rate of change and slope evaluate each function rule for x = –5. 1. y = x – 72. y = 7 – x 3. y...
TRANSCRIPT
Rate of Change and SlopeRate of Change and Slope
Evaluate each function rule for x = –5.
1. y = x – 7 2. y = 7 – x
3. y = 2x + 5 4. y = – x + 3
Write in simplest form.
5. 6. 7.
8. 9. 10.
25
7 – 33 – 1
3 – 56 – 0
8 – (–4) 3 – 7
–1 – 2 0 – 5
–6 – (–4) –2 – 6
0 – 11 – 0
(For help, go to Lessons 5-2 and 1-5.)
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
1. y = x – 7 for x = –5: y = –5 – 7 = –12
2. y = 7 – x for x = –5: y = 7 – (–5) = 12
3. y = 2x + 5 for x = –5: y = 2(–5) + 5 = –10 + 5 = –5
4. y = – x + 3 for x = –5: y = – (–5) + 3 = 2 + 3 = 5
5. = = 2 6. = – = –
7. = – = –3 8. = =
9. = = 10. = – = –1
7 – 33 – 1
3 – 56 – 0
8 – (–4) 3 – 7
–1 – 2 0 – 5
–6 – (–4) –2 – 6
0 – 11 – 0
25
25
42
26
13
12 4
–2–8
14
11
–3–5
35
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Solutions
Find the rate of change for each pair of consecutive mileage amounts.
For the data in the table, is the rate of change the
same for each pair of consecutive mileage amounts?
Fee for Miles Driven
Miles Fee
100 $30
150
200
250
$42
$54
$66
Rate of Change and SlopeRate of Change and SlopeALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
The rate of change for each pair of consecutive mileage amounts is $12 per 50 miles. The rate of change is the same for all the data.
(continued)
42 – 30 12 54 – 42 12 66 – 54 12 150–100 50 200–150 50 200–250 50
= = =
Rate of Change and SlopeRate of Change and SlopeALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
change in cost Cost depends on the change in number of miles number of miles.rate of change =
6-1
Rate of Change and SlopeRate of Change and Slope
Below is a graph of the distance traveled by a motorcycle from
its starting point. Find the rate of change. Explain what this rate of
change means.
Choose two pointsOn the graph (0,0) and (400,20)
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
The rate of change is 20 m/s.The motorcycle is traveling 20 meters each second.
vertical change change in distance horizontal change change in time
rate of change =
(continued)
Using the points (0,0) and (400,20), find the rate of change.
400 – 0
20 – 0
400
20
20
=
=
=
Use two points.
Divide the vertical change by the horizontal change.
Simplify.
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
The slope of the line is – .32
slope =riserun
Find the slope of each line.
a.
= – 32
= 3 –2
4 – 1
0 – 2
=
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
The slope of the line is 2.
slope =riserun
b. Find the slope of the line.
(continued)
= 2
= –2–1
–1 – 1
–2 – (–2)
=
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
slope =y2 – y1
x2 – x1
The slope of EF is – . 15
Find the slope of each line through E(3, –2) and F(–2, –1).
= – 15
= 1–5
Substitute (–2, –1) for (x2, y2) and(3, –2) for (x1, y1).
Simplify.
–1 – (–2) –2 – 3
=
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
2 – 2
1 – (–4)
=
= 0 5
Substitute (1, 2) for
(x2, y2) and (–4, 2)
for (x1, y1).
Simplify.
= 0
slope =y2 – y1
x2 – x1
The slope of the horizontal line is 0.
Find the slope of each line.
a.
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
Division by zero is undefined.
The slope of the vertical line in undefined.
b.
(continued)
Find the slope of the line.
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
= 5 0
Substitute (2, 1) for (x2, y2) and (2, –4) for (x1, y1).
= 1 – (–4)
2 – 2
Simplify.
slope =y2 – y1
x2 – x1
pages 286–289 Exercises
1. 3; the temperature increases 3°F each hour.
2. 3.95; the cost per person is $3.95.
