chapter 2bhhs.bhusd.org/ourpages/auto/2006/8/11/1155323019648/aat... · 2006. 8. 11. · algebra 2...

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51Algebra 2

Worked-Out Solution Key

Prerequisite Skills (p. 70)

1. A linear equation in one variable is an equation that can be written in the form y 5 ax 1 b where a and b are constants and a Þ 0.

2. The absolute value of a real number is the distance the number is from zero on a number line.

3. 22(x 1 1) when x 5 25

22(25 1 1) 5 22(24) 5 8

4. 11x 2 14 when x 5 23

11(23) 2 14 5 233 2 14 5 247

5. x2 1 x 1 1 when x 5 4

42 1 4 1 1 5 16 1 4 1 1 5 21

6. 2x2 2 3x 1 7 when x 5 1

2(12) 2 3(1) 1 7 5 21 2 3 1 7 5 37. 5x 2 2 5 8 8. 26x 2 10 5 20

5x 5 10 26x 5 30

x 5 2 x 5 25

Check: Check:

5x 2 2 5 8 26x 2 10 5 20

5(2) 2 2 0 8 26(25) 2 10 0 20

10 2 2 0 8 30 2 10 0 20

8 5 8 ✓ 20 5 20 ✓

9. 2x 1 9 5 2x 2 27 Check:

9 5 3x 2 27 2x 1 9 5 2x 2 27

36 5 3x 212 1 9 0 2(12) 2 27

12 5 x 23 0 24 2 27

23 5 23 ✓

10. 2x 1 3y 5 6 11. 2x 2 y 5 10

3y 5 6 2 2x 2y 5 10 1 x

y 5 6 2 2x

} 3 y 5 2(10 1 x)

y 5 6 } 3 2

2x } 3 y 5 210 2 x

y 5 2 2 2 } 3 x

12. x 1 4y 5 25

4y 5 25 2 x

y 5 25 2 x

} 4

y 5 2 5 } 4 2

x } 4

y 5 2 5 } 4 2

1 } 4 x

Lesson 2.1

2.1 Guided Practice (pp. 73–76)

1. a. The domain consists of all the x-coordinates: 24, 22, 0, and 1.

The range consists of all the y-coordinates: 24, 22, 1, and 3.

b. x y

24 3

22 24

22 1

0 3

1 22

2422

13

2422

01

Input Output

2. The relation is a function because each input is mapped onto exactly one output.

3. Yes it is still a function because Kevin Garnett’s age is the input and each age is mapped onto exactly one average point.

4. y 5 3x 2 2

x 22 21 0 1 2

y 28 25 22 1 4

1

x

y

21

5. The function f is not linear because it has an x3-term.

f (x) 5 x 2 1 2 x3

f (22) 5 (22) 2 1 2 (22)3 5 23 1 8 5 5

6. The function g is linear because it has the form g(x) 5 mx 1 b.

g(x) 5 24 2 2x

g(22) 5 24 2 2(22) 5 24 1 4 5 0

7. Because the depth varies from 0 feet to 35,800 feet, a reasonable domain is 0 ≤ d ≤ 35,800.

The minimum value of P(d) 5 1, and the maximum value of P(d) is P(35,800) 5 1075. So, a reasonable range is 1 ≤ p (d) ≤ 1075.

2.1 Exercises (pp. 76–79)

Skill Practice

1. In the equation y 5 x 1 5, x is the independent variable and y is the dependent variable.

2. The domain of a set of ordered pairs is all the x-coordinates, and the range is all the y-coordinates.

Chapter 2

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52Algebra 2Worked-Out Solution Key

3. (22, 3), (1, 2), (3, 21), (24, 23)

The domain consists of all the x-coordinates: 24, 22, 1, and 3.


y

21

1

x

2422

13

2321

23

Input Output

4. (5, 22), (23, 22), (3, 3), (21, 21)


The range consists of all the y-coordinates: 22, 21, and 3.

1

x

y

22

22

21

3

2321

35

Input Output

5. (6, 21), (22, 23), (1, 8), (22, 5)

The domain consists of all the x-coordinates: 22, 1, and 6.


y

22

2

x

2321

58

Input Output

22

1

6

6. (27, 4), (2, 25), (1, 22), (23, 6)



2

x

y

22

2522

46

2723

12

Input Output

7. (5, 20), (10, 20), (15, 30), (20, 30)


The range consists of all the y-coordinates: 20 and 30.

10

x

y

25

20

30

5101520

Input Output

8. (4, 22), (4, 2), (16, 24), (16, 4)

The domain consists of all the x-coordinates: 4 and 16.


1

x

y

2

4

16

Input Output

2422

24

9. B; (24, 2), (21, 23), (1, 4), (1, 23), and (2, 1)


10. Yes; The relation is a function because each input is mapped onto exactly one output.


12. No; The relation is not a function because the input 21 is mapped onto both 2 and 21, and the input 5 is mapped onto both 4 and 23.


14. The student incorrectly concludes that the relation is not a function because more than one input is mapped onto the same output.

The relation given by the ordered pairs (24, 2),(21, 5), (3, 6), and (7, 2) is a function because each input is mapped onto exactly one output.

15. The x-values are the inputs and the y-values are the outputs. There should be one value of y for each value of x.

The relation given by the table is not a function because the input 0 is mapped onto both 5 and 9, and input 1 is mapped onto both 6 and 8.

16. The relation given by the ordered pairs (3, 22), (0, 1), (1, 0), (22, 21), (2, 21) is a function because each imput is mapped onto exactly one output.

17. The relation given by the ordered pairs (2, 25), (22, 5), (21, 4), (22, 0), and (3, 24) is not a function because the input 22 is mapped onto both 0 and 5.

18. The relation given by the ordered pairs (0, 1), (1, 0), (2, 3), (3, 2), and (4, 4) is a function because each input is mapped onto exactly one output.


20. B; x 26 22 1 4 6

y 3 4 5 0 3

Of the ordered pairs to choose from, (6, 3) is the one possible option to make a new function because the input 6 is the only input not in the previous ordered pairs.

Chapter 2, continued

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53Algebra 2


21.

x

y

2

1

22.

x

y

1

2

The relation is a function. The relation is a function.

23.

x

y

1

1

The relation is not a function.

24. If a vertical line intersects the graph more than once, it means that for one x-value there is more than one y-value. Because x is the input variable, there must be only one y-value for each x-value for the relation to be a function.

25. y 5 x 1 2 y

21

1

x

x 22 21 0 1 2

y 0 1 2 3 4

26. y 5 2x 1 5

1x

y

21

x 22 21 0 1 2

y 7 6 5 4 3

27. y 5 3x 1 1 y

21

1

x

x 22 21 0 1 2

y 25 22 1 4 7

28. y 5 5x 2 3

1

x

y

21

x 22 21 0 1 2

y 213 28 23 2 7

29. y 5 2x 2 7 1

x

y

21

x 22 21 0 1 2

y 211 29 27 25 23

30. y 5 23x 1 2

1

x

y

21

x 22 21 0 1 2

y 8 5 2 21 24

31. y 5 22x 2

x

y

21

x 22 21 0 1 2

y 4 2 0 22 24

32. y 5 1 } 2 x 1 2

1

x

y

21

x 22 21 0 1 2

y 1 3 }

2 2

5 }

2 3

33. y 5 2 3 } 4 x 2 1

1

x

y

22

x 22 21 0 1 2

y 1 }

2 2

1 } 4 21 2

7 } 4 2

5 } 2

34. The function f is linear because it has the form f (x) 5 mx 1 b.

f (x) 5 x 1 15; f (8)

f (8) 5 8 1 15 5 23

35. The function f is not linear because it has an x2-term.

f (x) 5 x2 1 1; f (23)

f (23) 5 (23)2 1 1 5 9 1 1 5 10

36. The function f is not linear because it has an x-term. f (x) 5 x 1 10; f (24) f (24) 5 24 1 10 5 4 1 10 5 14 37. The function f is linear because it has the form

f (x) 5 mx 1 b, where m 5 0 and b 5 6.

f (x) 5 0x 1 6; f (2)

f (2) 5 0(2) 1 6 5 0 1 6 5 6

38. The function g is not linear because is has an x3- and x2-term.

g(x) 5 x3 2 2x2 1 5x 2 8; g(25)

g(25) 5 (25)3 2 2(25)2 1 5(25) 2 8

5 2125 2 50 2 25 2 8 5 2208

39. The function h is linear because it has the form h(x) 5 mx 1 b.

h(x) 5 7 2 2 } 3 x; h(15)

h(15) 5 7 2 2 } 3 (15) 5 7 2 10 5 23

40. The equation y 5 x is a function because each input has exactly one output.


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41. f (2a) 5 f (a 1 a) 5 f (a) 1 f (a) 5 2 p f (a)

f (0) 5 f (0 1 0) 5 f (0) 1 f (0) 5 2f (0)

2f (0) 2 f (0) 5 0

f (0) 5 0

Problem Solving

42. The ordered pairs do not represent a function. The x-values of 24, 25, and 26 each have two outputs.

43. The ordered pairs represent a function. For each x-value there is exactly one output.

44. V(s) 5 s3;

V(4) 5 43 5 64 units3

V(4) represents the volume of a cube when the length of its side is 4 units.

45. V(r) 5 4 } 3 π r

3

V(6) 5 4 } 3 π (6

3) 5 4 } 3 π(216) 5 288π ø 904.8 units3

V(6) represents the volume of a sphere with a radius of 6 units.

46. For 1974 t 5 0 and for 2004 t 5 30. Because t varies from 0 to 30, a reasonable domain is 0 ≤ t ≤ 30.

The minimum value of p(t) is p(0) 5 1.89, and the maximum value of p(t) is p(30) 5 6.21. So, a reasonable range is 1.89 ≤ p(t) ≤ 6.21.

The meaning of the range is that from 1974 to 2004 the average price of a theater ticket ranged from $1.89 to $6.21.

47. a. The graph of h(l) is

Hei

gh

t (i

nch

es)

0150 17 19 21 23

58

60

62

64

66

74

72

70

68

Length (inches)

l

h(l)

shown. Because the length varies from 15 inches to 24 inches, a reasonable domain is 15 ≤ l ≤ 24.

The minimum value of h(l) is h(15) 5 57.95, and the maximum value of h(l) is h(24) 5 75.5. So, a reasonable range is 57.95 ≤ h(l) ≤ 75.5.

b. At a length of 15.5 inches the height of the adult female was h(15.5) 5 1.95(15.5) 1 28.7 ø 58.9 inches, which you can verify by the graph.

c. 5 feet 11 inches 5 71 inches

At a height of 71 inches the length of the femur is

71 5 1.95l 1 28.7

42.3 5 1.95l

21.7 ø l The femur is about 21.7 inches long.

