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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.
The range consists of all the y-coordinates: 23, 21, 2, and 3.
y
21
1
x
2422
13
2321
23
Input Output
4. (5, 22), (23, 22), (3, 3), (21, 21)
The domain consists of all the x-coordinates: 23, 21, 3, and 5.
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.
The range consists of all the y-coordinates: 23, 21, 5, and 8.
y
22
2
x
2321
58
Input Output
22
1
6
6. (27, 4), (2, 25), (1, 22), (23, 6)
The domain consists of all the x-coordinates: 27, 23, 1, and 2.
The range consists of all the y-coordinates: 25, 22, 4, and 6.
2
x
y
22
2522
46
2723
12
Input Output
7. (5, 20), (10, 20), (15, 30), (20, 30)
The domain consists of all the x-coordinates: 5, 10, 15, and 20.
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.
The range consists of all the y-coordinates: 24, 22, 2, and 4.
1
x
y
2
4
16
Input Output
2422
24
9. B; (24, 2), (21, 23), (1, 4), (1, 23), and (2, 1)
The domain consists of all the x-coordinates: 24, 21, 1, and 2.
10. Yes; The relation is a function because each input is mapped onto exactly one output.
11. 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.
13. Yes; The relation is a function because each input is mapped onto exactly one output.
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.
19. The relation given by the ordered pairs (21, 21), (2, 5), (4, 8), (25, 29), and (21, 25) is not a function because the input 21 is mapped onto both 25 and 21.
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
Worked-Out Solution Key
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.
Chapter 2, continued
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54Algebra 2Worked-Out Solution Key
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
when x 5 3 and y 5 4
4 2 8
} 3 2 2
5 24
} 1 5 24
53. y 2 (25)
} x 2 1
when x 5 23 and y 5 23
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
when x 5 217 and y 5 8
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 ✓
Chapter 2, continued
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55Algebra 2
Worked-Out Solution Key
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.
Chapter 2, continued
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56Algebra 2Worked-Out Solution Key
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
2.2 Guided Practice (pp. 83–85)
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.
Chapter 2, continued
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57Algebra 2
Worked-Out Solution Key
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.
2.2 Exercises (pp. 86–88)
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
Because m > 0, the line rises.
4. m 5 3 2 9
} 24 2 8 5
26 }
212 5 1 } 2
Because m > 0, the line rises.
5. m 5 24 2 1
} 8 2 5 5 25
} 3 5 2 5 } 3
Because m < 0, the line falls.
6. m 5 22 2 (22)
} 3 2 (23)
5 0 } 6 5 0
Because m 5 0, the line is horizontal.
7. m 5 24 2 4
} 1 2 (21)
5 28
} 2 5 24
Because m < 0, the line falls.
8. m 5 25 2 5
} 26 2 (26)
5 210
} 0
Because m is undefi ned, the line is vertical.
9. m 5 3 2 (24)
} 21 2 (25)
5 7 } 4
Because m > 0, the line rises.
10. m 5 3 2 6 }
27 2 (23) 5
23 }
24 5 3 } 4
Because m > 0, the line rises.
11. m 5 9 2 4
} 4 2 4 5 5 } 0
Because m is undefi ned, the line is vertical.
12. m 5 3 2 5
} 7 2 5 5 22
} 2 5 21
Because m < 0, the line falls.
13. m 5 23 2 (23)
} 4 2 0 5 0 } 4 5 0
Because m 5 0, the line is horizontal.
14. m 5 24 2 (21)
} 21 2 1 5
23 }
22 5 3 } 2
Because m > 0, the line rises.
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
Because m > 0, the line rises.
18. Line 1: through (3, 21) and (6, 24)
Line 2: through (24, 5) and (22, 7)
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)
Line 2: through (23, 2) and (4, 0)
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.
Chapter 2, continued
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58Algebra 2Worked-Out Solution Key
20. Line 1: through (21, 4) and (2, 5)
Line 2: through (26, 2) and (0, 4)
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
reciprocals of each other. So, the lines are perpendicular.