3. – gal/mi
4. ; there are 2 lb of carbon
emissions for 3 h of television use.
5. –16 ; the skydiver descends 16 ft/s.
6. ; the cost of oregano is $1 for 4 ounces.
7.
8. –3
9.
2 3
2 3
2 31
41 2
2 3
10. 2
11. 2
12.
13. –
14. –1
15. –1
16.
17.
18. –
19. –
3 4 3
2
5 95 3 6
5
Rate of Change and SlopeRate of Change and Slope
1 15
20.
21. –5
22. 0
23. undefined
24. 0
25. undefined
26. 0
27. in./month
28. $5/person
29. 30 mi/hr
30.2 5
1 2
1 2
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
9 10
41. a.
b.
c. Answers may vary.
Sample: =
42. 2 ; 3
43. No; for example the line
passing through the points
such as (1, 6) and (2, 5) has
a slope of –1.
44. PQ: ; QR: 0; RS: 5; SP: –1
45. JK: – ; KL: 2; ML: – ;MJ: 2
y2 – y1
x2 – x1
y1 – y2
x2 – x125
35
38.
39.
40. a. C
b. C greatest;
A least; the slope
2323
31.
32.
33. –20
34. undefined
35. –3
36.
37.
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
Rate of Change and SlopeRate of Change and Slope
6-1
1 61 4
1212
12
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
Rate of Change and SlopeRate of Change and Slope
6-1
46. a.
b. c.
d. equal
47. a. Answers may vary.
Sample: (0, 0), (1,
)
b. Answers may vary.
Sample: (0, 0), (1,
– )
48. 0
49. –6
50. 6
51. 4
52. 12
53. 3
54–60. Counter examples may vary.
54. False; it can be neg.
or undefined.
55. true
56. False; y = x + 2
57. true
58. False; y = x
59. False; y = 0x
60. true
61. a.111; $111/h
b. 52 customers per h
34
23
23
12
69. No; PQ and QR do not
have the same slope.
70. Yes; GH and HI have
the same slope.
71. No; ST and TU do not
have the same slope.
72. C
73. G
74. [2] = ; the slope is .
= 12.5%; the grade
is 12.5%.
[1] finds grade only
62. Friend found instead
of .
63. 0
64. –
65.
66. Yes; AB and BC have
the same slope.
67. Yes; GH and HI have
the same slope.
68. No; DE and EF do not
have the same slope.
n 2m
2d – bc – 2a
runrise
75. A
76. c = 3.5n
77. p = 4.95q – 232
78. 0
79.
80.
81. 1
82. –3
83. –4
84.
85. 7
86. 5
87. –10
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
Rate of Change and SlopeRate of Change and Slope
6-1
riserun
3 24
18
18
18
1213
199
Rate of Change and SlopeRate of Change and Slope
1. Find the rate of change for the data in the table.
5 $75
6
7
8
$90
$105
$120
Tickets Cost
2. Find the rate of change for the data in the graph.
3. Find the slope of the line.
4. Find the slope of the line through (3, –2) and (–2, 5).
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
5. State whether the slope is zero or undefined.
a. b.
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
$15 per ticket400 calories per hour 4
3
75
–
1. Find the rate of change for the data in the table.
5 $75
6
7
8
$90
$105
$120
Tickets Cost
2. Find the rate of change for the data in the graph.
3. Find the slope of the line.
4. Find the slope of the line through (3, –2) and (–2, 5).
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
Rate of Change and SlopeRate of Change and Slope
undefined 0
5. State whether the slope is zero or undefined.
a. b.
ALGEBRA 1 LESSON 6-1ALGEBRA 1 LESSON 6-1
6-1
(For help, go to Lessons 1-6 and 2-6.)
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
Slope-Intercept FormSlope-Intercept Form
6-2
Evaluate each expression.
1. 6a + 3 for a = 2 2. –2x – 5 for x = 3
3. x + 2 for x = 16 4. 0.2x + 2 for x = 15
Solve each equation for y.