48. The time to get to

Ele

vati

on

(fe

et)

010 2 3 4

5000

6000

7000

9000

10,000

8000

Time (hours)

t

h(t)

Camp Muir is:

h(t) 5 1000t 1 5400

10,100 5 1000t 1 5400

4700 5 1000t

4.7 5 t

The graph of h(t) is shown. Because the time varies from 0 hours to 4.7 hours, a reasonable domain is 0 ≤ t ≤ 4.7.

The minimum value of h(t) is h(0) 5 5400, and the maximum value of h(t) is h(4.7) 5 10,100. So a reasonable range is 5400 ≤ h(t) ≤ 10,100.

At the time of 3.5 hours the elevation of the climber is h(3.5) 5 1000(3.5) 1 5400 5 8900 feet, which you can verify from the graph.

49. a. domain: 11,350,000; 12,280,000; 12,420,000; 15,980,000; 18,980,000; 20,850,000; 33,870,000

range: 20, 21, 27, 31, 34, 55

b. Yes, each input has exactly one output.

c. No, input 21 has more than one output.

50. a. Yes, it is a function, because each merchandise cost is mapped onto exactly one shipping cost.

b. No, it is not a function, because each shipping cost value is mapped onto a range of merchandise cost values.

Mixed Review

51. y 2 3

} x 2 4

when x 5 6 and y 5 2

2 2 3

} 6 2 4

5 21

} 2 5 2 1 } 2

52. y 2 8

} x 2 2


4 2 8

} 3 2 2

5 24

} 1 5 24

53. y 2 (25)

} x 2 1


23 2 (25)

} 23 2 1

5 23 1 5

} 23 2 1 5

2 }

24 5 2 1 } 2

54. 24 2 y

} 15 2 x


24 2 8

} 15 2 (217)

5 24 2 8

} 15 1 17 5 16

} 32 5 1 } 2

55. 3x 1 16 5 31 56. 24x 2 7 5 17

3x 5 15 24x 5 24

x 5 5 x 5 26

Check: Check:

3x 1 16 5 31 24x 2 7 5 17

3(5) 1 16 0 31 24(26) 2 7 0 17

15 1 16 0 31 24 2 7 0 17

31 5 31 ✓ 17 5 17 ✓


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55Algebra 2


57. 5x 1 12 5 23x 2 4 Check:

8x 1 12 5 24 5x 1 12 5 23x 2 4

8x 5 216 5(22) 1 12 0 23(22) 2 4

x 5 22 210 1 12 0 6 2 4

2 5 2 ✓

58. 5 2 8z 5 25 1 4z Check:

5 5 25 1 12z 5 2 8z 5 25 1 4z

220 5 12z 5 2 8 1 2 5 } 3 2 0 25 1 4 1 2 5 } 3 2

2 5 } 3 5 z 5 1

40 } 3 0 25 2

20 } 3

55

} 3 5

55 } 3 ✓

59. 5 }

2 (3v 2 4) 5 30 Check:

15

} 2 v 2 10 5 30

5 }

2 (3v 2 4) 5 30

15

} 2 v 5 40

5 }

2 1 3 1 16 } 3 2 2 4 2 0 30

v 5 40 1 2 } 15 2 5 }

2 (16 2 4) 0 30

v 5 16

} 3 5 }

2 (12) 0 30

30 5 30 ✓

60. 6(4w 1 1) 5 1.5(8w 1 18)

24w 1 6 5 12w 1 27

12w 1 6 5 27

12w 5 21

w 5 7 } 4

Check:

6(4w 1 1) 5 1.5(8w 1 18)

6 1 4 1 7 } 4 2 1 1 2 0 1.5 1 8 1 7 } 4 2 1 18 2 6 (7 1 1) 0 1.5(14 1 18)

6(8) 0 1.5(32)

48 5 48 ✓

61. 2x 2 6 > 8 121064 8

2x > 14

x > 7

62. 1 }

4 x 1 7 > 0

222224226228230

1 }

4 x > 27

x > 27 1 4 } 1 2 x > 228

63. 15 2 2x ≤ 7 86420

22x ≤ 28 x ≥ 4 64. 4 2 x < 3

8640 222 2x < 21

x > 1

65. 27 < 6x 2 1 < 5

27 1 1 < 6x 2 1 1 1 < 5 1 1

26 < 6x < 6

21 < x < 1

22 21 210

66. x 2 2 ≤ 1 or 4x 1 3 ≥ 19 x ≤ 3 or 4x ≥ 16 x ≥ 4

8640 222

2.1 Extension (p. 81)

1. y 5 2x 1 3; domain: 22, 21, 0, 1, 2

x 22 21 0 1 2

y 21 1 3 5 7

y

21

1

x

The graph consists of separate points, so the function is discrete. Its range is 21, 1, 3, 5, and 7.

2. f (x) 5 0.5x 2 4; domain: 24, 22, 0, 2, 4

x 24 22 0 2 4

y 26 25 24 23 22

1

x

y

21

The graph consists of separate points, so the function is discrete. Its range is 26, 25, 24, 23, and 22.

3. y 5 23x 1 9; domain: x < 5

x 21 0 1 2 5

y 12 9 6 3 26

y

21

1

x

The graph is unbroken, so the function is continuous. Its range is y > 26.


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4. f (x) 5 1 } 3 x 1 6; domain: x ≥ 26

x 26 23 0 3 6

y 4 5 6 7 8

1

x

y

21

The graph is unbroken, so the function is continuous. Its range is y ≥ 4.

5. The function is

Dis

tan

ce (

mile

s)

010 2 3 4

2

6

10

14

Hours

x

d(x)

d(x) 5 3.5x. Amandacan walk any nonnegative amount of time, so the domain is x ≥ 0. From the graph you can see that the range is y ≥ 0. The graph is unbroken, so the function is continuous.

6. The function is

Co

st (

do

llars

)

020 4 6 8

2

4

6

10

8

Number of rides

x

s(x)

s(x) 5 1.25x. The fi rst nine points of the graph s(x) are shown. The number of times the subway is ridden must be a whole number, so the domain is the set of whole numbers 0, 1, 2, 3, . . . From the graph, you can see that the range is 0, 1.25, 2.50, 3.75, 5.00, . . . The graph consists of separate points, so the function is discrete.

7. The function is

Gal

lon

s o

f m

ilk

010 32 5 64 7 8 9 10

3

6

9

12

15

27

30

24

21

18

Weeks

x

m(x)

m(x) 5 3x. The fi rst eleven points of the graph m(x) are shown. Because milk is only deliveredonce a week, only whole numbers canbe used for each week. The domain isthe set of whole numbers 0, 1, 2, 3, . . . From the graph, you can see that the range is 0, 3, 6, 9, . . . The graph consists of separate points, so the function is discrete.

8. The function is

Wei

gh

t (p

ou

nd

s)

0100 20 30 40

4

8

12

Length of cable (feet)

x

f(x)

w (x) 5 0.24x. The steel cable can be any nonnegative amount of length in feet, so the domain is x ≥ 0. From the graph, you can see that the range is y ≥ 0. The graph is unbroken, so the function is continuous.

9. The function is d(x) 5 x 2 3.

1

x

y

21

x can be any real number on a number line, so the domain is the set of all real numbers. From the graph, you can seethat the range is the set of all real numbers. The graph is unbroken, so the function is continuous.

Lesson 2.2


1. The skateboard ramp has a rise of 12 inches and a run of 54 inches.

Slope 5 rise

} run 5 12

} 54 5 2 } 9

The slope of the ramp is 2 }

9 .

2. D;

Let (x1, y1)5 (24, 9) and (x2, y2) 5 (28, 3).

m 5 y2 2 y1

} x2 2 x1 5

3 2 9 }

28 2 (24) 5

26 }

24 = 3 }

2

3. Let (x1, y1)5 (0, 3) and (x2, y2) 5 (4, 8).

m 5 y2 2 y1

} x2 2 x1 5

8 2 3 } 4 2 0 5

5 } 4

4. Let (x1, y1) 5 (25, 1) and (x2, y2) 5 (5, 24).

m 5 y2 2 y1

} x2 2 x1 5

24 2 1 }

5 2 (25) 5

25 } 10 5 2

1 } 2

5. Let (x1, y1) 5 (23, 22) and (x2, y2) 5 (6, 1).

m 5 y2 2 y1

} x2 2 x1 5

1 2 (22) }

6 2 (23) 5

3 } 9 5

1 } 3

6. Let (x1, y1) 5 (7, 3) and (x2, y2) 5 (21, 7).

m 5 y2 2 y1

} x2 2 x1 5

7 2 3 }

21 2 7 5 4 }

28 5 2 1 } 2

7. (24, 3), (2, 26)

m 5 26 2 3

} 2 2 (24)

5 29

} 6 5 2 3 } 2

Because m < 0, the line falls.

8. (7, 1), (7, 21)

m 5 21 2 1

} 7 2 7 5 22

} 0

Because m is undefi ned, the line is vertical.

9. (3, 22), (5, 22)

m 5 22 2 (22)

} 5 2 3 5 0 } 2 5 0

Because m 5 0, the line is horizontal.

10. (5, 6), (1, 24)

m 5 24 2 6

} 1 2 5 5 210

} 24 5

5 } 2

Because m > 0, the line rises.


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57Algebra 2


11. Line 1: through (22, 8) and (2, 24)

Line 2: through (25, 1) and (22, 2)

m1 5 24 2 8

} 2 2 (22)

5 212

} 4 5 23

m2 5 2 2 1 }

22 2 (25) 5

1 } 3

Because m1m2 5 23 p 1 }

3 5 21, m1 and m2 are negative

reciprocals of each other. So, the lines are perpendicular.

12. Line 1: through (24, 22) and (1, 7)

Line 2: through (21, 24) and( 3, 5)

m1 5 7 2 (22)

} 1 2 (24)

5 9 } 5

m2 5 5 2 (24)

} 3 2 (21)

5 9 } 4

Because m1m2 5 9 } 5 p

9 }

4 Þ 21 and m1 Þ m2, the lines are

neither perpendicular nor parallel.

13. Average rate of change 5 change in diameter

}} change in time

5 251 in. 2 248 in.

}} 2005 2 1965

5 3 in.

} 40 years

5 0.075 inch per year

Find the number of years from 2005 to 2105, multiply this number by the average rate of change to fi nd the total increase in diameter during the period 2005 - 2105.Number of years 5 2105 2 2005 5 100

Increase in diameter 5 (100 years) 1 0.075 inch } year 2 5 7.5 inches. In 2105, the diameter of the sequoia will be

about 251 1 7.5 5 258.5 inches.


