1. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
(For help, go to Lesson 1-9.)
Use the graph at the right.
Name the point with the given coordinates.
1. (4, –2) 2. (4, 3)
3. (2, –4) 4. (–2, 1)
Name the coordinates of each given point.
5. B 6. F 7. G
5-1

2. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
Solutions
1. (4, –2) is point C 2. (4, 3) is point D
3. (2, –4) is point E 4. (–2, 1) is point A
5. B coordinates: (0, 0)
6. F coordinates: (–4, –2)
7. G coordinates: (–3, 3)
5-1

3. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
This graph shows someone taking a walk in
the neighborhood. Describe what it shows by labeling each part.
5-1

4. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
A pelican flies above the water searching for fish. Sketch a graph of its altitude from takeoff from shore to diving to the water to catch a fish. Label each section.
5-1

5. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
Which graph in Example 3 could show a car sitting at a stoplight?
Graph I indicates a quantity that does not change with time.
Graph II shows an increase over time.
A car sitting at a stoplight would stay in the same place over time.
Graph I could show a car sitting at a stoplight.
5-1

6. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
pages 238–240  Exercises
1–4. Labels may vary. Samples are given. 1. 2. 3. 4. 5. 6.
5-1

7. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1 7. 8.
9. C; the temperature increases steadily and then
alternates cooling and warming as the oven turns
off and on during a cooking cycle.
10. The pressure dropped from 7 A.M. to 3 P.M.,
stayed about the same until 9 P.M., and then
generally rose until 7 A.M. the next day.
11. a. b.
No; the graphs are different because you have a
constant speed traveling up but not down.
5-1

8. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
12. a. blue; red
b. The baby weighs more
at first and gains weight
steadily for a number of
years. The puppy’s weight
levels off at an earlier age.
13. Answers may vary. Sample:
14. It shows a person bicycling down and then
up a hill because speed increases on the
way down and decreases on the way up.
15. a.
b. section showing the distance decreasing
c. first 2 sections
16. A; one’s growth rate is not steady, as
shown in B. Also, as one gets older,
one’s height decreases slightly.
5-1

9. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
17. a. Check students’ work.
b. A graph of temperatures
at the equator would show
little change for daily
high temperatures.
18. $3
19. $6
20. more than 2 h up to 4 h
21. Answers may vary. Sample:
Yes, the line segments make
the graph look like steps.
22. a. Answers may vary. Sample:
The student started skating and
got to cruising speed. After a while,
the student sped up going downhill,
lost control, and crashed. After
getting up, the student decided
not to go as fast.
b. Answers may vary. Sample:
I–speeding up;
II–cruising; III–crash;
IV–slower speed
5-1

10. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
29.
30.
31.
32. x < 7
33. x < 8
34. x –2
35. x > –4
36. x –13
37. x <
38. 11
39. 17
40. 45
23. D
24. H
25. [2] IV; the graph rises sharply
at first and then slows. This
indicates that the container
is narrow at the bottom and
wide at the top.
[1] chooses correct graph but
explanation is incomplete
OR chooses the wrong
graph but supports choice
26.
27.
28. 1 12
41. –15
42. –16
44. 6
45. 3
46. 7
47. 41
49. 1 4 1 6
> –
< – 1 4 2 3 1 36 1 36 1 18
5-1

11. Relating Graphs to Events
ALGEBRA 1 LESSON 5-1
Use the graph for questions 1–3.              
1. Why does this graph not start at zero?
2. Why is the graph not a straight slanted line?
3. What does the flat part at the top of the graph represent?
A person’s height is not zero when born.
A straight line represents a constant rate of growth and people do not grow that way.
The person has stayed the same height over a long period of time.
4. Would a horizontal line, a straight slanted line, a curve, or a jagged curve best represent the calories you burn during the day?
a jagged curve
5-1

12. Relations and Functions
ALGEBRA 1 LESSON 5-2
(For help, go to Lessons 1–9 and 1–2.)
Graph each point on a coordinate plane.
1. (2, –4) 2. (0, 3) 3. (–1, –2) 4. (–3, 0)
Evaluate each expression.
5. 3a – 2 for a = – for x = x2 for x = 6
x + 3 –6
5-2

13. Relations and Functions
ALGEBRA 1 LESSON 5-2
Solutions 1. 2. 3. 4.
5. 3a – 2 for a = –5: 3(–5) – 2 = –15 – 2 = –17
for x = 3: = = –1
7. 3x2 for x = 6: 3 • 62 = 3 • 36 = 108
x + 3 –6
3 + 3 6
5-2

14. Relations and Functions
ALGEBRA 1 LESSON 5-2
Find the domain and the range of the ordered pairs.
age weight
domain: {12, 14, 16, 18} List the values in order. Do not repeat values.
range: {110, 120, 124, 125, 126}
5-2