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)
Line 2: through (8, 1) and (14, 5)
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
Chapter 2, continued
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59Algebra 2
Worked-Out Solution Key
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
Chapter 2, continued
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60Algebra 2Worked-Out Solution Key
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
Chapter 2, continued
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61Algebra 2
Worked-Out Solution Key
Lesson 2.3
2.3 Guided Practice (pp. 90–92)
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-intercept: 3x 2 2(0) 5 12
x 5 4
The x-intercept is 4. So, plot the point (4, 0).
y-intercept: 3(0) 2 2y 5 12
y 5 26
The y-intercept is 26. So, plot the point (0, 26).
1
x
y
21
Chapter 2, continued
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62Algebra 2Worked-Out Solution Key
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
2.3 Exercises (pp. 93–96)
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
The graphs of y 5 x 1 5 and y 5 x both have a slope of 1, but the graph of y 5 x 1 5 has a y-intercept of 5 instead of 0.
6.
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.
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 y-intercept is 26, so plot the point (0, 26).
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
Chapter 2, continued
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63Algebra 2
Worked-Out Solution Key
11. y 5 2x 1 6
The y-intercept is 6, so plot the point (0, 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 y-intercept is 24, so plot the point (0, 24).
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 y-intercept is 21, so plot the point (0, 21).
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 y-intercept is 22, so plot the point (0, 22).
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 y-intercept is 21, so plot the point (0, 21).
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 y-intercept is 1, so plot the point (0, 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 y-intercept is 23, so plot the point (0, 23).
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 y-intercept is 4, so plot the point (0, 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
Chapter 2, continued
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64Algebra 2Worked-Out Solution Key
19. f (x) 5 21.5x 1 2
The y-intercept is 2, so plot the point (0, 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
The x-intercept is 215.
y-intercept: 0 1 5y 5 215
y 5 23
The y-intercept is 23.
26. 3x 2 4y 5 212
x-intercept: 3x 2 4(0) 5 212
x 5 24
The x-intercept is 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
The x-intercept is 5.
y-intercept: 2(0) 2 y 5 10
y 5 210
The y-intercept is 210.
28. 4x 2 5y 5 20
x-intercept: 4x 2 5(0) 5 20
x 5 5
The x-intercept is 5.
y-intercept: 4(0) 2 5y 5 20
y 5 24
The y-intercept is 24.
29. 26x 1 8y 5 236
x-intercept: 26x 1 8(0) 5 236
x 5 6
The x-intercept is 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
The x-intercept is 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
Chapter 2, continued
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65Algebra 2
Worked-Out Solution Key
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-intercept: 3x 1 4(0) 5 12
x 5 4
y-intercept: 3(0) 1 4y 5 12
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
The graph of x 5 3 is the vertical line that passes through the point (3, 0).
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
Chapter 2, continued
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66Algebra 2Worked-Out Solution Key
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
The graph of x 5 3 is the vertical line that passes through the point (3, 0).
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-intercept: 3x 1 7(0) 5 14
x 5 14
} 3
y-intercept: 3(0) 1 7y 5 14
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
Worked-Out Solution Key
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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68Algebra 2Worked-Out Solution Key
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
Worked-Out Solution Key
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
76. The relation given by the ordered pairs (22, 27), (0, 3), (1, 22), (22, 13), and (3, 212) is not a function because the input 22 is mapped onto both 27 and 13.
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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70Algebra 2Worked-Out Solution Key
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
reciprocals of each other. So, the lines are perpendicular.
6. y 5 25x 1 3
The y-intercept is 3, so plot the point (0, 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
The graph of x 5 10 is the vertical line that passes through the point (10, 0).
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
Worked-Out Solution Key
Chapter 2, continued
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
2.4 Guided Practice (pp. 98–101)
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
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 (23, 28).
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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72Algebra 2Worked-Out Solution Key
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
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 (21, 2).
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.
2.4 Exercises (pp. 101–104)
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
5. The slope is m 5 6 and the y-intercept is b 5 0. Use slope-intercept form to write an equation of the line.
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
8. The slope is m 5 25 and the y-intercept is b 5 21. Use slope-intercept form to write an equation of the line.
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
Worked-Out Solution Key
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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74Algebra 2Worked-Out Solution Key
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
Worked-Out Solution Key
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
Chapter 2, continued
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76Algebra 2Worked-Out Solution Key
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
Worked-Out Solution Key
Chapter 2, continued
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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78Algebra 2Worked-Out Solution Key
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