Skill Practice

1. The slope m of a nonvertical line is the ratio of vertical change to horizontal change.

2. If the slopes of both lines are equal then the two lines are parallel. If the product of the slopes of the lines equals 21 then the slopes are negative reciprocals of each other and the lines are perpendicular.

3. m 5 21 2 (24)

} 4 2 2 5 3 } 2


4. m 5 3 2 9

} 24 2 8 5

26 }

212 5 1 } 2


5. m 5 24 2 1

} 8 2 5 5 25

} 3 5 2 5 } 3


6. m 5 22 2 (22)

} 3 2 (23)

5 0 } 6 5 0


7. m 5 24 2 4

} 1 2 (21)

5 28

} 2 5 24


8. m 5 25 2 5

} 26 2 (26)

5 210

} 0


9. m 5 3 2 (24)

} 21 2 (25)

5 7 } 4


10. m 5 3 2 6 }

27 2 (23) 5

23 }

24 5 3 } 4


11. m 5 9 2 4

} 4 2 4 5 5 } 0


12. m 5 3 2 5

} 7 2 5 5 22

} 2 5 21


13. m 5 23 2 (23)

} 4 2 0 5 0 } 4 5 0


14. m 5 24 2 (21)

} 21 2 1 5

23 }

22 5 3 } 2


15. The error is found in the run of the slope. The run should be x2 2 x1.

(24, 23), (2, 21)

m 5 21 2 (23)

} 2 2 (24)

5 2 } 6 5

1 } 3

16. The error is found in the rise and the run. The rise should be y2 2 y1, and the run should be x2 2 x1.

m 5 1 2 4

} 5 2 (21)

5 23

} 6 5 2 1 } 2

17. A; m 5 1 2 (24)

} 5 2 2 5 5 } 3


18. Line 1: through (3, 21) and (6, 24)


m1 5 24 2 (21)

} 6 2 3 5 23

} 3 5 21

m2 5 7 2 5 }

22 2 (24) 5

2 } 2 5 1

Because m1m2 5 21 p 1 5 21, m1 and m2 are negative reciprocals of each other. So, the lines are perpendicular.

19. Line 1: through (1, 5) and (3, 22)


m1 5 22 2 5

} 3 2 1 5 27

} 2

m2 5 0 2 2

} 4 2 (23)

5 22

} 7

Because m1m2 5 27

} 2 p 22

} 7 5 1 Þ 21 and m1 Þ m2,the lines are neither perpendicular nor parallel.


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20. Line 1: through (21, 4) and (2, 5)


m1 5 5 2 4

} 2 2 (21)

5 1 } 3

m2 5 4 2 2

} 0 2 (26)

5 2 } 6 5

1 } 3

Because m1 5 m2 (and the lines are different), you can conclude that the lines are parallel.

21. Line 1: through (5, 8) and (7, 2)

Line 2: through (27, 22) and (24, 21)

m1 5 2 2 8

} 7 2 5 5 26

} 2 5 23

m2 5 21 2 (22)

} 24 2 (27)

5 1 } 3

Because m1m2 5 23 p 1 }

3 5 21, m1 and m2 are negative


22. Line 1: through (23, 2) and (5, 0)

Line 2: through (21, 24) and (3, 23)

m1 5 0 2 2

} 5 2 (23)

5 22

} 8 5 2 1 } 4

m 2 5 23 2 (24)

} 3 2 (21)

5 1 } 4

Because m1m2 5 2 1 } 4 p

1 }

4 5 2

1 } 16 Þ 21 and m1 Þ m2 ,

the lines are neither perpendicular nor parallel.

23. Line 1: through (1, 24) and (4, 22)


m1 5 22 2 (24)

} 4 2 1 5 2 } 3

m2 5 5 2 1

} 14 2 8 5 4 } 6 5

2 } 3

Because m1 5 m2 (and the lines are different), you can conclude that the lines are parallel.

24. (2, 12), (5, 30) x is measured in hours and y is measured in dollars.

Average rate of change 5 change in dollars

}} change in hours

5 30 dollars 2 12 dollars

}} 5 hours 2 2 hours

5 18 dollars

} 3 hours

5 6 dollars per hour

25. (0, 11), (3, 50) x is measured in gallons and y is measured in miles.

Average rate of change 5 change in miles

}} change in gallons

5 50 miles 2 11 miles

}} 3 gallons 2 0 gallons

5 39 miles

} 3 gallons

5 13 miles per gallon

26. (3, 10), (5, 18) x is measured in seconds and y is measured in feet.

Average rate of change 5 change in feet

}} change in seconds

5 18 feet 2 10 feet

}} 5 seconds 2 3 seconds

5 8 feet

} 2 seconds

5 4 feet per second

27. (1, 8), (7, 20) x is measured in seconds and y is measured in meters.

Average rate of change 5 change in meters

}} change in seconds

5 20 meters 2 8 meters

}} 7 seconds 2 1 second

5 12 meters

} 6 seconds

5 2 meters per second

28. If lines l1 and/or l2 are vertical, their slopes would be undefi ned.

29. Let (x1, y1) 5 1 21, 3 } 2 2 and (x2, y2) 5 1 0, 7 }

2 2 .

m 5 7 } 2 2

3 } 2 }

0 2 (21) 5

4 } 2 } 1 5 2

30. Let (x1, y1) 5 1 2 3 } 4 , 22 2 and (x2, y2) 5 1 5 } 4 , 23 2 .

m 5 23 2 (22)

} 5 }

4 2 1 2 3 } 4 2

5 21

} 8 }

4 5 2

1 } 2

31. Let (x1, y1) 5 1 2 1 } 2 , 5 }

2 2 and (x2, y2) 5 1 5 } 2 , 3 2 .

m 5 3 2

5 } 2 }

5 }

2 2 1 2 1 } 2 2

5 1 } 2 }

6 }

2 5

1 } 6

32. Let (x1, y1) 5 (24.2, 0.1) and (x2, y2) 5 (23.2, 0.1).

m 5 0.1 2 0.1

}} 23.2 2 (24.2)

5 0 } 1 5 0

33. Let (x1, y1) 5 (20.3, 2.2) and (x2, y2) 5 (1.7, 20.8).

m 5 20.8 2 2.2

} 1.7 2 (20.3)

5 23

} 2 5 2 3 } 2

34. Let (x1, y1) 5 (3.5, 22) and (x2, y2) 5 (4.5, 0.5).

m 5 0.5 2 (22)

} 4.5 2 (3.5)

5 2.5

} 1 5 2.5

35. No; no

When fi nding the slope of a line, it does not matter which points are picked on the line.

Sample answer:

P (23, 2) to R (1, 0) m 5 0 2 2

} 1 2 (23)

5 22

} 4 5 2 1 } 2

Q(21, 1) to T(5, 22) m 5 22 2 1

} 5 2 (21)

5 23

} 6 5 2 1 } 2

R (1, 0) to S (3, 21) m 5 21 2 0

} 3 2 1 5 21

} 2 5 2 1 } 2


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59Algebra 2


It also does not make a difference which point is 1 x1 y1 2 and which point is 1 x2 y2 2 .

from before

P (23, 2) to R (1, 0) m 5 2 1 } 2

Q(21, 1) to T(5, 22) m 5 2 1 } 2

R (1, 0) to S (3, 21) m 5 2 1 } 2

reverse

R (1, 0) to P (23, 2) m 5 2 2 0

} 23 2 1 5

2 }

24 5 2 1 } 2

T(5, 22) to Q(21, 1) m 5 1 2 (22)

} 21 2 5 5

3 }

26 5 2 1 } 2

S (3, 21) to R(1, 0) m 5 0 2 (21)

} 1 2 3 5 1 }

22 5 2 1 } 2

36. Sample answer:

x

y

212

(0, 3)

(1, 21)1

24

The points (21, 7) and (1, 21) also lie on the line.

37. (2, 23) and (k, 7); m 5 22

m 5 7 2 (23)

} k 2 2

5 10 }

k 2 2 5 22

10

} 22

5 k 2 2

25 5 k 2 2

23 5 k

Check: m 5 7 2 (23)

} 23 2 2 5

10 }

25 5 22

38. (0, k) and (3, 4); m 5 1

m 5 4 2 k

} 3 2 0 5 4 2 k

} 3 5 1

4 2 k 5 3

k 5 1

Check: m 5 4 2 1

} 3 2 0 5 3 } 3 5 1

39. (24, 2k) and (k, 25); m 5 21

m 5 25 2 2k

} k 2 (24)

5 25 2 2k

} k 1 4

5 21

25 2 2k 5 2k 2 4

25 5 k 2 4

21 5 k

Check: m 5 25 2 2(21)

} 21 2 (24)

5 25 1 2

} 21 1 4 5

23 } 3 5 21

40. (22, k) and (2k, 2); m 5 20.25

m 5 2 2 k

} 2k 2 (22)

5 2 2 k

} 2k 1 2

5 20.25

2 2 k 5 20.25(2k 1 2)

2 2 k 5 20.5k 2 0.5

2 5 0.5k 2 0.5

2.5 5 0.5k

5 5 k

Check: m 5 2 2 5 }

2(5) 2 (22) 5

23 } 10 1 2 5

23 } 12 5 20.25

Problem Solving

41. rise 5 28 ft

run 5 48 ft

Slope 5 rise

} run 5 28

} 48 5 7 } 12

42.

rise 5 400 ft

run 5 685 ft

Slope 5 rise

} run 5 400

} 685 5 80

} 137

43. rise 5 195 ftrun 5 3000 ft

Slope 5 rise

} run 5 195

} 3000 5 13

} 200

grade 5 13

} 200 5 0.065 5 6.5%

44. The rise of each ramp of the three-section ramp is

5.25 ft

} 3 5 1.75 ft. The slope of each ramp of the three-

section ramp is: slope 5 rise

} run 5 1.75

} 28 5 1 } 16 .

If there were only a single-section ramp, the slope would

be: slope 5 rise

} run 5 3(1.75)

} 28 5 5.25

} 28 5 3 } 16 .

The slope of a single-section ramp would be 3 times as steep as the three-section ramp. The benefi t of the three-section ramp is that people can walk at a slope that doesn’t require so much work. It is also easier for those who use a wheelchair.

45. A;

Average rate of change 5 change in gallons

}} change in days

5 214 gallons 2 400 gallons

}} 30 days 2 0 days

5 2186 gallons

} 30 days

5 26.2 gallons per day


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46. Average rate of change 5 change in centimeters

}} change in years

5 15.5 cm 2 11.9 cm

}} 110 years 2 30 years

5 3.6 cm

} 80 years

5 0.045 centimeters per year

47. a. slope 5 rise

} run 5 15 ft

} 40 ft

5 3 } 8

b. minimum pitch 5 3 ft

} 12 ft

5 1 } 4

Because 3 }

8 >

1 }

4 , the roof satisfi es the building code.

c. Let r represent the minimum rise which satisfi es the

code, then r }

40 5

3 }

12 → r 5 10. Therefore the rise

exceeds the code minimum 15 2 10 5 5 ft.