5. y – 5 = 4x 6. y + 2x = 7
7. 2y + 6 = –8x
14
Slope-Intercept FormSlope-Intercept Form
1. 6a + 3 for a = 2: 6(2) + 3 = 12 + 3 = 15
2. –2x – 5 for x = 3: –2(3) – 5 = –6 – 5 = –11
3. x + 2 for x = 16: (16) + 2 = 4 + 2 = 6
4. 0.2x + 2 for x = 15: 0.2(15) + 2 = 3 + 2 = 5
5. y – 5 = 4x 6. y + 2x = 7 y – 5 + 5 = 4x + 5 y + 2x – 2x = 7 – 2x y = 4x + 5 y = –2x + 7
7. 2y + 6 = –8x 2y + 6 – 6 = –8x – 6 2y = –8x – 6
=
y = –4x – 3
–8x – 6 2
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
Solutions
6-2
14
14
2y y
Slope-Intercept FormSlope-Intercept Form
The slope is 2; the y-intercept is –3.
What are the slope and y-intercept of y = 2x – 3?
mx + b=
2x – 3=
Use the Slope-Intercept form.
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
6-2
25
Use the slope-intercept form.
Substitute for m and 4 for b.
y = mx + b
y = x + 425
Slope-Intercept FormSlope-Intercept Form
Write an equation of the line with slope and
y-intercept 4.
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
6-2
25
Slope-Intercept FormSlope-Intercept Form
23
y = mx + b
y = – x + 1
Substitute – for m and 1 for b.
23
Step 2 Write an equation in slope–
intercept form. The y-intercept is 1.
Write an equation for the line.
Step 1 Find the slope. Two points on
the line are (0, 1) and (3, –1).
–1 – 1 3 – 0
slope =
= – 23
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
6-2
Step 2
The slope is . Use slope
to plot a second point.
13
Slope-Intercept FormSlope-Intercept Form
Step 1
The y-intercept is –2.
So plot a point at (0,–2).
Step 3
Draw a line through
the two points.
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
Graph y = x – 2.13
6-2
Slope-Intercept FormSlope-Intercept Form
The base pay for a used car salesperson is $300 per week.
The salesperson also earns 15% commission on sales made. The
equation t = 300 + 0.15s relates total earnings t to sales s. Graph the
equations.
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
6-2
Step 1 Identify the slope and y-intercept.
t = 300 + 0.15s
t = 0.15s + 300 Rewrite the equation in slope intercept form.
slope y-intercept
Step 2 Plot two points. First plot a point at the y-intercept. Then use the
slope to plot a second point. The slope is 0.15, which equals .
Plot a second point 15 units above and 100 units to the right of
the y-intercept.
Slope-Intercept FormSlope-Intercept Form
(continued)
Step 3 Draw a line through the points.
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
6-2
15 100
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
Slope-Intercept FormSlope-Intercept Form
6-2
pages 294–296 Exercises
1. –2; 1
2. – ; 2
3. 1; –
4. 5; 8
5. ; 1
6. –4; 0
7. –1; –7
8. –0.7; –9
9. – ; –5
10. y = x + 3
11. y = 3x +
12. y = x + 3
13. y = 1
14. y = –x – 6
15. y = – x + 5
16. y = 0.3x + 4
17. y = 0.4x + 0.6
18. y = –7x +
19. y = – x –
20. y = – x +
21. y = x +
12
54
23
34
29
29
92
23
13
22. y = – x + 1
23. y = x + 2
24. y = 2x – 2
25. y = x +
26. y = – x + 2.8
27. y = x –
28.
y = x + 4
15
25
14
54
83
23
23
12
12
25
54
12
12
34
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
Slope-Intercept FormSlope-Intercept Form
6-2
29.
y = x – 1
30.
y = –5x + 231.
y = 2x + 5
32.
y = x + 433.
y = –x + 2
34.
y = 4x – 3
35.
y = – x
36.
y = x – 3
37.
y = – x + 2
32
38.
y = – x + 4
39.
y = –0.5x + 240. a.
t = 14c – 4
b. $80
45
23
25
23
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
Slope-Intercept FormSlope-Intercept Form
6-2
51.
y = –2x
52.
y = – x – 2
53.
y = 5x – 6
41. –3; 2
42. – ; 0
43. 9;
44. 3; –9
45. ; 3
46. 9; –15
47. c; d
48. 2 – a; a
49. –3; –2n
50.
y = 7 – 3x
54.
y = – x –
55.
y = 6x – 856. The slope was used for
the y-int., and the y-int. was used for the slope.
