordered pairs in the table to find that the slope is 22. If y is the missing y-value, substitute (23, 12) and (0, y) into the slope formula:

y 2 12 5 22; solve this equation to 0 2 (23)

find y = 6.

y 5 75x 1250

67. B 0

1 2 3 4 5 6 7 8 9 10 Months

x

3 68. y 5 x , or y 5 1.5x 2 69. a. Yes; explanations may vary. The cost of 0 bags of soil is $0 and each bag of soil costs the same amount, so the equation can be written in the form y = kx, where y is the total cost and x is the number of bags purchased. Three bags cost 3.75 $3.75, so k 5 5 $1.25. 3 b. $8.75

c. $75/mo d. The rate of change is $75/mo, which is the amount that Carlos pays each month. It is the slope of the graph and the coefficient of x in the equation. 60. B 61. Answers will vary. An elevator descends at a rate of 15 feet per second.

70. a. Answers will vary. Students should write a linear equation that cannot be written in the form y 5 kx, such as y 5 3x 1 2.

62. a. Positive; the slope is 10.

b. Answers will vary. Students should graph a line that does not pass through (0, 0).

b. Negative; the slope is 26. c. Positive; the slope is 60.

71. Explanations may vary. Although the equation is not in the form y = kx, it can be written in that form by solving for y: y 5 2x.

63. B 64. Explanations may vary. The x-coordinates for every point on a vertical line are equal. This means that the denominator in the slope formula will be 0. Division by 0 is undefined, so the slope is undefined, not 0.

72. C 73. y 5

y 50

65. Answers will vary; (21, 5). The slope of the line containing the points (22, 0) and (4, 10) is 10 2 0 10 5 5 5  . Find another point (x, y) for 4 2 (22) 6 3 5 which the slope is . Using the slope formula and 3 y 20 5 5 ; y 5 5 and x 5 21 the point (22, 0), x 2 (22) 3 is one possible solution.

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40 x

Time (hours)

40 30 20 10 0

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1 2 3 4 5 6 7 8 9 10 Painters

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SpringBoard Algebra 1, Unit 2 Practice

80. a. Answers will vary. Students may recommend that Mariah not participate, because she would have to donate $2 for each item, and if each item sells for $3, Mariah will be donating most of what she earns for each item.

74. Color Your World; with 7 painters, Color Your World could paint walls in about 5.7 hours, which is faster than the 6 hours Good Hues would require. 75. x =

1 2

b. Answers will vary. Knowing the amount it costs Mariah to make each magnet would be helpful, because you could then determine whether she will lose money by donating $2 per magnet. If she still profits by making $1 per magnet, then she may want to participate in the fair. Students may also want to know whether Mariah wants to advertise her products; if so, she may want to participate regardless of profits.

76. a. x

y5

1 2 4 8

120 x

120 60 30 15

y5

160 x

160 80 40 20

y5

192 x

192 96 48 24

81. B

b. When x is doubled, y is divided in half.

82. No. Use the equation to show that there are y-values that correspond to more than one x-value; for example, both f(2) and f(22) are equal to 16. The graph fails the horizontal line test.

77. a. p(x) 5 15x 2 450 y 500

x 9.25 b. The total amount spent on tickets

400

83. a. f 21 (x) 5

300 Profit ($)

200 100 0 2100

4 8 12 16 20 24 28 32 36 40 44 48

2200 2300

c. 5

x

84. a. {(January, 31), (February, 28), (March, 31), (April, 30), (May, 31), (June, 30), (July, 31), (August, 31), (September, 30), (October, 31), (November, 30), (December, 31)}

y 5 15x 2450

2400 2500

b. No; the function is not one-to-one. For example, April and June are both paired with 30.

Hours Worked

c. Answers will vary. A function that pairs each person with their Social Security Number is one-to-one, because each person has a unique Social Security Number.

b. (0, 2450); it represents Erin’s profit (2$450) when she has worked 0 hours. c. 20 jobs; use the equation or the graph to find that Erin must work 30 hours to break even. If a job takes 1.5 hours, then Erin must work 30 5 20 jobs to break even. 1.5 d. Answers may vary. Erin could double her hourly rate to $30. Or Erin could keep her hourly rate at $15 but shop around for less expensive supplies; if she can reduce her initial costs to $225, she will break even after 10 jobs.

85. No; if two functions are inverses, then they undo each other. Because f(0) 5 3, if g(x) were its inverse then g(3) would be 0. However, g(3) 5 21. Therefore the functions are not inverses. 86. B 87. No; in an arithmetic sequence, each term is found by adding, a common difference to the term before, not multiplying.

78. f(x) 5 2x 1 25 79. B

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SpringBoard Algebra 1, Unit 2 Practice

88. a. 26 1 b. 2 c. 10

99. a. f –1(n) 5

n 11 3

b. The inputs are terms in the sequence; the outputs are the corresponding term numbers. c. Answers will vary. The value 56 corresponds to what term in the sequence? Substitute 56 into the inverse function: 56 1 1 f –1(56) 5 5 19. 56 is the value of the 19th 3 term in the sequence.

89. Yes; the common difference is 0. 90. Sequence 1 91. Sequence 3 92. D

100. D

93. an 5 3 1 2(n 2 1), or 2n 1 1; an is the number of tiles in the nth stage and d 5 2 is the number of tiles added to each stage.