48. a. Let x represent the horizontal distance you cover when descending the slide.

slope 5 rise

} run 5 80

} x 5 1 } 3

x 5 80(3) 5 240

You cover a horizontal distance of 240 feet when descending the slide.

b. rise 5 80 ft

run 5 240 ft

a2 1 b2 5 c2

802 1 2402 5 c2

6400 1 57,600 5 c2

64,000 5 c2

c ø 253 The slide is about 253 feet long.

c. The slope will be increased because the run will be decreased and the vertical distance remains the same.

49. Average highway rate 5 36 miles per gallon 5 36 miles

} 1 gallon

Average city rate 5 24 miles per gallon

5 24 miles

} 1 gallon

5 36 miles

} 1.5 gallons

36 miles 1 36 miles

}} 1 gallon 1 1.5 gallons

5 72 miles

} 2.5 gallons

5 28.8 miles per gallon

Mixed Review

50. 5(8 1 12) 5 5(8) 1 5(12)

Distributive Property

51. (7 1 9) 1 13 5 7 1 (9 1 13)

Associative Property of Addition

52. 4 1 (24) 5 0

Inverse Property of Addition

53. 5 p 10 5 10 p 5

Commutative Property of Multiplication

54. 15 p 1 } 15

5 1

Inverse Property of Multiplication

55. 2 }

3 p 1 5 2 } 3

Identity Property of Multiplication

56. 3x 1 y 5 7 57. 2x 2 y 5 3

y 5 7 2 3x 2y 5 3 2 2x

y 5 2x 2 3

58. y 2 4x 5 26 59. 2x 1 3y 5 212

y 5 4x 2 6 3y 5 22x 2 12

y 5 2 2 } 3 x 2 4

60. 7x 2 4y 5 10 61. 2x 1 2y 5 9

24y 5 10 2 7x 2y 5 x 1 9

y 5 10

} 24 1

7 } 4 x y 5

1 } 2 x 1

9 } 2

y 5 7 }

4 x 2

5 } 2

62. 5 1 2x 5 7 5 1 2x 5 7 or 5 1 2x 5 27

2x 5 2 or 2x 5 212

x 5 1 or x 5 26

63. 4x 2 9 5 5 4x 2 9 5 5 or 4x 2 9 5 25

4x 5 14 or 4x 5 4

x 5 7 } 2 or x 5 1

64. 6 2 5x 5 9 6 2 5x 5 9 or 6 2 5x 5 29

25x 5 3 or 25x 5 215

x 5 2 3 } 5 or x 5 3

65. 3 2 7x < 10 210 < 3 2 7x < 10

213 < 27x < 7

13

} 7 > x > 21

66. 3x 1 1 > 25 3x 1 1 > 25 or 3x 1 1 < 225

3x > 24 or 3x < 226

x > 8 or x < 2 26

} 3

67. 3 2 4x ≥ 7 3 2 4x ≥ 7 or 3 2 4x ≤ 27 24x ≥ 4 or 24x ≤ 210

x ≤ 21 or x ≥ 5 } 2


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61Algebra 2


Lesson 2.3


1.

2

x

y

21

The graphs of y 5 22x and y 5 x both have a y-intercept of 0, but the graph of y 5 22x has a slope of 22 insteadof 1.

2.

1

x

y

21

The graphs of y 5 x 2 2 and y 5 x both have a slope of 1, but the graph of y 5 x 2 2 has a y-intercept of 22 instead of 0.

3.

1

x

y

21

The graphs of y 5 4x and y 5 x both have a y-intercept of 0, but the graph of y 5 4x has a slope of 4 instead of 1.

4.

1

x

y

21

5.

1

x

y

21

6. 1

x

y

21

7.

1

x

y

21

8.

1

x

y

21

9.

2

x

y

21

10. Step 1: Graph the equation y 5 6x 1 48.

Bo

dy

len

gth

(in

.)

040 8 12 16

30

60

90

150

120

Age (months)

x

y

Step 2: Interpret the slope and y-intercept. The slope 6 represents the calf’s rate of growth in inches per month. The y-intercept 48 represents a newborn calf’s body length in inches.

Step 3: Estimate the body length of the calf at age 10 months by starting at 10 on the x-axis and moving up until you reach the graph. Then move left to the y-axis. At age 10 months, the body length of the calf is about 108 inches.

11. 2x 1 5y 5 10

x-intercept: 2x 1 5(0) 5 10

x 5 5

The x-intercept is 5. So, plot the point (5, 0).

y-intercept: 2(0) 1 5y 5 10

y 5 2

The y-intercept is 2. So, plot the point (0, 2).

1x

y

21

12. 3x 2 2y 5 12


x 5 4

The x-intercept is 4. So, plot the point (4, 0).


y 5 26

The y-intercept is 26. So, plot the point (0, 26).

1

x

y

21


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13. x 5 1

The graph of x 5 1 is the vertical line that passes through the point (1, 0).

1

x

y

21

14. y 5 24

The graph of y 5 24 is the horizontal line that passes through the point (0, 24).

1

x

y

21


Skill Practice

1. The linear equation y 5 2x 1 5 is written in slope-intercept form.

2. Identify the x-intercept: let y 5 0, solve for x, then plot the point.

Identify the y-intercept: let x 5 0, solve for y, then plot the point.

Draw a line through the two points.

3. y

21

1

x

The graphs of y 5 3x and y 5 x both have a y-intercept of 0, but the graph of y 5 3x has a slope of 3 instead of 1.

4.

1

x

y

21

The graphs of y 5 2x and y 5 x both have a y-intercept of 0, but the graph of y 5 x has a slope of 21 instead of 1.

5. y

21

1

x


6.

1

x

y

21


7.

1

x21

f(x)

The graph of y 5 2x 2 1 has a slope of 2 and a y-intercept of 21 instead of having a slope of 1 and a y-intercept of 0.

8.

1

x21

f(x)

The graph of y 5 23x 1 2 has a slope of 23 and a y-intercept of 2 instead of having a slope of 1 and a y-intercept of 0.

9. y 5 2x 2 3

The y-intercept is 23, so plot the point (0, 23).

The slope is 21, so plot a second point by starting at (0, 23) and then moving down 1 unit and right 1 unit. The second point is (1, 24).

y

21

1

x

10. y 5 x 2 6


The slope is 1, so plot a second point by starting at (0, 26) and then moving up 1 unit and right 1 unit. The second point is (1, 25).

1

x

y

21


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63Algebra 2


11. y 5 2x 1 6


The slope is 2, so plot a second point by starting at (0, 6) and then moving up 2 units and right 1 unit. The second point is (1, 8).

y

21

1

x

12. y 5 3x 2 4


The slope is 3, so plot a second point by starting at (0, 24) and then moving up 3 units and right 1 unit. The second point is (1, 21).

1

x

y

21

13. y 5 4x 2 1


The slope is 4, so plot a second point on the line by starting at (0, 21) and then moving up 4 units and right 1 unit. The second point is (1, 3).

1

x

y

21

14. y 5 2 } 3 x 2 2


The slope is 2 }

3 , so plot a second point by starting

at (0, 22) and then moving up 2 units and right 3 units. The second point is (3, 0).

1

x

y

21

15. f (x) 5 2 1 } 2 x 2 1


The slope is 2 1 } 2 , so plot a second point by starting at

(0, 21) and then moving down 1 unit and right 2 units. The second point is (2, 22).

1

x

y

22

16. f (x) 5 2 5 } 4 x 1 1


The slope is 2 5 } 4 , so plot a second point by starting at

(0, 1) and then moving down 5 units and right 4 units. The second point is (4, 24).

1

x

y

21

17. f (x) 5 3 } 2 x 2 3


The slope is 3 }

2 , so plot a second point by starting at

(0, 23) and then moving up 3 units and right 2 units. The second point is (2, 0).

1

x

y

21

18. f (x) 5 5 } 3 x 1 4


The slope is 5 }

3 , so plot a second point by starting at (0, 4)

and then moving up 5 units and right 3 units. The second point is (3, 9).

1

x

y

21


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19. f (x) 5 21.5x 1 2


The slope is 21.5, so plot a second point by starting at (0, 2) and then moving down 1.5 units and right 1 unit. The second point is (1, 0.5).

1

x

y

21

20. f (x) 5 3x 2 1.5

The y-intercept is 21.5, so plot the point (0, 21.5).

The slope is 3, so plot a second point by starting at (0, 21.5) and then moving up 3 units and right 1 unit. The second point is (1, 1.5).

1

x

y

21

21. The y-intercept was incorrectly plotted at (0, 2) instead of (0, 3), and the slope was incorrectly graphed as 3 instead of 2.

y

1

1

x

22. The slope was incorrectly graphed as 1 }

4 instead of 4.

1

x

y

21

23. C; 4x 2 3y 5 18

23y 5 18 2 4x

y 5 4 } 3 x 2 6

24. x 2 y 5 4

x-intercept: x 2 (0) 5 4

x 5 4

The x-intercept is 4.

y-intercept: (0) 2 y 5 4

y 5 24

The y-intercept is 24.

25. x 1 5y 5 215

x-intercept: x 1 5(0) 5 215

x 5 215


y-intercept: 0 1 5y 5 215

y 5 23


26. 3x 2 4y 5 212

x-intercept: 3x 2 4(0) 5 212

x 5 24


y-intercept: 3(0) 2 4y 5 212

y 5 3

y-intercept is 3.

27. 2x 2 y 5 10

x-intercept: 2x 2 0 5 10

x 5 5


y-intercept: 2(0) 2 y 5 10

y 5 210


28. 4x 2 5y 5 20


x 5 5



y 5 24


29. 26x 1 8y 5 236

x-intercept: 26x 1 8(0) 5 236

x 5 6


y-intercept: 26(0) 1 8y 5 236

y 5 24.5

The y-intercept is 24.5.