23
23
13
1212
32
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
Slope-Intercept FormSlope-Intercept Form
6-2
68. A; slope in A = > =
2 = slope in B.
69. y = 2x – 170. y = –4x + 7
71. y = – x + 8
72. y = x + 5
73. y = –x – 374. y = 3x – 675. a–b.
c. Both; check students’ work.
57. a.
b. h = – t + 12
c. 90 min58. a. Slope represents the
weight of a gallon of fuel.
b. 2662 lb59. no60. yes61. no62. III63. I64. II
84
104.5
65. a. d = 7pb. 84 dog years
66. a. t = 5d + 15b.
t = 5d + 15c. Answers may vary.
Sample: no neg. charges and no neg. number of days
67. Answers may vary. Sample: Plot point (0, 5), then move up 3 and right 4. Plot (4, 8) and connect the two points.
12
14
2 15
b. Rectangle; check students’ work.
c. y = x OR y = – x;
explanations may vary.
82. a. r = –15d + 265
b.
r = –15d + 265
c. 18 days
83. D
84. G
85. A
76. a. ;
b. 2; –2c. same slopes
77. Check students’ work.
78. –
79. –5
80.
81. a.
x = –2; x = 2y = 3; y = –3
14
14
12
34
32
32
86. [2]
slope: = = 2y-intercept: 3equation: y = 2x + 3
[1] correct graph and minor error in finding function
87. –
88. –
89.90. –191. 4.8 billion
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
Slope-Intercept FormSlope-Intercept Form
6-2
9756
92
5 – 31 – 0
Slope-Intercept FormSlope-Intercept Form
5. Write an equation for the line.
6. Graph y = –x + 3.
m = ; b = –334
m = –2; b = 0
y = – x – 435
y = 0.5x + 1
y = x – 312
ALGEBRA 1 LESSON 6-2ALGEBRA 1 LESSON 6-2
6-2
Find the slope and y-intercept of each equation.
1. y = x – 3
2. y = –2x
Write an equation of a line with the given slope and y-intercept.
3. m = – , b = –4
4. m = 0.5, b = 1
34
35
Solve each equation for y.
1. 3x + y = 5 2. y – 2x = 10 3. x – y = 6
4. 20x + 4y = 8 5. 9y + 3x = 1 6. 5y – 2x = 4
Clear each equation of decimals.
7. 6.25x + 8.5 = 7.75 8. 0.4 = 0.2x – 5
9. 0.9 – 0.222x = 1
Standard FormStandard Form
(For help, go to Lessons 2-3 and 2-6.)
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
6-3
1. 3x + y = 5 2. y – 2x = 10
3x – 3x + y = 5 – 3x y – 2x + 2x = 10 + 2x y = –3x + 5 y = 2x + 10
3. x – y = 6 4. 20x + 4y = 8 x = 6 + y 4y = –20x + 8
x – 6 = y y =
y = x – 6 y = –5x + 2
5. 9y + 3x = 1 6. 5y – 2x = 4 9y = –3x + 1 5y = 2x + 4
y = y =
y = – x + y = x +
–20x + 8 4
–3x+ 1 9
19
Standard FormStandard FormALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
6-3
Solutions
7. Multiply each term by 100:100(6.25x)+100(8.5) = 100(7.75)
Simplify:625x + 850 = 775
8. Multiply each term by 10:10(0.4) = 10(0.2x) – 10(5)
Simplify:4 = 2x – 50
9. Multiply each term by 1000:
1000(0.9)–1000(0.222x)=1000(1) Simplify:900 – 222x = 1000
13
2x+ 4 525
45
Find the x- and y-intercepts of 2x + 5y = 6.