15. Relations and Functions
ALGEBRA 1 LESSON 5-2
Use the vertical-line test to determine whether the relation {(3, 2), (5, –1), (–5, 3), (–2, 2)} is a function.
Step 1: Graph the ordered pairs on a coordinate plane.
Step 2: Pass a pencil across the graph. Keep your pencil straight to represent a vertical line.
A vertical line would not pass through more than one point, so the relation is a function.
5-2

16. Relations and Functions
ALGEBRA 1 LESSON 5-2
Determine whether each relation is a function.
a. {(4, 3), (2, –1), (–3, –3), (2, 4)}
The domain value 2 corresponds to two range values, –1 and 4.
The relation is not a function.
b. {(–4, 0), (2, 12), (–1, –3), (1, 5)}
There is no value in the domain that corresponds to more than one value of the range.
The relation is a function.
5-2

17. Relations and Functions
ALGEBRA 1 LESSON 5-2
a. Evaluate ƒ(x) = –5x + 25 for x = –2.
ƒ(x) = –5x + 25
ƒ(–2) = –5(–2) + 25 Substitute –2 for x.
ƒ(–2) = Simplify.
ƒ(–2) = 35
b. Evaluate y = 4x2 + 2 for x = 3.
y = 4x2 + 2
y = 4(3)2 + 2 Substitute 3 for x.
y = 4(9) + 2 Simplify the powers first.
y = Simplify.
y = 38
5-2

18. Relations and Functions
ALGEBRA 1 LESSON 5-2
Evaluate the function rule ƒ(g) = –2g + 4 to find the range for the domain {–1, 3, 5}.
ƒ(g) = –2g + 4
ƒ(–1) = –2(–1) + 4
ƒ(–1) = 2 + 4
ƒ(–1) = 6
ƒ(g) = –2g + 4
ƒ(3) = –2(3) + 4
ƒ(3) = –6 + 4
ƒ(3) = –2
ƒ(g) = –2g + 4
ƒ(5) = –2(5) + 4
ƒ(5) = –10 + 4
ƒ(5) = –6
The range is {–6, –2, 6}.
5-2

19. Relations and Functions
ALGEBRA 1 LESSON 5-2
pages 244–246  Exercises
1. {4, 5, 6}, {3, 6, 7, 19}
2. {–3, –2, 0, 4}, {5, 7, 8, 22}
3. {–2, 2, 3}, {–3, –2, 3}
4. {1}, {–7, 0, 5, 6.1, 10}
5. {–3.1, 1.2, 8.4}, {–5.2, 0, 4}
6. {– , , 4, 5}, {–1, 0, }
7. yes
8. no
9. no
10. yes
11. no
12. no
13. yes
14. no
15. 4
16. –34
17. 9
18. 12
19. 18
20. –7
21. –
22. 5
23. {0.5, 53}
24. {–8, –2, 18}
25. {–27, –7, –2, 8, 48}
26. {–4 , – , 0}
27. no
28. no
29. yes; {–4, –1, 0, 3}; {–4}
30. No; two 4-year-old iguanas
may have different lengths.
31. Answers may vary. Sample:
A relation is not a function if
two range values have the
same domain value. 1 2 3 4 2 3 1 2 3 5 3 4
5-2

20. Relations and Functions
ALGEBRA 1 LESSON 5-2
38. yes
39. no
40. yes
41. a. Answers may vary. Sample:
The cost appears to be far too little.
b. Answers may vary. Sample:
The student failed to convert
hours to minutes.
c. $10.80
d. whole numbers; positive numbers
42. a. 3,720,000 mi
b. 11,160,000 mi
32. a. {–300, –210, 0, 72}
b. Domain is the number
of cameras sold, and
range is the profit.
33. Answers may vary. Sample:
Data represent the ages (x)
and heights (y) of 4 students.
34. {–3, 3, 15.8}
35. {–13.8, –1, 5}
36. {–0.5, 0, 2.7}
37. {–0.75, 0, 12.69}
5-2

21. Relations and Functions
ALGEBRA 1 LESSON 5-2
43. 23
44. 18
45. 6
46. 20
47. Yes, it passes the vertical
line test; no, it doesn’t pass
the vertical line test.
48. a. 0, –1, –2, –6
b. all integers
51. –3
52. 3
53. [2]
Yes, the data represent
a function. (OR graph shown)
[1] shows calculation but no
mapping diagram or graph
54. Labels may vary. Sample:
5-2

22. Relations and Functions
ALGEBRA 1 LESSON 5-2 mi mi mi mi mi mi
, 33.5, none, 3
62. – , 0, –2 and 1, 4
, 5, 5, 11
, 14, 13, 11
65. – , –7, –7, 8
, 46.5, 43 and 48, 10 1 5 52 9
117 8 47 8
5-2

23. Relations and Functions
ALGEBRA 1 LESSON 5-2
1. a. Find the domain and range of the ordered pairs (1, 3), (–4, 0), (3, 1), (0, 4), (2, 3).
b. Use mapping to determine whether the relation is a function.
domain: {–4, 0, 1, 2, 3} range: {0, 1, 3, 4}
The relation is a function.
2. Use the vertical-line test to determine whether each relation is a function. a. b. no
yes
3. Find the range of the function ƒ(g) = 3g – 5 for the domain {–1.5, 2, 4}.
{–9.5, 1, 7}
5-2

24. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
(For help, go to Lesson 5-2.)
Graph the data in each table.
x y
–3 –7
–1 –1
0 2
2 8 1.
x y
–3 4
–2 0
0 –2
2 4 2.
x y
–4 –3
0 –2
2 –1.5
4 –1 3.
5-3

25. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
1. Graph the points: 2. Graph the points:
(–3, –7) (–3, 4)
(–1, –1) (–2, 0)
(0, 2) (0, –2)
(2, 8) (2, 4)
Solutions
3. Graph the points:
(–4, –3)
(0, –2)
(2, –1.5)
(4, –1)
5-3

26. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
Model the function rule y = using a table of values and a graph. 1 3 x
Step 1: Choose input value for x. Evaluate to find y
x (x, y)
–3 y = (–3) + 2 = 1 (–3, 1)
 0 y = (0) + 2 = 2 (0, 2)
 3 y = (3) + 2 = 3 (3, 3)
y = x + 2 1 3
Step 2: Plot the points for the ordered pairs.
Step 3: Join the points to form a line.
5-3

27. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
At the local video store you can rent a video game for $3. It costs you $5 a month to operate your video game player. The total monthly cost C(v) depends on the number of video games v you rent. Use the function rule C(v) = 5 + 3v to make a table of values and a graph.
v C(v) = 5 + 3v (v, C(v))
0 C(0) = 5 + 3(0) = 5 (0, 5)
1 C(1) = 5 + 3(1) = 8 (1, 8)
2 C(2) = 5 + 3(2) = 11 (2, 11)
5-3

28. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
a. Graph the function y = |x| + 2.
Make a table of values.
x y = |x| + 2 (x, y)
–3 y = |–3| + 2 = 5 (–3, 5)
–1 y = |–1| + 2 = 3 (–1, 3)
  0 y = |0| + 2 = 2 (0, 2)
  1 y = |1| + 2 = 3 (1, 3)
3 y = |3| + 2 = 5 (3, 5)
Then graph the data.
5-3

29. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
(continued)
b. Graph the function ƒ(x) = x2 + 2.
x ƒ(x) = x2 + 2 (x, y)
–2 ƒ(–2) = = 6 (–2, 6)
–1 ƒ(–1) = = 3 (–1, 3)
0 ƒ(0) = = 2 ( 0, 2)
1 ƒ(1) = = 3 ( 1, 3)
2 ƒ(2) = = 6 ( 2, 6)
Make a table of values.
Then graph the data.
5-3

30. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
pages 249–252  Exercises
1. C
2. A
3. B
4–12. Tables may vary.
Samples are given. 4. 5. 6. 7. 8.
5-3

31. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3 9.
10.
11.
12.
13. a. M = 3.5 h
b–d. Answers may vary.
Samples are given. b. c.
d. about 8.5 h
5-3

32. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
14. a. b.
15.
16.
17.
18.
19.
20.
5-3

33. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
21.
22.
23.
24. a.
b. P(t) = 2t + 2,
25. Answers may vary. Sample:
Make a table to find values
for ƒ(x) when x = –2, 0,
and 2. Then graph the ordered pairs (x, ƒ(x)) and join the graphed points with a line.
26.
5-3

34. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
27.
28.
29.
30.
31.
32.
33.
34.
5-3

35. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
35.
36.
37.
38. a gal b.
c. Check students’ work.
d. Check students’ work.
39. a. dependent variable
b. Input is to output as
domain is to range.
40. a. b.
41. a. $.71
b. about 12 min
5-3

36. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
42. a.
b. x-axis
c. y = –|x| – 1
43. B
44. a.
b. It changes the y-intercept.
45. a.
b. It makes the graph wider or narrower.
46. a. 1, 1, –1, –1
b. {–1, 0, 1}
c. Tables may vary.
Sample:
d. No; s(3 + 5) = s(8) = 1 and
s(3) + s(5) = = 2; 1 = 2. /
47. C
48. F
49. B
50. I
5-3

37. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
51. [2]
The graphs of all equations go
through the point (0, 3). y = x + 3
and y = x2 + 3 share points on the
right side of the y-axis. Both sides
of the graphs y = x2 + 3 and y = |x|+ 3
go upward, but y = |x| + 3 has straight lines
and y = x2 + 3 has curved lines.
[1] at least two of the graphs drawn correctly
52. [4] a. Tables may vary. Sample: b.
[3] appropriate methods but one
calculation error
[2] correct table but an error
in graph
[1] table with no graph OR graph
with no table
5-3

38. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
53. {–5, 1, 11.5}
54. {–11, –5, 5.5}
55. {–6.5, 4, 10}
56. {–5, –3, –1.5}
57. {8.5, 12, 14}
58. {–20, –10, 7.5}
59. {–11, 1, 22}
60. {–9, –8, –6.25}
61. {–29.5, 9, 31}
62. 4, – 4
63. No sol.; 9 = 10 – 1
and |b| cannot be –1.
64. –3.6, 3.6
65. –7, 7
66. –7, 7
67. 4, –2
68. –7, –15
, –0.5
70. –1, –7 km km km km km
76. a. = , ft
b. = , ft
c. = , ft 7 8 x 29
35.833
18.5
5-3

39. Function Rules, Tables, and Graphs
ALGEBRA 1 LESSON 5-3
1. Model y = –2x + 4 with a table of values and a graph.
x y = –2x + 4 (x, y)
–1 y = –2(–1) + 4 = 6 (–1, 6)
0 y = –2(0) + 4 = 4 (0, 4)
1 y = –2(1) + 4 = 2 (1, 2)
2 y = –2(2) + 4 = 0 (2, 0)
2. Graph y = |x| – 2.
3. Graph ƒ(x) = 2x2 – 2.
5-3

40. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
(For help, go to Lesson 5-3.)
Model each rule with a table of values.
1. f(x) = 5x – 1 2. y = –3x g(t) = 0.2t – 7
4. y = 4x f(x) = 6 – x 6. c(d) = d + 0.9
Evaluate each function rule for n = 2.
7. A(n) = 2n – 1 8. f(n) = –3 + n – 1 9. g(n) = 6 – n
5-4

41. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
Solutions 1. 2. 3. 4. 5. 6.
5-4

42. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
Solutions (continued)
7. A(n) = 2n – 1 for n = 2:
A(2) = 2(2) – 1 = 4 – 1 = 3
8. f(n) = –3 + n – 1 for n = 2:
f(2) = –3 + 2 – 1 = –1 – 1 = –2
9. g(n) = 6 – n for n = 2:
g(2) = 6 – 2 = 4
5-4

43. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
Write a function rule for each table.
x ƒ(x)
2 8
4 10
6 12
8 14 a.
Ask yourself, “What can I do to 2 to get 8, 4 to get 10, ...?”
You add 6 to each x-value to get the ƒ(x) value.
Relate:    equals plus 6
Write: =
ƒ(x) x
A rule for the function is ƒ(x) = x + 6.
5-4

44. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
(continued) b.
x y
1 2
2 5
3 10
4 17
Ask yourself, “What can I do to 1 to get 2, 2 to get 5, ?”
You multiply each x-value times itself and add 1 to get the y value.
Relate:    equals plus 1
Write: = x2 y
x times itself
A rule for the function is y = x2 + 1.
5-4

45. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
The journalism class makes $25 per page of advertising in the yearbook. If the class sells n pages, how much money will it earn?
a. Write a function rule to describe this relationship.
Relate:    is  25  times  
money earned
number of pages sold
Define: Let = number of pages sold. n
Let = money earned.
P(n)
Write: = 25 •
The function rule P(n) = 25n describes the relationship between the number of pages sold and the money earned.
5-4

46. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
(continued)
b. The class sold 6 pages of advertising. How much money did the class make?
P(6) = 25 • 6 Substitute 6 for n.
P(6) = 150 Simplify.
P(n) = 25 • n
The class made $150.
5-4

47. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
The choir spent $100 producing audio tapes of its last performance and will sell the tapes for $5.50 each. Write a rule to describe the choir’s profit as a function of the number of tapes sold.
Relate: is $5.50 times minus cost of tape production
total profit
tapes sold
Define: Let = number of tapes sold. t
Let = total profit.
P(t)
Write: = • – 100
The function rule P(t) = 5.5t – 100 describes the profit as a function of the number of tapes sold.
5-4

48. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
pages 256–259  Exercises
1. B
2. A
3. C
4. ƒ(x) = 3x
5. ƒ(x) = x – 0.5
6. ƒ(x) = 0.5x
7. ƒ(x) = –3x
8. y = 4x
9. y = x 2
10. t(c) = 0.79c
11. d(n) = 45n
12. ƒ(h) = h
13. e(n) = 6.37n
14. A(n) = n2
15. V(n) = n3
16. A(r) = r 2
17. a.ƒ(x) = 0.19x
b. $1.52
18. a. ƒ(x) = (x – 1)
b. $.97
19. ƒ(x) = 1000x
20. ƒ(x) = 2.54x 1 12
21. a. C(a) = 10a + 1
b. $31
c. 61; the total cost of
12 books
22. a. C(b) = 6b
b. 72; the total cost of
c. about $5.08
d. Club; $61 is less
than $72.
23. Answers may vary.
Sample: The input
values you need
may not be in the table.
5-4

49. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
24. a. gal of water, number
of loads
b. w(n) = 34n
c. 238 gal
d. 13 loads
25–28. Tables may vary.
Samples are given.
25.
ƒ(x) = x
26.
y = x + 3
27.
y = –x + 2
28.
y = x
29. Answers may vary. Sample:
ƒ(x) = 60x; ƒ(3) = 180,
180 mi in 3 h, the distance
you can travel at a constant
speed of 60 mi/h
30. a. 0.15b
b. c(b) = 1.15b
c. $20.70
31. ƒ(x) = x3
32. ƒ(x) = –x3
33. ƒ(x) = –x3 – 1
34. a. c(m) = m
b. $70.60, $89.60
c. 38 mi
d. $202 3 2 1 2 2 3
5-4

50. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
35. a. B(v) = 6.93v
b. B(w) = w
36. A
37. H
38. D
39. B
40. C
41. [2] D = 0.1w 2 + 5w
D = 0.1(60)2 + 5(60)
D = 0.1(3600) + 300
D = 660  
The dosage is 660 mg.
[1] correctly substitutes
60 for w but makes a
minor calculation error
42–47. Tables may vary. Samples are given.
42.
43. 7 10
5-4

51. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
44.
45.
46.
47.
48. 17%; increase
49. 2%; increase
50. 8%; decrease
51. 83%; decrease
52. 29%; increase
53. 42%; decrease
54. a. 2 oranges;
3 oranges
b. 28 tbsp
5-4

52. Writing a Function Rule
ALGEBRA 1 LESSON 5-4
1. Write a function rule for each table.
x y
–1 –5
0 0
1 5
2 10 a.
x ƒ(x)
–1 –4
0 –3
1 –2
2 –1 b.
y = 5x
ƒ(x) = x – 3
2. Write a function rule to describe each relationship.
a. the total cost T(c) of c pounds of apples at $.82 a pound
b. a scale model s of a moth m that is 6 times the actual size of the moth
T(c) = 0.82c
s(m) = 6m
3. You borrow $60 to buy a bread-making machine. You charge customers $1.50 a loaf for your special bread. Write a rule to describe your profit as a function of the number n of loaves sold.
P(n) = 1.5n - 60
5-4

53. Solve each equation for the given variable.
Direct Variation
ALGEBRA 1 LESSON 5-5
(For help, go to Lessons 2–7 and 4–1.)
Solve each equation for the given variable.
1. nq = m; q 2. d = rt; r 3. ax + by = 0; y
Solve each proportion.
= = =
= = = 5 8 x 12 4 9 n 45 25 15 y 3 7 n 35 50 8 d 20 36 14 18 63 n
5-5

54. Direct Variation Solutions 1. nq = m 2. d = rt 3. ax + by = 0 = =
ALGEBRA 1 LESSON 5-5
Solutions
1. nq = m 2. d = rt
= =
q = = r
r = nq n m d t rt
ax + by = 0
ax – ax + by = 0 – ax
by = –ax =
y = – by b
–ax ax =
8x = 5(12)
8x = 60
x = 7.5 5 8 x 12
= =
9n = 4(45) 15y = 25(3)
9n = y = 75
n = y = 5 4 9 n 45 25 15 y 3
5-5

55. Solutions (continued)
Direct Variation
ALGEBRA 1 LESSON 5-5
Solutions (continued)
= =
35n = 7(50) 20d = 8(36)
35n = d = 288
n = d = 14.4 7 n 35 50 8 d 20 36 14 18 63 =
14n = 18(63)
14n = 1134
n = 81
5-5

56. –3y = 1 – 2x Subtract 2x from each side.
Direct Variation
ALGEBRA 1 LESSON 5-5
Is each equation a direct variation? If it is, find the constant of variation.
a. 2x – 3y = 1
–3y = 1 – 2x Subtract 2x from each side.
y = – x Divide each side by –3. 1 3 2
The equation does not have the form y = kx. It is not a direct variation.
b. 2x – 3y = 0
–3y = –2x Subtract 2x from each side.
y = x Divide each side by –3. 2 3
The equation has the form y = kx, so the equation is a direct variation. The constant of variation is . 2 3
5-5

57. y = kx Use the general form of a direct variation.
ALGEBRA 1 LESSON 5-5
Write an equation for the direct variation that includes the point (–3, 2).
y = kx Use the general form of a direct variation.
2 = k(–3) Substitute –3 for x and 2 for y.
– = k Divide each side by –3 to solve for k. 2 3
y = – x Write an equation. Substitute – for k in y = kx. 2 3
The equation of the direct variation is y = – x . 2 3
5-5

58. Relate: The weight varies directly with the mass. When x = 6, y = 59.
Direct Variation
ALGEBRA 1 LESSON 5-5
The weight an object exerts on a scale varies directly with the mass of the object. If a bowling ball has a mass of 6 kg, the scale reads 59. Write an equation for the relationship between weight and mass.
Relate: The weight varies directly with the mass.
When x = 6, y = 59.
Define: Let = the mass of an object.
Let = the weight of an object. x y
5-5