30. C;

5x 2 6y 5 30

5x 2 6(0) 5 30

x 5 6


31. x 1 4y 5 8 y

22

1

x

(0, 2)

(8, 0)

x-intercept: x 1 4(0) 5 8

x 5 8

y-intercept: 0 1 4y 5 8

y 5 2


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65Algebra 2


32. 2x 2 6y 5 212

1

x

y

21(26, 0)

(0, 2) x-intercept: 2x 2 6(0) 5 212

x 5 26

y-intercept: 2(0) 2 6y 5 212

y 5 2

33. The graph of x 5 4 is the vertical line that passes through the point (4, 0).

y

21

1

x

(4, 0)

34. The graph of y 5 22 is the horizontal line that passes through the point (0, 22).

(0, 22)

1

x

y

21

35. 5x 2 y 5 3

(0, 23)

1

x

y

21 ( , 0)35 x-intercept: 5x 2 0 5 3

x 5 3 } 5

y-intercept: 5(0) 2 y 5 3

y 5 23

36. 3x 1 4y 5 12

(4, 0)

(0, 3)

1

x

y

21


x 5 4


y 5 3

37. 25x 1 10y 5 20

1

x

y

1

(0, 2)

(24, 0)

x-intercept: 25x 1 10(0) 5 20

x 5 24

y-intercept: 25(0) 1 10y 5 20

y 5 2

38. 2x 2 y 5 6

(0, 26)

(26, 0)1

x

y

21 x-intercept: 2x 2 0 5 6

x 5 26

y-intercept: 20 2 y 5 6

y 5 6

39. The graph of y 5 1.5 is the horizontal line that passes through the point (0, 1.5).

(0, 1.5)1

x

y

21

40. 2.5x 2 5y 5 215

1

x

y

21

(0, 3)

(26, 0)

x-intercept: 2.5x 2 5(0) 5 215

x 5 26

y-intercept: 2.5(0) 2 5y 5 215

y 5 3

41. The graph of x 5 2 5 } 2 is the vertical line that passes

through the point 1 2 5 } 2 , 0 2 .

1

x

y

21(2 , 0)52

42. 1 }

2 x 1 2y 5 22

1

x

y

24 (0, 21)(24, 0)

x-intercept: 1 }

2 x 1 2(0) 5 22

x 5 24

y-intercept: 1 }

2 (0) 1 2y 5 22

y 5 21

43. 6y 5 3x 1 6 y

21

2

x

6y 2 3x 5 6

x-intercept: 6(0) 2 3x 5 6

x 5 22

y-intercept: 6y 2 3(0) 5 6

y 5 1

44. 23 1 x 5 0

x 5 3


1

x

y

21

45. y 1 7 5 22x 1

x

y

21 y 1 2x 5 7

x-intercept: 0 1 2x 5 27

x 5 2 7 } 2

y-intercept: y 1 2(0) 5 27

y 5 27

46. 4y 5 16

y 5 4

The graph of y 5 4 is the horizontal line that passes through the point (0, 4).

1

x

y

21


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Chapter 2, continued 47. 8y 5 22x 1 20

1

x

y

21

8y 1 2x 5 20

x-intercept: 8(0) 1 2x 5 20

x 5 10

y-intercept: 8y 1 2(0) 5 20

y 5 5 } 2

48. 4x 5 2 1 } 2 y 2 1

4

x

y

21

4x 1 1 } 2 y 5 21

x-intercept: 4x 1 1 } 2 (0) 5 21

x 5 2 1 } 4

y-intercept: 24(0) 1 1 } 2 y 5 21

y 5 22

49. 24x 5 8y 1 12 1

x

y

1 24x 2 8y 5 12

x-intercept: 24x 2 8(0) 5 12

x 5 23

y-intercept: 24(0) 2 8y 5 12

y 5 2 3 } 2

50. 3.5x 5 10.5

x 5 3


1

x

y

21

51. y 2 5.5x 5 6 y 5 5.5x 1 6 The y-intercept is 6, so plot the point (0, 6). The slope is 5.5, so plot a second point by starting at

(0, 6) and then moving up 5.5 units and right 1 unit. The second point is (1, 11.5).

6

x

y

22

52. 14 2 3x 5 7y

1x

21

y

3x 1 7y 5 14


x 5 14

} 3


y 5 2

53. 2y 2 5 5 0

y 5 5 } 2

The graph of y 5 5 } 2 is the horizontal line that passes

through the point 1 0, 5 } 2 2 .

1

x

y

21

54. 5y 5 7.5 2 2.5x

1x

y

1

5y 1 2.5x 5 7.5

x-intercept: 5(0) 1 2.5x 5 7.5

x 5 3

y-intercept: 5y 1 2.5(0) 5 7.5

y 5 1.5

55. Sample answer:

A line that has an x-intercept but no y-intercept is a vertical line. One example is the line x 5 3.

x

y

1

1

A line that has a y-intercept but no x-intercept is a horizontal line. One example is the line y 5 2

x

y

1

1

56. Sample answer:

For positive values of m, as m increases, the steepness of the line increases. For negative values of m, as m decreases, the steepness of the line increases. You can conclude that the further away m gets from zero, the steeper the line will be.

3

x

y

22

14y 5 2 x

y 5 4xy 5 23x

12y 5 x

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67Algebra 2


Chapter 2, continued57. Ax 1 By 5 C

x-intercept: Ax 1 B(0) 5 C

x 5 C

} A

The point 1 C } A , 0 2 is on the line. y-intercept: A(0) 1 By 5 C

y 5 C

} B

The point 1 0, C } B 2 is on the line.

The slope is C

} B 2 0 }

0 2 C

} A

5

C(A) }

B(21)(C) 5 2

A } B .

58. Sample answer:

Two points on the line are (0, b) and (1, m 1 b).

Using the slope formula gives m 1 b 2 b

} 1 2 0

5 m

} 1 5 m.

Problem Solving

59. To fi nd the total cost of the membership after 9 months, start at 9 on the x-axis and move up until you reach the graph. Then move left to the y-axis. The total cost of the membership after 9 months is about $480.

Months

Co

st

0 1 2 3 4 5 6 7 8 x

y

0

75

150

225

300

375

450

525

600

60. y 5 5x 1 35

The slope, 5, represents how much it costs to stay per night, $5. The y-intercept, 35, represents the annual membership fee, $35, without staying any nights.

Co

st (

do

llars

)

020 1 3 5 7 94 6 8

20

10

40

30

60

50

80

70

Number of nights

x

y

61. C(g) 5 3g 1 1.5

The y-intercept, 1.5, represents the cost to rent shoes, $1.50. The slope, 3, represents the cost per game, $3.

Number of games

Co

st

0 1 2 3 4 5 6 7 8 g

C(g)

0

4

8

12

16

20

24

28

32

36

62. To determine a reasonable domain, fi nd the minimum and maximum values of w. The minimum value of w is 0 because the number of weeks cannot be negative. To fi nd the maximum value of w, let M(w) 5 0 and solve for w.

m(w) 5 230w 1 300

0 5 230w 1 300

30w 5 300

w 5 10

So, a reasonable domain is 0 ≤ w ≤ 10. To determine a reasonable range, substitute the minimum

and maximum values for w into the equation.

w 5 0: m(w) 5 230(0) 1 300

m(w) 5 300

w 5 10: m(w) 5 230(10) 1 300

m(w) 5 0

So, a reasonable range is 0 ≤ m(w) ≤ 300. The slope, 230, represents the number of minutes per

week you use the card. So, you use 30 minutes per week.

Nu

mb

er o

f m

inu

tes

020 4 6 8 10

100

200

300

Number of weeks

w

M(w)

63. The y-intercept will be 30, the amount of money on the card ($30) before any smoothies were purchased. The line will fall from left to right because the more smoothies you buy, the less the amount on the card will be.

64. a. Domain: 0 ≤ x ≤ 60; Range: 0 ≤ y ≤ 20 The y-intercept represents the amount of time it would

take to go 1800 yards down the river if there were no drifting time, 20 minutes.

The x-intercept represents the amount of time it would take to go 1800 yards down the river if there were no paddling time, 60 minutes.

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Chapter 2, continued b. If you paddle for 5 minutes, let y 5 5.

30x 1 90(5) 5 1800

30x 1 450 5 1800

30x 5 1350

x 5 45

When y 5 5, x 5 45. So, the total trip time is x 1 y 5 50 minutes.

c. If you paddle and drift equal amounts of time, then x 5 y. 30x 1 90x 5 1800

120x 5 1800

x 5 15

When x 5 15 and y 5 15, the total trip time is x 1 y 5 30 minutes.

Pad

dlin

g t

ime

(min

.)

0140 28 42 56

4

8

12

16

20

Drifting time (min.)

x

y

65. Sample answer:

Let r 5 0: 6(0) 1 3.5w 5 14

3.5w 5 14

w 5 4

You could walk 4 hours and run 0 hours.

Let w 5 0: 6r 1 3.5(0) 5 14

6r 5 14

r 5 2 1 }

3

You could walk 0 hours and run 2 1 }

3 hours.

Let w 5 1: 6r 1 3.5(1) 5 14

6r 5 10.5

r 5 1.75 5 1 3 }

4

You could walk 1 hour and run 1 3 }

4 hours.

Running time (hours)

Wal

kin

g t

ime

(ho

urs

)

0 1 2 r

w

0

1

2

3

4

66. Sample answer:

Let a 5 10: 5s 1 7(10) 5 150

5s 1 70 5 150

5s 5 80

s 5 16

The honor society could buy 16 science museum tickets and 10 art museum tickets.

Let a 5 5: 5s 1 7(5) 5 150

5s 1 35 5 150

5s 5 115

s 5 23

The honor society could buy 23 science museum tickets and 5 art museum tickets.

Art

mu

seu

m t

icke

ts

060 12 18 24 30

4

8

12

16

20

Science museum tickets

s

a

67. a. t (min) 0 1 2 3 4 5

h (ft) 200 350 500 650 800 950

b.

Time (minutes)

Hei

gh

t (f

eet)

0 1 2 3 4 t

h

0100200300400500600700800900

c. Balloon’sheight

5

Initialheight

1 Ascentrate

p Time

h 5 200 1 150 p t

h 5 200 1 150t

68. a. The graphs of y 5 1400 2 50x and y 5 1200 2 50x have the same slope but y 5 1400 2 50x has a y-intercept of 1400 and y 5 1200 2 50x has a y-intercept of 1200. The graphs are parallel.

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69Algebra 2


Chapter 2, continued b. The y-intercepts represent how many words each

research paper contains, the x-intercepts represent how long it will take each person to type the research paper, and the slopes represent the rate of typing for each person. Your paper has 1400 words, it will take you 28 minutes to fi nish, and you type at a rate of 50 words per minute.Your friend’s paper has 1200 words, it will take him/her 24 minutes to fi nish, and he/she types at a rate of 50 words per minute.

c. Although you both type at the same rate, your friend will fi nish fi rst, because your friend’s paper is shorter. It will only take him/her 24 minutes to fi nish, while it takes you 28 minutes to fi nish.