Step 1 To find the x-intercept, substitute 0 for y and solve for x.
2x + 5y = 6
2x + 5(0) = 6
2x = 6
x = 3
The x-intercept is 3.
Step 2 To find the y-intercept, substitute 0 for x and solve for y.
2x + 5y = 6
2(0) + 5y = 6
5y = 6
y =
The y-intercept is .
65
65
Standard FormStandard FormALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
6-3
Step 2 Plot (5, 0) and (0, 3). Step 1 Find the intercepts.
3x + 5y = 15
3x + 5(0) = 15 Substitute 0 for y.
3x = 15 Solve for x.
x = 5
3x + 5y = 15
3(0) + 5y = 15 Substitute 0 for x.
5y = 15 Solve for y.
y = 3
Graph 3x + 5y = 15 using intercepts.
Standard FormStandard Form
Draw a line through the points.
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
6-3
a. Graph y = 4 b. Graph x = –3.
Standard FormStandard Form
0 • x + 1 • y = 4 Write in standard form.For all values of x, y = 4.
1 • x + 0 • y = –3 Write in standard form.For all values of y, x = –3.
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
6-3
The equation in standard form is –2x + 3y = 18.
–2x + 3y = 18 Subtract 2x from each side.
y = x + 623
23
3y = 3( x + 6 ) Multiply each side by 3.
3y = 2x + 18 Use the Distributive Property.
Write y = x + 6 in standard form using integers.23
Standard FormStandard FormALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
6-3
Define: Let = the hours mowing lawns.
Let = the hours delivering newspapers.
x
y
Write an equation in standard form to find the number of hours
you would need to work at each job to make a total of $130.
Amount Paid per hour
Mowing lawns $12
Deliveringnewspapers $5 Relate: $12 per h plus $5 per h equals $130
mowing delivering
Write: 12 + 5 = 130x y
The equation standard form is 12x + 5y = 130.
Standard FormStandard FormALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
6-3
Job
pages 301–303 Exercises
1. 18; 9
2. 3; –9
3. –6; 30
4. ; –3
5. – ; –
6. –8; 12
7. 6; 4
8. ; –2
9. –5; 4
10. B
11. C
12. A
13.
14.
15.
3 2 3
2
9 2
4 7
16.
17.
18.
19. horizontal
20. vertical
21. horizontal
22. vertical
23.
24.
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
Standard FormStandard Form
6-3
25.
26.
27. –3x + y = 128. 4x – y = 729. x – 2y = 630. –2x + 3y = 1531. –3x – 4y = 1632. –4x – 5y = 3533. –14x + 4y = 134. 4x + 10y = 135. 3x + y = 0
36. a. Answers may vary. Sample: x = no. of cars;y = no. of vans or trucks
b. 5x + 6.5y = 80037. a. Answers may vary.
Sample: x = time walking; y = time running
b. 3x + 8y = 1538.
39.
40.
41.
42.
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
Standard FormStandard Form
6-3
43.
44.
45.
47. a. 3x + 7y = 28b. 7 oz
48. 4.29x + 3.99y = 30
49.
50.
51. y = – x – 3
52. y = x –
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
Standard FormStandard Form
6-3
y = x + 1045
46.
y = x + 367
45
59
53
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
Standard FormStandard Form
6-3
53. y = – x – 8
54. y = x –
1611
19
23
55. Answers may vary.
Sample: slope-intercept
form when comparing the
steepness of two lines;
standard form when
making quick graphs
56. Answers may vary.
Sample: 0x + 0y = 0,
no linear equation exists.
57. –3x instead of 3x
58. y = 2
59. y = –2
60. x = 1
61. x = –2
62. a. 200s + 150a = 1200b.