59. Write: = k Use the general form of a direct variation.
ALGEBRA 1 LESSON 5-5
(continued)
Write: = k Use the general form of a direct variation.
59 = k(6) Solve for k. Substitute 6 for x and 59 for y. x y
= k Divide each side by 6 to solve for k. 59 6
y = x Write an equation. Substitute for k in y = kx. 59 6
The equation y = x relates the weight of an object to its mass. 59 6
5-5

60. same for each pair of data.
Direct Variation
ALGEBRA 1 LESSON 5-5
For the data in each table, tell whether y varies directly with x. If it does, write an equation for the direct variation. 1 –2
= –0.5 –1 2 4 y x
x y
–2 1
2 –1 4 –2 2 –1
= –2 1 –4
= 2 y x
x y
–1 2 –4
No, the ratio is not the
same for each pair of data. x y
Yes, the constant of variation is –0.5. The equation is y = –0.5x.
5-5

61. Define: Let n = the force you need to lift 210 lb.
Direct Variation
ALGEBRA 1 LESSON 5-5
Suppose a windlass requires 0.75 lb of force to lift an object that weighs 48 lb. How much force would you need to lift 210 lb?
Relate: The force of 0.75 lb lifts 48 lb. The force of n lb lifts 210 lb.
Define: Let n = the force you need to lift 210 lb.
Write: = Use a proportion.
force1
weight1
force2
weight2
Substitute 0.75 for force1, 48 for weight1, and 210 for weight2.
0.75 48 n
210 =
0.75(210) = 48n
Use cross products.
Solve for n. n
You need about 3.3 lb of force to lift 210 lb.
5-5

62. 22–23. Choices of variables may vary. 22. P( ) = 8 23. E(h) = 7.10h
Direct Variation
ALGEBRA 1 LESSON 5-5
pages 264–267  Exercises
1. no
2. no
3. yes; –2
4. no
5. yes;
6. yes;
7. yes; –
8. yes; 0.5
9. yes; –
10. y = 5x
11. y = x
22–23. Choices of variables may vary.
22. P( ) = 8
23. E(h) = 7.10h
24. yes; y = 1.8x
25. no
26. yes; y = –1.5x
27. a. or
b. = , 52 lb
28. = , 9 mi
29. y = x
30. y = –20x
12. y = – x
13. y = x
14. y = – x
15. y = – x
16. y = – x
17. y = – x
18. y = – x
19. y = 2x
20. y = – x
21. y = x 5 4 9 5 3 2 1 6 5 6 4 3 7 3 50 20 5 2 4 3 1 10 50 20
130 x 4 3 10 3 30 x 3 2 1 6 2 3 1 5 1 5
5-5

63. b. A line through the origin that is neither vertical nor
Direct Variation
ALGEBRA 1 LESSON 5-5
31. y = – x
32. y = 6x
33. y = 9x
34. y = – x
35. y = – x
36. y = x
37. a. The ratio is the
same for each pair
of values.
b. A line through the origin
that is neither vertical nor
horizontal is the graph of
a direct variation. 36 25
38. True; a line that is neither horizontal
nor vertical can pass through
(0, 0) and (–2, 4).
39. False; the line through (0, 3) and (0, 0)
is vertical, so it is not a function and is
therefore not a direct variation.
40. True; for the equation y = kx, if one side
is multiplied by 3, then the other side must
be multiplied by 3.
41.
y = x 1 32 15 52 27 64 y x 5 2
5-5

64. Direct Variation 44. y = x 45. a. b. b = w c. Check students’ work.
ALGEBRA 1 LESSON 5-5
44.
y = x
45. a.
b. b = w
c. Check students’ work.
46. a. 48 volts
b ohms
47. 12
48. –8
49. 8
50. –6
51. 5
52. 2
53. a. c = 1.83g; yes
b. c = m or
c = m
54. A
55. H
56. A
57. B
58. C
42.
y = – x
43. 5 2 5 2
1.83 24 1 32 1 32 5 2
5-5

65. [1] correct answer but no work shown 60. y = 3x 61. y = x
Direct Variation
ALGEBRA 1 LESSON 5-5
59. [2] y = kx
–4 = k(–1)
4 = k
y = 4x
[1] correct answer but no work shown
60. y = 3x
61. y = x
62. y = 12 – x
63. y = x
64. r > –18
65. c –1.8
66. m < –3
67. a
68. n < –80
69. t
70. v –
71. b >
72. 15,600 ships 5 2
> –
> – 1 2 1 3 1 6
< –
> –
5-5

66. Direct Variation
ALGEBRA 1 LESSON 5-5
1. Is each equation a direct variation? If it is, find the constant of variation.
a. x + 5y = 10
b. 3y + 8x = 0 no
yes; – 8 3
2. Write an equation of the direct variation that includes the point (–5, –4).
y = x 4 5
3. For each table, tell whether y varies directly with x. If it does, write an equation for the direct variation.
x y
–1 3
0 0
1 –6
2 –9 a.
x y
–1 –2
0 0
1 2
3 –6 b.
yes; y = –3x no
5-5

67. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
(For help, go to Lessons 1–2 and 1–5.)
Evaluate each expression for x = 2, 3, 4.
(x – 1) (x – 1) – 3(x – 1)
Subtract.
4. 8 – (–6) 5. –7 – – 3.4
5-6

68. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
Solutions
(x – 1) for x = 2: 9 + 3(2 – 1) = 9 + 3(1) = = 12 for x = 3: 9 + 3(3 – 1) = 9 + 3(2) = = 15 for x = 4: 9 + 3(4 – 1) = 9 + 3(3) = = 18
(x – 1) for x = 2: 8 + 7(2 – 1) = 8 + 7(1) = = 15 for x = 3: 8 + 7(3 – 1) = 8 + 7(2) = = 22 for x = 4: 8 + 7(4 – 1) = 8 + 7(3) = = 29
– 3(x – 1) for x = 2: (2 – 1) = 0.4 – 3(1) = 0.4 – 3 = –2.6 for x = 3: (3 – 1) = 0.4 – 3(2) = 0.4 – 6 = –5.6 for x = 4: (4 – 1) = 0.4 – 3(3) = 0.4 – 9 = –8.6
4. 8 – (–6) = = 14
5. –7 – 10 = –7 + (–10) = –17
– 3.4 = (–3.4) = –1.9
5-6

69. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
Use inductive reasoning to describe the pattern. Then find the next two numbers in the pattern.
a.  1, 5, 9
The pattern is “add 4 to the previous term.” +4
To find the next two numbers, you add 4 to each previous term: = 13 and = 17.
b.  1, 3, 9
The pattern is “multiply the previous term by 3.”
 3
To find the next two numbers, you multiply each previous term by 3: 9  3 = 27 and 27  3 = 81.
c. 1, 9, 25, 49, . . .
The pattern is “square of consecutive odd integers.”
To find the next two numbers, square the next two consecutive integers: 92 = 81, 112 = 121.
5-6

70. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
Find the common difference of each arithmetic sequence.
a. 5, 2, –1, –4
b. 8, 11, 14, 17
The common difference is –3. –3
The common difference is 3. +3
5-6

71. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
Find the first, fifth, and tenth terms of the sequence that has the rule A(n) = 15 + (n – 1)(5).
first term: A(1) = 15 + (1 – 1)(5) = (5) = 15
fifth term: A(5) = 15 + (5 – 1)(5) = (5) = 35
tenth term: A(10) = 15 + (10 – 1)(5) = (5) = 60
5-6

72. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
pages 270–273  Exercises
1. “Add 2 to the previous term”; 12, 14.
2. “Multiply the previous term by 1 ”; 20 ,
3. “Add 2 to the first term, 3 to the second term
and continue, adding 1 more each time”; 18, 24.
4. “Add 0.04 to the previous term”; 3.16, 3.20.
5. “Multiply the previous term by 1.1”; ,
6. “Add –2 to the previous term”; –5, –7.
7. “Add 1.1 to the previous term”; 5.5, 6.6.
8. “Multiply the previous term by 10”; 10, 100.
9. “Multiply the previous term by 4”; 512, 2048.
10. “Square the reciprocals of consecutive integers”; , .
11. “Add –14 to the previous term”; –47, –61.
12. “Multiply the previous term by 5”; 937.5,
13. 3
14. –4
15. –11
16. 13
17. –
19. –2
20. 12
21. 5
22. 5, 14, 26
23. –3, 15, 39
24. –3, 9, 25 1 2 1 4 3 8 1 6 1 25 1 36
5-6

73. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
25. 17, 44, 80
, 12.5, 24.5
27. 2, 23, 51
28. 3, –15, –39
29. –7.1, –22.1, –42.1
30. 58, 37, 9
31. 17, 5, –11
32. –8, –17, –29
33. –0.8, –3.8, –7.8
34. –4, –10
, 3
36. 26, 37 ,
38. 35, 48
39. 31, 40
, 1.25
41. 8, 8
, –
43. a. Answers may vary.
Sample: Inductive
reasoning is making
conclusions based on
patterns, while deductive
reasoning is making conclusions
based on given facts.
b. Answers may vary.
Check students’ work. 4 27 4 81
44. 5 min
45. Answers may vary.
Sample: A(n) = 2 – 4n
46. 7 lb 4 oz, 7 lb 9 oz,
7 lb 14 oz, 8 lb 3 oz,
8 lb 8 oz; the baby’s weight
at the end of the 4th week
47. $4500, $4350, $4200,
$4050, $3900; the balance
after 4 payments 1 4 4 27 4 81 1 4 1 2
5-6

74. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
, 12, 13
, –4.5, –22.5
58. –2, –5.2, –11.6
59. 1, 2 , 5
60. a. 11, 14 b.
c. The points lie on a line. 1 3 1 3
48. a. 1   2   4; 7
b. = 2; = 2; 8
c. When there are more than three
terms you can test the pattern to
make sure it is reasonable.
49. No; there is no common difference.
50. Yes; the common difference is –4.
51. No; there is no common difference.
52. No; there is no common difference.
53. Yes; the common difference is –15.
54. Yes; the common difference is –0.8.
55. a. 1, 5, 10, 10, 5, 1
b. 1, 2, 4, 8, 16; 32 ) ) 2 1 4 2 3 5 4 5
5-6

75. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
66. value of new term =
value of previous term + 4
67. value of new term =
value of previous term ÷ 7
68. value of new term =
value of previous term • (–2.5)
69. x; 4x + 4
70. 3a + 2b; 10a + 7b + c
71. a. 10
b. –6
c. A(n) = 10 + (n – 1)(–6)
61. a. Yes; for each input there
is only one output value.
b. For every increase of 7 in
the key position,
the frequency doubles.
62. a. 21
b. 89
c. Answers may vary.
Sample: 3, 3, 6, 9, 15, 24, 39
63. value of new term =
value of previous term + 6
64. value of new term =
value of previous term • 1.5
65. value of new term =
value of previous term – 2.5
5-6

76. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
72. a.
b. Blue; the colors rotate red, blue, and purple. Every third figure is purple.
Since 21 is divisible by 3, the 21st figure is purple. The figure just before
a purple figure is blue.
c. 12 sides; the figures show this pattern for number of sides.
Figure  Number of Sides
  1–3 3
  4–6 4
  7–9 5
10–12 6
13–15 7
16–18 8
19–21 9
22–24 10
25–27 11
28 12
5-6

77. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
73. a. –5
b. 6
c. A(n) = –5 + (n – 1)(6)
74. C
75. F
76. A
77. H
78. C
79. H
80. [2] The common difference is –3.
The seventh term is
24 + (7 – 1)(–3) = 6.
[1] explanation incomplete
81. [4] a. ƒ(n) = 26, n
b. 2008 is 7 years after 2001.
ƒ(7) = 26,500 + (2880)(7) = $46,660
Marta’s 2008 salary is $46,660.
[3] function and salary found but
work not shown
[2] minor computation error in
finding salary
[1] function shown but salary not found
82. y = – x
83. y = 24x
84. y = –14x
85. y = 0.14x
86. y = – x
87. y = – x 5 4 31 11 31
110
5-6

78. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
88. y = – x
89. y = x
90. {–20, –4, 8}
91. {–19, –3, 9}
92. {–2, 7, 19}
93. {2, 4, 10}
94. {2, 4, 10}
95. {–6 , –4 , –1 }
96. a. about 84%
b. about 11% 3 2 2 3 1 2 1 4 1 4
5-6

79. Describing Number Patterns
ALGEBRA 1 LESSON 5-6
1. Use inductive reasoning to describe each pattern. Then find the next two numbers in each pattern.
a. 1, 2.5, 4, ...    b.  , , , ... 1 2 1 4 1 8
add 1.5 to the previous term; 5.5, 7 1 2
multiply the previous term by ; 16 32 ,
2. Find the common difference of each arithmetic sequence.
a. –1, – , – , 0, ...    b. 46, 34, 22, 10, ... 2 3 1 3 1 3
–12
3. Find the second, sixth, and ninth terms of the sequence that has the rule A(n) = –3 + (n – 1)(6).
3, 27, 45
4. Is –2, 3, 8, 10, ... an arithmetic sequence? Explain.
No; the difference between the first two terms is 5, but the difference between the fourth and third terms is 2.
5-6

80. 1–2. Answers may vary. Samples: 1.
Graphs and Functions
ALGEBRA 1 CHAPTER 5
1–2. Answers may vary. Samples: 1. 2.
3. yes; domain: {–2, 3, 5, 8}; range: {5, 6, 12}
4. no
5. If any vertical line crosses the graph in
two or more places, the graph does not
represent a function.
6. {5, 9, 14, 41, 69}
7. {–14, –5, –2, 2.5, 7} 8.
5-A

81. 15. Answers may vary. Sample: x = number of cars washed
Graphs and Functions
ALGEBRA 1 CHAPTER 5
15. Answers may vary. Sample:
x = number of cars washed
at $5 each and y = money earned.
16. a. p(t) = 0.59t
b. $8.26
c. about 16.9 lb
17. y = x
18. y = x
19. y = – x
20. y = – x
21. yes; y = 3x
22. no
23. 9.
10–12. Choice of variables may vary.
10. c(d) = 0.038d
11. m(n) = 15n
12. p(n) = 0.7n
13. y = 2x + 1
14. ƒ(x) = –4.5x 1 2 1 3 3 5 13 10
5-A

82. Graphs and Functions 24. 4.5 25. –1 26. 150 mL 27. 5; –35, –30, –25
ALGEBRA 1 CHAPTER 5
25. –1 mL
27. 5; –35, –30, –25
28. 1; 5.7, 6.7, 7.7
29. –8
30. 3
31. No; there is no
common difference.
32. Yes; the common
difference is 0.25.
33. C(p) = 3p
5-A