Wo

rds

left

060 12 18 24 x

300

600

900

1200

y

Minutes

69. a. Area 5 length 3 width. The area of the 3 by 1 rectangle is 3, the area of the 4 by 1 rectangle is 4, and the area of the 5 by 5 rectangle is 25. Let x 5 the number of 3 by 1 rectangles you need and y 5 the number of 4 by 1 rectangles you need. Therefore, 3x 1 4y 5 25. x and y must be whole numbers because you cannot use partial rectangles.

b. 3x 1 4y 5 25

Start by making a table of values.

x 0 1 2 3 4 5 6 7 8

y 6.25 5.5 4.75 4 3.25 2.5 1.75 1 0.25

From the table, it appears that (3, 4) and (7, 1) are the only solutions where x and y are whole numbers.

Check:

(3, 4): 3(3) 1 4(4) 0 25 (7, 1): 3(7) 1 4(1) 0 25

9 1 16 0 25 21 1 4 0 25

25 5 25 ✓ 25 5 25 ✓

c. Not all the solutions from part (b) represent combinations of rectangles that can actually cover the grid. (3, 4) works, but (7, 1) does not work.

(3, 4)

not possible

(7, 1)

Mixed Review

70. When n 5 5:

3n 2 10 5 3(5) 2 10 5 15 2 10 5 5

71. When x 5 22:

24x 1 16 5 24(22) 1 16 5 8 1 16 5 24

72. When p 5 4:

2(11 2 5p) 5 22 2 10p 5 22 2 10(4)

5 22 2 40

5 218

73. When q 5 21:

(4q 1 5)(2q) 5 8q2 1 10q

5 8(21)2 1 10(21)

5 8 2 10

5 22

74. When m 5 23:

m2 2 4m 5 (23)2 2 4(23) 5 9 1 12 5 21

75. When d 5 6:

(d 1 1)2 2 d 5 d2 1 2d 1 1 2 d

5 d2 1 d 1 1

5 (6)2 1 6 1 1

5 36 1 6 1 1

5 43


77. The relation given by the ordered pairs (1, 3), (0, 0), (2, 22), (23, 6), and (22, 22) is a function because each input is mapped onto exactly one output.

78. Let (x1, y1) 5 (1, 23) and (x2, y2) 5 (5, 0).

m 5 0 2 (23)

} 5 2 1 5 3 } 4

79. Let (x1, y1) 5 (22, 1) and (x2, y2) 5 (6, 27).

m 5 27 2 1

} 6 2 (22)

5 28

} 8 5 21

80. Let (x1, y1) 5 (4, 4) and (x2, y2) 5 (8, 4).

m 5 4 2 4

} 8 2 4 5 0 } 4 5 0

81. Let (x1, y1) 5 (2, 5) and (x2, y2) 5 (25, 8).

m 5 8 2 5

} 25 2 2 5

3 }

27 5 2 3 } 7

82. Let (x1, y1) 5 (6, 23) and (x2, y2) 5 (1, 213).

m 5 213 2 (23)

} 1 2 6 5 210

} 25 5 2

83. Let (x1, y1) 5 (2.5, 0) and (x2, y2) 5 (23.5, 24).

m 5 24 2 0

} 23.5 2 2.5 5

24 }

26 5 2 } 3

Quiz 2.1–2.3 (p. 96)

1. The relation is a function because each input value is mapped onto exactly one output value.

2. The relation is a function because each input value is mappped onto exactly one output value.

3. The relation is not a function because the input 0 is mapped onto both 4 and 5.

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Chapter 2, continued4. Line 1: let (x1, y1) 5 (23, 27) and (x2, y2) 5 (1, 9)

Line 2: let (x1, y1)5 (21, 24) and (x2, y2) 5 (0, 22)

m1 5 9 2 (27)

} 1 2 (23)

5 16

} 4 5 4

m2 5 22 2 (24)

} 0 2 (21)

5 2 } 1 5 2

Because m1m2 5 4(2) 5 8 Þ 21 and m1 Þ m2, the lines are neither perpendicular nor parallel.

5. Line 1: let (x1, y1) 5 (2, 7) and (x2, y2) 5 (21, 22)

Line 2: let (x1, y1) 5 (3, 26) and (x2, y2) 5 (26, 23)

m1 5 22 2 7

} 21 2 2 5

29 }

23 5 3

m2 5 23 2 (26)

} 26 2 3 5

3 }

29 5 2 1 } 3

Because m1m2 5 3 p 2 1 } 3 5 21, m1 and m2 are negative


6. y 5 25x 1 3


The slope is 25, so plot a second point on the line by starting at (0, 3) and then moving down 5 units and right 1 unit.

1

x

y

21

7. x 5 10


2

x

y

22

8. 4x 1 3y 5 224

x-intercept: 4x 1 3(0) 5 224

x 5 26

y-intercept: 4(0) 1 3y 5 224

y 5 28

1

x

y

21

9. Because she rowed a total of 3333 miles, a reasonable range is 0 ≤ d ≤ 3333. To determine a reasonable domain, fi nd the values of

t when d is at its minimum and maximum values.

d 5 0: 0 5 41t d 5 3333: 3333 5 41t

0 5 t 81.3 ø t So, a reasonable domain is 0 ≤ t ≤ 82. To estimate how long it took Tori Murden to row 1000

miles, start at 1000 on the vertical axis and move right until you reach the graph. Then move down to the horizontal axis. It took Tori Murden about 24 days to row 1000 miles.

50

t

d

22

Graphing Calculator Activity 2.3 (p. 97)

1. y 1 14 5 17 2 2x

y 5 3 2 2x

2. 3x 2 y 5 4

2y 5 4 2 3x

y 5 3x 2 4

3. 3x 2 6y 5 218

26y 5 23x 2 18

y 5 1 } 2 x 1 3

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71Algebra 2



4. 8x 5 5y 1 16

8x 2 16 5 5y

8 } 5 x 2

16 } 5 5 y

5. 4x 5 25y 2 240

4x 1 240 5 25y

4 }

25 x 1

240 } 25 5 y

290 0 30

28

0

16

6. 1.25x 1 4.2y 5 28.7

4.2y 5 28.7 2 1.25x

y 5 41

} 6 2 1.25

} 4.2 x

260 0 60

28

0

16

Lesson 2.4


1. The slope is m 5 3 and the y-intercept is b 5 1. Use the slope-intercept form to write an equation of the line.

y 5 mx 1 b

y 5 3x 1 1

2. The slope is m 5 22 and the y-intercept is b 5 24. Use the slope-intercept form to write an equation of the line.

y 5 mx 1 b

y 5 (22)x 1 (24)

y 5 22x 2 4

3. The slope is m 5 2 3 } 4 and the y-intercept is b 5

7 } 2 .

Use the slope-intercept form to write an equation of the line.

y 5 mx 1 b

y 5 2 3 } 4 x 1

7 } 2

4. Because you know the slope and a point on the line, use point-slope form to write an equation of the line. Let (x1, y1) 5 (21, 6) and m 5 4.

y 2 y1 5 m(x 2 x1)

y 2 6 5 4(x 2 (21))

y 2 6 5 4(x 1 1)

y 2 6 5 4x 1 4

y 5 4x 1 10

5. a. The given line has a slope of m1 5 3. So, a line parallel to it has a slope of m2 5 m1 5 3. You know the slope and a point on the line, so use the point-slope form with (x1, y1) 5 (4, 22) to write an equation of the line.

y 2 y1 5 m2(x 2 x1)

y 2 (22) 5 3(x 2 4)

y 1 2 5 3(x 2 4)

y 1 2 5 3x 2 12

y 5 3x 2 14

b. A line perpendicular to a line with slope m1 5 3 has a

slope of m2 5 2 1 } m1 5 2

1 } 3 . Use point-slope form

with (x1, y1) 5 (4, 22).

y 2 y1 5 m2(x 2 x1)

y 2 (22) 5 2 1 } 3 (x 2 4)

y 1 2 5 2 1 } 3 (x 2 4)

y 1 2 5 2 1 } 3 x 1

4 } 3

y 5 2 1 } 3 x 2

2 } 3

6. The line passes through (x1, y1) 5 (22, 5) and(x2, y2) 5 (4, 27). Find its slope.

m 5 y2 2 y1

} x2 2 x1 5

27 2 5 }

4 2 (22) 5

212 } 6 5 22

You know the slope and a point on the line, so use point-slope form with either given point to write an equation of the line. Choose (x1, y1) 5 (22, 5).

y 2 y1 5 m(x 2 x1)

y 2 5 5 22(x 2 (22))

y 2 5 5 22(x 1 2)

y 2 5 5 22x 2 4

y 5 22x 1 1

7. The line passes through (x1, y1) 5 (6, 1) and(x2, y2) 5 (23, 28). Find its slope.

m 5 y2 2 y1

} x2 2 x1

5 28 2 1

} 23 2 6 5

29 }

29 5 1


y 2 y1 5 m(x 2 x1)

y 2 (28) 5 1 1 x 2 (23) 2 y 1 8 5 1(x 1 3)

y 1 8 5 x 1 3

y 5 x 2 5

8. The line passes through (x1, y1) 5 (21, 2) and (x2, y2) 5 (10, 0). Find its slope.

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Chapter 2, continued m 5

y2 2 y1 } x2 2 x1 5

0 2 2 }

10 2 (21) 5

22 } 11 5 2

2 } 11


y 2 y1 5 m(x 2 x1)

y 2 2 5 2 2 } 11 (x 2 (21))

y 2 2 5 2 2 } 11 (x 1 1)

y 2 2 5 2 2 } 11 x 2

2 } 11

y 5 2 2 } 11 x 1

20 } 11

9. Let x represent the time (in years) since 1993 and let y represent the number of participants (in millions).

The initial value is 3.42. The rate of change is the slope m. Use (x1, y1) 5 (0, 3.42) and (x2, y2) 5 (10, 3.99).

m 5 y2 2 y1

} x2 2 x1 5

3.99 2 3.42 } 10 2 0 5

0.57 } 10 5 0.057

Participants 5

Initialnumber

1 Rate of change

p Yearssince1993

y 5 3.42 1 0.057 p x

In slope-intercept form, a linear model is y 5 0.057x 1 3.42

10. Company A song price p

Songs from Company A

1 Company B song price

p

Songs from Company B 5

Your budget

0.69 p x 1 0.89 p y 5 30

An equation for this situation is 0.69x 1 0.89y 5 30.


Skill Practice

1. The linear equation 6x 1 8y 5 72 is written in standard form.

2. With two points on a line a slope can be calculated. Once the slope is calculated, simply substitute it in the point-slope form along with either one of the two points.