Answers may vary. Sample: s = $2.00, a = $5.33; s = $3.00, a = $4.00; s = $6.00, a = $0.00s = $3.00 and a = $4.00 because they are whole dollar amounts, and adults pay more.
63. y = x + 435
66. C67. H
68. [2] y = x + 1;
x – 4y = –4[1] correct slope-
intercept equation69. [4] a. 48x + 56y = 2008
b.
c. Answers will include three of the following: 1 Supreme, 35 Prestige; 8 Supreme, 29 Prestige; 15 Supreme, 23 Prestige; 22 Supreme, 17 Prestige; 29 Supreme, 11 Prestige;36 Supreme, 5 Prestige
64. square
65. a.
b. – ; –
c. Answers may vary. Sample: The x- and y-intercepts of 2x + 3y = 18 are 3 times those of 2x + 3y = 6.
23
23
14
[3] correct equation and graph
[2] wrong equation, but a corresponding correct graph
[1] Correct equation only70. yes71. no72. no
73.
74.75. 476. 2.2577. –0.578. –21
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
Standard FormStandard Form
6-3
1 36 1 12
Find each x- and y-intercepts of each equation.
1. 3x + y = 12 2. –4x – 3y = 9
3. Graph 2x – y = 6 using x- and y-intercepts.
For each equation, tell whether its graph is horizontal or vertical.
4. y = 3 5. x = –8
6. Write y = x – 3 in standard form using integers.52
Standard FormStandard Form
x = 4, y = 12
horizontal vertical
–5x + 2y = –6 or 5x – 2y = 6
x = – , y = –394
ALGEBRA 1 LESSON 6-3ALGEBRA 1 LESSON 6-3
6-3
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
(For help, go to Lessons 6-1 and 1-7.)
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
Find the rate of change of the data in each table.
1. 2. 3.
Simplify each expression.
4. –3(x – 5) 5. 5(x + 2) 6. – (x – 6)49
2 4
5
8
11
–2
–8
–14
x y
–3 –5
–1
1
3
–4
–3
–2
x y
10 4
7.5
5
2.5
–1
–6
–11
x y
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear EquationsALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
1. Use points (2, 4) and (5, –2). rate of change = = = –2
2. Use points (–3, –5) and (–1, –4). rate of change = = = =
3. Use points (10, 4) and (7.5, –1). rate of change = = = = 2
4. –3(x – 5) = –3x – (–3)(5) = –3x + 15
5. 5(x + 2) = 5x + 5(2) = 5x + 10
6. – (x – 6) = – x – (– )(6) = – x + 49
83
–5 + 4–3 + 1
4 – (–2) 2 – 5
4 – (–1 ) 10 – 7.5
–5 – (–4)–3 – (–1)
4 + 12.5
52.5
–1–2
6 –3
12
49
49
49
Solutions
13
The equation of a line that passes
through (1, 2) with slope .
Graph the equation y – 2 = (x – 1).
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
Start at (1, 2). Using the slope, go up 1 unit and right 3 units to (4, 3).
Draw a line through the two points.
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
13
Simplify the grouping symbols.y + 3 = –2(x – 3)
Write the equation of the line with slope –2 that passes
through the point (3, –3).
y – y1 = m(x – x1)
Substitute (3, –3) for (x1, y1) and –2 for m.
y – (–3) = –2(x – 3)
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear EquationsALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
The slope is – .13
Step 1 Find the slope.
= my2 – y1
x2 – x1
4 – 3 –1 – 2
= – 13
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
Write equations for the line in point-slope form and in
slope-intercept form.
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
Step 3 Rewrite the equationfrom Step 2 in slope–intercept form.
y – 4 = – (x + 1)
y – 4 = – x –
y = – x + 3
13
13
13
13
23
(continued)
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear EquationsALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
Step 2 Use either point to write the the equation in point-slope form.
Use (–1, 4).
y – y1 = m(x – x1)
y – 4 = – (x + 1)13
Step 1 Find the rate of change for consecutive ordered pairs.