 1  1 a1 5 2  f (1) 5 2 101. a.  ; 3 3 a 5 a 1   n n21 4  f (n) 5 f (n 2 1) 1 4  f (1) 52 3 a 523 ; b.  1 an 5 an21 1 2  f (n) 5 f (n 2 1) 1 2 a1 5 5  f (1) 5 5 c.  ; an 5 an21 1 4  f (n) 5 f (n 2 1) 1 4

94. If a4 5 11 and a9 5 36, then 11 1 5d 5 36. Solve this equation to find d 5 5. Then a1 1 3(5) 5 11; solve this equation to find a1 5 24. The explicit formula is an 5 24 1 5(n 2 1), or 5n 2 9. y

a1 52 8  f (1) 528 ; d.  an 5 an21 1 6  f (n) 5 f (n 2 1) 1 6

50 45 40 35 30 25 20 15 10 5 0 25

102. No; explanations may vary. The first 4 terms of this sequence are 1, 8, 22, and 48. There is no common difference between consecutive terms, so the sequence is not arithmetic.

1 2 3 4 5 6 7 8 9 10

103. Answers may vary. The first term must be defined to determine where the sequence begins. For example, an 5 an21 2 3 indicates a sequence with a common difference of 23, but there are infinitely many such sequences; two examples are 0, 23, 26, 29, … and 4, 1, 22, 25, … . Without knowing the first term, it is impossible to determine the sequence.

x

95. f(n) 5 5n 2 8 96. f(2) 5 0; f(11) 5 18 97. C 98. a. Yes; a graph of the sequence includes the points (3, 212) and (7, 0). Luisa knows that the slope of the graph is the common difference of the sequence, so she is using slope to find the value of d.

104. D 105. y 5 24x 1 1 1 106. y 52 x 2 1 2

b. f (n) 5 3n 2 21

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SpringBoard Algebra 1, Unit 2 Practice

118. h, g, e, d, b, a, f, c

107. a. y 5 50x 1 200 b. 9 months; find the value of x for which f(x) 5 650. Because 50(9) 1 200 5 650, x 5 9.

119. 4x 1 3.75y 5 50

c. The equations will have the same constant term, 200, and the graphs will have the same y-intercept, (0, 200), because both Laura and her brother began with $200. The equations will have different coefficients for x and the graphs will have different slopes because each person deposits a different amount of money each month.

120. B 121. a 5 4; explanations may vary. The graph 2 of 2x 1 3y 5 6 has slope  . A line that is 3 3 perpendicular will have slope . The graph of 2 6 3 ay 5 6x 1 10 has slope ; for this to be equal to , a 2 a must equal 4. 19 122. a. Answers will vary; y 2 x 5 19 19 b. Yes; any line with slope (except y 5 x 2 7 , 5 5 the equation of the given line) is a possible

108. From the given equation, the y-intercept is (0, 25). Use this point and (22, 3) to find the slope. The equation is y 5 24x 2 5. 109. y 1 1 5 15(x 2 3)

answer, and there are infinitely many such lines. 19 Another example is y 5 x 1 2. 5

110. Answers may vary; y 2 1 5 0.5(x 2 3) 111. Infinitely many; any point on the line can be substituted into the point-slope form.

1 23 1 123. a. y 2 12 52 ( x 11), or y 5 2 x 1 2 2 2 b. No; a given point and slope define exactly one line.

112. One; the line has only one slope and one y-intercept, so only one equation can be written in slope-intercept form.

124. C

113. C

125. The finishing times decrease as the months of training increase.

114. You cannot enter an equation in point-slope form into a graphing calculator because the equation is not solved for y. To enter an equation in pointslope form into a calculator, you must first solve the equation for y. For example, the equation y 2 2 5 23(x 1 5) can be entered into a graphing calculator as y 5 23(x 1 5) 1 2, or y 5 23x 2 13.

126.

y 100 80 60 40

115. x-intercept 5 (27, 0); y-intercept 5 (0, 24); 4 slope 5 7

20

116. C

0

1 2 3 4 5 6 7 8

x

117. Chase has found the equation of the vertical line through (3, 5), not the horizontal line. The correct equation is 0x 1 y 5 5, or y 5 5.

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SpringBoard Algebra 1, Unit 2 Practice

132. Answers will vary; the number of people at the public pool and the number of frozen yogurt cups sold at a frozen yogurt shop across town. It is reasonable to expect that there might be a positive correlation between these two variables, but it is probably not the case that an increase in one variable causes an increase in the other. Rather, increases in these variables are both most likely due to an increase in temperature.

127. Trend lines and equations may vary. y 80 60 40 20

0

1 2 3 4 5 6 7 8

x

133. B

Using the points (1, 65) and (6, 40), the equation is y 5 25x 1 70.

134. Yes; an increase in the age of the painting causes the value of the painting to increase.

128. Yes; the point (8, 30) lies near or on the trend line.

135. a. y 5 0.21 x2 1 12.07x 1 660; $1788.50 b. y 5 653.44 (1.02)x ; $1758.79

129. D 130. a. There is a positive correlation. As the temperature increases, the number of people at the pool increases. b. Yes; it is reasonable to assume that higher temperatures cause more people to visit the pool. 131. a. y 5 11.05x 2 798.62 b. 30

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SpringBoard Algebra 1, Unit 2 Practice