3. The slope is m 5 0 and the y-intercept is b 5 2. Use slope-intercept form to write an equation of the line.

y 5 mx 1 b

y 5 0x 1 2

y 5 2

4. The slope is m 5 3 and the y-intercept is b 5 24. Use slope-intercept form to write in equation of the line.

y 5 mx 1 b

y 5 3x 1 (24)

y 5 3x 2 4


y 5 mx 1 b

y 5 6x 1 0

y 5 6x

6. The slope is m 5 2 } 3 and the y-intercept is b 5 4. Use

slope-intercept form to write an equation of the line.

y 5 mx 1 b

y 5 2 } 3 x 1 4

7. The slope is m 5 2 5 } 4 and the y-intercept is b 5 7. Use

slope-intercept form to write an equation of the line.

y 5 mx 1 b

y 5 2 5 } 4 x 1 7


y 5 mx 1 b

y 5 25x 1 (21)

y 5 25x 2 1

9. Let (x1, y1) 5 (0, 22) and m 5 4.

y 2 y1 5 m(x 2 x1)

y 2 (22) 5 4(x 2 0)

y 1 2 5 4x

y 5 4x 2 2

10. Let (x1, y1) 5 (3, 21) and m 5 23.

y 2 y1 5 m(x 2 x1)

y 2 (21) 5 23(x 2 3)

y 1 1 5 23(x 2 3)

y 1 1 5 23x 1 9

y 5 23x 1 8

11. Let (x1, y1) 5 (24, 3) and m 5 2.

y 2 y1 5 m(x 2 x1)

y 2 3 5 2(x 2 (24))

y 2 3 5 2(x 1 4)

y 2 3 5 2x 1 8

y 5 2x 1 11

12. Let (x1, y1) 5 (25, 26) and m 5 0.

y 2 y1 5 m(x 2 x1)

y 2 (26) 5 0(x 2 (25))

y 1 6 5 0(x 1 5)

y 1 6 5 0

y 5 26

13. Let (x1, y1) 5 (8, 13) and m 529.

y 2 y1 5 m(x 2 x1)

y 2 13 5 29(x 2 8)

y 2 13 5 29x 1 72

y 5 29x 1 85

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73Algebra 2


Chapter 2, continued14. Let (x1, y1) 5 (12, 0) and m 5

3 } 4 .

y 2 y1 5 m(x 2 x1)

y 2 0 5 3 } 4 (x 2 12)

y 5 3 } 4 x 2 9

15. Let (x1, y1) 5 (7, 23) and m 5 2 4 } 7 .

y 2 y1 5 m(x 2 x1)

y 2 (23) 5 2 4 } 7 (x 2 7)

y 1 3 5 2 4 } 7 (x 2 7)

y 1 3 5 2 4 } 7 x 1 4

y 5 2 4 } 7 x 1 1

16. Let (x1, y1) 5 (24, 2) and m 5 3 } 2 .

y 2 y1 5 m(x 2 x1)

y 2 2 5 3 } 2 (x 2 (24))

y 2 2 5 3 } 2 (x 1 4)

y 2 2 5 3 } 2 x 1 6

y 5 3 } 2 x 1 8

17. Let (x1, y1) 5 (9, 25) and m 5 2 1 } 3 .

y 2 y1 5 m(x 2 x1)

y 2 (25) 5 2 1 } 3 (x 2 9)

y 1 5 5 2 1 } 3 (x 2 9)

y 1 5 5 2 1 } 3 x 1 3

y 5 2 1 } 3 x 2 2

18. The error was made when substituting for x1. It should be 24, not 4.

y 2 y1 5 m(x 2 x1)

y 2 2 5 3(x 2 (24))

y 2 2 5 3(x 1 4)

y 2 2 5 3x 1 12

y 5 3x 1 14

19. The x1 and y1 values were transposed.

y 2 y1 5 m(x 2 x1)

y 2 1 5 22(x 2 5)

y 2 1 5 22x 1 10

y 5 22x 1 11

20. (23, 25); parallel to y 5 24x 1 1.

m1 5 24

m2 5 m1 5 24

Let (x1, y1) 5 (23, 25).

y 2 y1 5 m2(x 2 x1)

y 2 (25) 5 24(x 2 (23))

y 1 5 5 24(x 1 3)

y 1 5 5 24x 2 12

y 5 24x 2 17

21. (7, 1); parellel to y 5 2x 1 3.

m1 5 21

m2 5 m1 5 21

Let (x1, y1) 5 (7, 1).

y 2 y1 5 m2(x 2 x1)

y 2 1 5 21(x 2 7)

y 2 1 5 2x 1 7

y 5 2x 1 8

22. (2, 8); parallel to y 5 3x 2 2.

m1 5 3

m2 5 m1 5 3

Let (x1, y1) 5 (2, 8).

y 2 y1 5 m2(x 2 x1)

y 2 8 5 3(x 2 2)

y 2 8 5 3x 2 6

y 5 3x 1 2

23. (4, 1); perpendicular to y 5 1 } 3 x 1 3.

m1 5 1 } 3

m2 5 2 1 } m1 5 23

Let (x1, y1) 5 (4, 1).

y 2 y1 5 m2(x 2 x1)

y 2 1 5 23(x 2 4)

y 2 1 5 23x 1 12

y 5 23x 1 13

24. (26, 2); perpendicular to y 5 22.

m1 5 0

m2 5 2 1 } m1 5 undefi ned

Because the slope is undefi ned, you know the line is vertical. Because it passes through (26, 2), its equation is x 5 26.

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Chapter 2, continued 25. (3, 21); perpendicular to y 5 4x 1 1

m1 5 4

m2 5 2 1 } m1 5 2

1 } 4

Let (x1, y1) 5 (3, 21).

y 2 y1 5 m2(x 2 x1)

y 2 (21) 5 2 1 } 4 (x 2 3)

y 1 1 5 2 1 } 4 (x 2 3)

y 1 1 5 2 1 } 4 x 1

3 } 4

y 5 2 1 } 4 x 2

1 } 4

26. C; (1, 4); perpendicular to y 5 2x 2 3.

m1 5 2

m2 5 2 1 } m1 5 2

1 } 2

Let (x1, y1) 5 (1, 4).

y 2 y1 5 m2(x 2 x1)

y 2 4 5 2 1 } 2 (x 2 1)

y 2 4 5 2 1 } 2 x 1

1 } 2

y 5 22x 1 9 } 2

27. From the graph, you can see that the slope is m 5 22. Choose (x1, y1) 5 (3, 0).

y 2 y1 5 m(x 2 x1)

y 2 0 5 22(x 2 3)

y 5 22x 1 6

28. From the graph, you can see that the slope is m 5 5. Choose (x1, y1) 5 (4, 4).

y 2 y1 5 m(x 2 x1)

y 2 4 5 5(x 2 4)

y 2 4 5 5x 2 20

y 5 5x 2 16

29. From the graph, you can see that the slope is m 5 2 1 } 4 .

Choose (x1, y1) 5 (21, 5).

y 2 y1 5 m(x 2 x1)

y 2 5 5 2 1 } 4 (x 2 (21))

y 2 5 5 2 1 } 4 (x 1 1)

y 2 5 5 2 1 } 4 x 2

1 } 4

y 5 2 1 } 4 x 1

19 } 4

30. Let (x1, y1) 5 (21, 3) and (x2, y2) 5 (2, 9).

m 5 y2 2 y1

} x2 2 x1 5

9 2 3 }

2 2 (21) 5

6 } 3 5 2

Choose (x1, y1) 5 (2, 9).

y 2 y1 5 m(x 2 x1)

y 2 9 5 2(x 2 2)

y 2 9 5 2x 2 4

y 5 2x 1 5

31. Let (x1, y1) 5 (4, 21) and (x2, y2) 5 (6, 27).

m 5 y2 2 y1

} x2 2 x1 5

27 2 (21) } 6 2 4 5

26 } 2 5 23

Choose (x1, y1) 5 (4, 21).

y 2 y1 5 m(x 2 x1)

y 2 (21) 5 23(x 2 4)

y 1 1 5 23(x 2 4)

y 1 1 5 23x 1 12

y 5 23x 1 11

32. Let (x1, y1) 5 (22, 23) and (x2, y2) 5 (2, 21).

m 5 y2 2 y1

} x2 2 x1 5

21 2 (23) }

2 2 (22) 5

2 } 4 5

1 } 2

Choose (x1, y1) 5 (2, 21).

y 2 y1 5 m(x 2 x1)

y 2 (21) 5 1 } 2 (x 2 2)

y 1 1 5 1 } 2 (x 2 2)

y 1 1 5 1 } 2 x 2 1

y 5 1 } 2 x 2 2

33. Let (x1, y1) 5 (0, 7) and (x2, y2) 5 (3, 5).

m 5 y2 2 y1

} x2 2 x1 5

5 2 7 } 3 2 0 5

22 } 3 5 2

2 } 3

Choose (x1, y1) 5 (3, 5).

y 2 y1 5 m(x 2 x1)

y 2 5 5 2 2 } 3 (x 2 3)

y 2 5 5 2 2 } 3 x 1 2

y 5 2 2 } 3 x 1 7

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75Algebra 2


34. Let (x1, y1) 5 (21, 2) and (x2, y2) 5 (3, 24).

m 5 y2 2 y1

} x2 2 x1 5

24 2 2 }

3 2 (21) 5

26 } 4 5 2

3 } 2

Choose (x1, y1) 5 (21, 2).

y 2 y1 5 m(x 2 x1)

y 2 2 5 2 3 } 2 (x 2 (21))

y 2 2 5 2 3 } 2 (x 1 1)

y 2 2 = 2 3 } 2 x 2

3 } 2

y 5 2 3 } 2 x 1

1 } 2

35. Let (x1, y1) 5 (25, 22) and (x2, y2) 5 (23, 8).

m 5 y2 2 y1

} x2 2 x1 5

8 2 (22) }

23 2 (25) 5

10 } 2 5 5

Choose (x1, y1) 5 (25, 22).

y 2 y1 5 m(x 2 x1)

y 2 (22) 5 5(x 2 (25))

y 1 2 5 5(x 1 5)

y 1 2 5 5x 1 25

y 5 5x 1 23

36. Let (x1, y1) 5 (15, 20) and (x2, y2) 5 (212, 29).

m 5 y2 2 y1

} x2 2 x1 5

29 2 20 }

212 2 15 5 9 }

227 5 2 1 } 3

Choose (x1, y1) 5 (15, 20).

y 2 y1 5 m(x 2 x1)

y 2 20 5 2 1 } 3 (x 2 15)

y 2 20 5 2 1 } 3 x 1 5

y 5 2 1 } 3 x 1 25

37. Let (x1, y1) 5 (3.5, 7) and (x2, y2) 5 (21, 20.5).

m 5 y2 2 y1

} x2 2 x1 5

20.5 2 7 }

21 2 3.5 5 13.5

} 24.5 5 23

Choose (x1, y1) 5 (21, 20.5).

y 2 y1 5 m(x 2 x1)

y 2 20.5 5 23(x 2 (21))

y 2 20.5 5 23(x 1 1)

y 2 20.5 5 23x 2 3

y 5 23x 1 17.5

38. Let (x1, y1) 5 (0.6, 0.9) and (x2, y2) 5 (3.4, 22.6).

m 5 22.6 2 0.9

} 3.4 2 0.6 5 23.5

} 2.8 5 21.25

Choose (x1, y1) 5 (0.6, 0.9).

y 2 y1 5 m(x 2 x1)

y 2 0.9 5 21.25(x 2 0.6)

y 2 0.9 5 21.25x 1 0.75

y 5 21.25x 1 1.65

39. C;

Let (x1, y1) 5 (9, 25) and m 5 26.

y 2 y1 5 m(x 2 x1)

y 2 (25) 5 26(x 2 9)

y 1 5 5 26(x 2 9)

y 1 5 5 26x 1 54

y 5 26x 1 49

So, an equation of the line is y 5 26x 1 49. The point (7, 7) is a solution of this equation, so it lies on the line.