–2–1
–6–3
–2–1
= 2
= 2
= 2
The relationship is linear. The rate of change is 2.
Is the relationship shown by the data linear? If so, model the
data with an equation.
–1( ) –2
–3( ) –6–2( ) –4
3 6
2
–1
–3
4
–2
–6
x y
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear EquationsALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
y – y1 = m(x – x1) Use the point-slope form.
y – 4 = 2(x – 2) Substitute (2, 4) for (x1, y1) and 2 for m.
Step 2 Use the slope 2 and a point (2,4) to write an equation.
(continued)
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear EquationsALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
The relationship is not linear.
Is the relationship shown by the data linear? If so, model the
data with an equation.
1 ( ) 1
2 ( ) 11 ( ) 1
–2 –2
–1
1
2
–1
0
1
x y
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear EquationsALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
Step 1 Find the rate of change for consecutive ordered pairs.
11
12
11
= 1
= 1
= 1–
–
–
/
pages 307–309 Exercises
1.
2.
3.
4.
5.
6.
7.
8.
9.
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
6-4
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
6-4
10. y + 4 = 6(x – 3)
11. y – 2 = – (x – 4)
12. y – 2 = (x)
13. y + 7 = – (x + 2)
14. y = 1(x – 4)15. y + 8 = –3(x – 5)16. y – 2 = 0(x + 5) or
y = 2
17. y + 8 = – (x – 1)
18. y – 1 = (x + 6)
19–30. Answers may vary for the point indicated by the equation.
19. y = 1(x + 1); y = x + 1
20. y – 5 = (x – 3);
y = x
21. y + 2 = – (x – 4);
y = – x +
22. y + 4 = –1(x – 6); y = –x + 2
23. y + 5 = (x + 1);
y = x –
24. y + 4 = (x + 3);
y = x – 3
25. y – 7 = 11(x – 2);y = 11x – 15
26. y – 6 = – (x + 2);
y = – x + 4
27. y + 8 = – (x – 3);
y = – x –
28. y – = (x – 1);
y = x –
29. y – 2 = –1(x – );
y = –x +
30. y – 1.1 = (x – 0.2);
y = x +
53
45
32
15
23
53
53
65
65
14 5
16
16
29 6
13
13
57
57
47
13 5
13 5
15
1234
14
12
521.96.8
1.96.8
7.16.8
34
31. Yes; answers may vary. Sample: y – 9 = –2(x + 4)
32. Yes; answers may vary. Sample: y – 40 = 3(x – 5)
33. no34. Yes; answers may vary.
Sample: y – 75 = 10(x – 10)35. no
36–53. Answers may vary for point indicated by the equation.
36. y – 2 = (x – 1)
37. y + 3 = (x – 1)
38. y = – (x – 5)
39. y – 4 = (x – 1);
–3x + 2y = 540. y + 3 = 0(x – 6);
y = –341. y + 2 = 2(x + 1);
–2x + y = 042. y – 2 = 0(x – 0);
y = 2
43. y – 6 = – (x + 6);
x + 3y = 12
44. y – 3 = – (x – 2);
2x + 3y = 1345. y + 3 = – (x – 5);
7x + 2y = 29
3425
57
32
13
23
46. y – 2 = – (x – 2);
5x + 3y = 16
47. y – 1 = – (x + 7);
x + 6y = –1
48. y – 4 = – (x + 8);
3x + 2y = –1649. y – 4 = 2(x – 2);
–2x + y = 050. y – 3 = –2(x – 5);
2x + y = 13
51. y – 1 = x; –x + 3y = 3
52. y – 4 = – (x + 2);
9x + 2y = –10
53
16
32
13
92
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
72
6-4
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
6-4
53. y – 2 = (x – 6);
–3x + 5y = –8
54. a. y = – x + 1
b. about 4 atmospheres55. y = –2.6x + 315.656. a. Answers may vary.