40. When m 5 23 and b 5 5:

y 5 mx 1 b

y 5 23x 1 5

3x 1 y 5 5

41. When m 5 4 and b 5 23:

y 5 mx 1 b

y 5 4x 1 (23)

y 5 4x 2 3

24x 1 y 5 23

42. Using m 5 2 3 } 2 and (x1, y1) 5 (4, 27):

y 2 y1 5 m(x 2 x1)

y 2 (27) 5 2 3 } 2 (x 2 4)

y 1 7 5 2 3 } 2 (x 2 4)

y 1 7 5 2 3 } 2 x 1 6

y 5 2 3 } 2 x 2 1

2y 5 23x 2 2

3x 1 2y 5 22

43. Using m 5 4 } 5 and (x1, y1) 5 (2, 3):

y 2 y1 5 m(x 2 x1)

y 2 3 5 4 } 5 (x 2 2)

y 2 3 5 4 } 5 x 2

8 } 5

y 5 4 } 5 x 1

7 } 5

5y 5 4x 1 7

24x 1 5y 5 7

44. Let (x1, y1) 5 (21, 3) and (x2, y2) 5 (26, 27).

m 5 y2 2 y1

} x2 2 x1 5

27 2 3 }

26 2 (21) 5

210 }

25 5 2

Choose (x1, y1) 5 (21, 3).

y 2 y1 5 m(x 2 x1)

y 2 3 5 2(x 2 (21))

y 2 3 5 2(x 1 1)

y 2 3 5 2x 1 2

y 5 2x 1 5

22x 1 y 5 5


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Chapter 2, continued45. Let (x1, y1) 5 (2, 8) and (x2, y2) 5 (24, 16).

m 5 y2 2 y1

} x2 2 x1 5

16 2 8 }

24 2 2 5 8 }

26 5 2 4 } 3

Choose (x1, y1) 5 (24, 16).

y 2 y1 5 m(x 2 x1)

y 2 16 5 2 4 } 3 (x 2 (24))

y 2 16 5 2 4 } 3 (x 1 4)

y 2 16 5 2 4 } 3 x 2

16 } 3

y 5 2 4 } 3 x 1

32 } 3

3y 5 24x 1 32

4x 1 3y 5 32

46. a. The line y 5 22 is horizontal. So, a line parallel to it is also horizontal. Because it passes through (3, 4), its equation is y 5 4.

b. The line y 5 22 is horizontal. So, a line perpendicular to it is vertical. Because it passes through (3, 4), its equation is x 5 3.

c. The line x 5 22 is vertical. So a line parallel to it is also vertical. Because it passes through (3, 4), its equation is x 5 3.

d. The line x 5 22 is vertical. So, a line perpendicular to it is horizontal. Because it passes through (3, 4), its equation is y 5 4.

47. Sample answer:

y 5 23x 1 5; m1 5 23, b1 5 5

y 5 2x 1 1; m2 5 2, b2 5 1

Because m2 Þ 2 1 } m1 , the two lines are not perpendicular.

To form a right triangle line l must be perpendicular to either y 5 23x 1 5 or y 5 2x 1 1. Because the

triangle can be any size, line l can be placed anywhere except the point of intersection of y 5 23x 1 5 and y 5 2x 1 1.

One possible line l could be perpendicular to y 5 23x 1 5 through the point (0, 0).

m1 5 23, m2 5 2 1 } m1 5 2

1 }

23 5 1 } 3

y 2 0 5 1 } 3 (x 2 0)

y 5 1 } 3 x

x

y

1

1

y 5 23x 1 5

y 5 2x 1 1

y 5 x 13

48. Begin by fi nding the slope of each line.

A1x 1 B1y 5 C1 A2x 1 B2y 5 C2 B1y 5 C1 2 A1x B2y 5 C2 2 A2x

y 5 C1

} B1 2

A1 } B1 x y 5

C2 } B2 2

A2 } B2 x

y 5 2 A1

} B1 1 x 2 C1 } A1 2 y 5 2

A2 } B2 1 x 2 C2 } A2 2

m1 5 2 A1

} B1 m2 5 2

A2 } B2

a. If A1x 1 B1y 5 C1 and A2x 1 B2y 5 C2 are

parallel, then m2 5 m1. Using the slopes you found above, substitute for m1 and m2 as shown.

m1 5 m2

2 A1

} B1 5 2

A2 } B2

2A1B2 5 2A2B1 A1B2 5 A2B1 b. If A1x 1 B1y 5 C1 and A2x 1 B2y 5 C2 are

perpendicular, then m2 5 2 1 } m1 . Using the slopes you

found above, substitute for m1 and m2 as shown.

m2 5 21

} m1

2 A2

} B2 5

21 }

1 2 A1 } B1 2

2 A2

} B2 5

B1 } A1

2A1A2 5 B1B2 0 5 A1A2 1 B1B2 49. The line has an x-intercept at point (a, 0) and a

y-intercept at point (0, b).

Let (x1, y1) 5 (a, 0) and (x2, y2) 5 (0, b).

m 5 y2 2 y1

} x2 2 x1 5

b 2 0 } 0 2 a 5 2

b } a

Choose (x1, y1) 5 (a, 0).

y 2 0 5 2 b } a (x 2 a)

y 5 2 b } a x 1 b

bx

} a 1 y 5 b

x }

a 1

y }

b 5 1

Problem Solving

50. Let x represent the time (in months) since buying the car and let y represent the total cost (in dollars).

TotalCost 5

Initial number 1

Rate of change p

Months from now

y 5 6500 1 350 p x

In slope-intercept form, a linear model isy 5 350x 1 6500.

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77Algebra 2



51. Number of houses 5

Initial number 1

Rate of change p

Years from now

n 5 50 1 15 p t

In slope-intercept form, a linear model is n 5 15t 1 50.

52. Area of plot 5 25 p 16 5 400 square feet.

Space of

tomato plant p Number of

tomato plants

1 Space of

pepper plant p

Number of pepper plants 5

Area of plot

8 p x 1 5 p y 5 400

An equation for this situation is 8x 1 5y 5 400.

Let x 5 15.

8(15) 1 5y 5 400

120 1 5y 5 400

5y 5 280

y 5 56

If you grow 15 tomato plants, you can grow 56 pepper plants.

53. General admission p

General tickets 1

Student admission p

Student tickets 5

Total in ticket sales

15 p x 1 9 p y 5 4500

An equation that models this situation is 15x 1 9y 5 4500.

Start at 200 on the genral admission axis and move up until you reach the graph. Then fi nd the point that it corresponds to on the student admission axis. If 200 general admission tickets were sold, about 167 student tickets were sold.

0 50 100 150 200 x

y

0

100

200

300

400

500

54. a. First building rate p

Number ofsquare feet 1

Second building rate

p

Number of square feet

5 Budget

21.75 p x 1 17 p y 5 86,000

21.75x 1 17y 5 86,000

b. Let y 5 2500.

21.75y 1 17(2500) 5 86,000

21.75x 1 42,500 5 86,000

21.75y 5 43,500

x 5 2000

2000 square feet can be rented in the fi rst building.

c. Let x 5 y.

21.75y 1 17y 5 86,000

38.75y 5 86,000

y ø 2219 x 1 y ø 4438 The total number of square feet that can be rented is

about 4438 square feet.

55. Let y represent the average monthly cost (in dollars), and let x represent the number of years since 1994. Use (x1, y1) 5 (0, 21.62) and (x2, y2) 5 (10, 38.23).

Rate of change:

m 5 y2 2 y1

} x2 2 x1 5

38.23 2 21.62 }} 10 2 0 5

16.61 } 10 ø 1.66

Monthly

cost 5 Initial

number

1 Rate of change p

Years since 1994

y 5 21.62 1 1.66x

In 2010, x 5 16.

y 5 21.62 1 1.66(16)

y 5 21.62 1 26.56

y 5 48.18

The predicted average monthly cost for basic cable in 2010 is $48.18.

56. Let y represent the tire’s pressure (in pounds per square inch) and let x represent the air temperature (8F).

Rate of change 5 1 psi

} 108F

m 5 0.1

Let (x1, y1) 5 (55, 30).

y 2 y1 5 m(x 2 x1)

y 2 30 5 0.1(x 2 55)

y 2 30 5 0.1x 2 5.5

y 5 0.1x 1 24.5

57. a. Length of rectangle

1 Width of rectangle 5 Perimeter

2l 1 2w 5 24

l 1 w 5 12

b. w 5 2l 1 12

0 4 8 12 16

w

0

4

8

12

16

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Chapter 2, continued c.

w 6 5 4 3 2

l 6 7 8 9 10

58. Total Money raised

5 Total fi xed

amount

1 Total

amount per hour

p Hours danced

y 5 (15 1 35 1 20) 1 (4 1 8 1 3) p x

y 5 70 1 15x

Mixed Review

59. 9x 5 27 60. 5x 5 20

x 5 27 1 1 } 9 2 x 5 20 1 1 } 5 2

x 5 3 x 5 4

Check: Check:

9x 5 27 5x 5 20

9(3) 0 27 5(4) 0 20

27 5 27 ✓ 20 5 20 ✓

61. 23x 5 21 62. 8x 5 6

x 5 21 1 2 1 } 3 2 x 5 6 1 1 }

8 2

x 5 27 x 5 3 } 4

Chec

chapter 2bhhs.bhusd.org/ourpages/auto/2006/8/11/1155323019648/aat... · 2006. 8. 11. · algebra 2...

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