Sample: y + 6 = 2(x + 4); chose slope and substituted into y – y1 = m(x – x1)
b. infinite; infinite number of slopes
57. y-intercept changes
58. Yes; the point satisfies the equation.
59. Answers may vary. Sample:a. y = x + 1b. –x + y = 1c. y – 1 = 1(x – 0)
60. a. Answers may vary.
Sample: y – 332 = (x – 0)b. 341 m/sc. 368 m/s
61. y = 7x + 1662. y = 3
63. y = x –
64. a. 14.75b. 57.5c. –4d. 100
35
1 33
35
25
14 5
65.
66. 9
67.
68. 4
69. 6
70.
71.
12
73
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
6-4
72.
73.
74.
75.
76. 5; 3; 8
77. ; ;
78. 0.07; 2.66; 2.73
79. –0.05; –3.35; –3.4
80. 17; 69; 86
81. 4; 5; 9
26
32
1116
–10 –705
20
–3–15
x y
1. Graph the equation y + 1 = –(x – 3).
2. Write an equation of the line with
slope – that passes through
the point (0, 4).
3. Write an equation for the line that passes through (3, –5) and (–2, 1) in Point-Slope form and Slope-Intercept form.
4. Is the relationship shown by the data linear? If so, model that data with an equation.
23
Point-Slope Form and Writing Linear EquationsPoint-Slope Form and Writing Linear Equations
y – 4 = – (x – 0), or y = – x + 423
23
65
65
75
y + 5 = – (x – 3); y = – x –
25yes; y + 3 = (x – 0)
ALGEBRA 1 LESSON 6-4ALGEBRA 1 LESSON 6-4
6-4
What is the reciprocal of each fraction?
1. 2. 3. – 4. –
What are the slope and y-intercept of each equation?
5. y = x + 4 6. y = x – 8 7. y = 6x 8. y = 6x – 2
12
43
25
75
53
53
Parallel and Perpendicular LinesParallel and Perpendicular Lines
(For help, go to Lessons 1-6 and 6-2.)
ALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
6-5
Parallel and Perpendicular LinesParallel and Perpendicular LinesALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
6-5
1. The reciprocal of is or 2. 2. The reciprocal of is .
3. The reciprocal of – is – . 4. The reciprocal of – is – .
5. y = x + 4 6. y = x – 8
slope: slope:
y-intercept: 4 y-intercept: –8
7. y = 6x 8. y = 6x – 2 slope: 6 slope: 6 y-intercept: 0 y-intercept: – 2
12
43
25
21
34
52
75
57
53
53
53
53
Solutions
The lines are parallel. The equations have the same slope, –2, and different y-intercepts.
Are the graphs of y = –2x – 1 and 4x + 2y = 6 parallel?
Write 4x + 2y = 6 in slope-intercept form. Then compare with y = –2x – 1.
Parallel and Perpendicular LinesParallel and Perpendicular LinesALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
2y = –4x + 6 Subtract 4x from each side.
y = –2x + 3 Simplify.
Divide each side by 2.2y –4x + 6 2 2
=
6-5
Step 1 Identify the slope
of the given line.
y = x – 4
slope
52
Write an equation for the line that contains (–2, 3) and is
parallel to y = x – 4.
y – 3 = x + 5 Simplify.52
y = x + 8 Add 3 to each side and simplify.
52
Step 2 Write the equation of the line through (–2, 3) using slope-intercept form.
y – y1 = m(x – x1) Use point-slope form.
y – 3 = (x + 2) Substitute (–2, 3) for
(x1, y1) and for m.52
52
y – 3 = x + (2) Use the Distributive Property.
52
Parallel and Perpendicular LinesParallel and Perpendicular LinesALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
6-5
52
Step 1 Identify the slope of the given line.
y = – 2 x + 7
slope
Step 2 Find the negative reciprocal of the slope.
The negative reciprocal of –2 is . 12
Write an equation of the line that contains (6, 2) and is
perpendicular to y = –2x + 7.
Parallel and Perpendicular LinesParallel and Perpendicular LinesALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
6-5
Step 3 Use the slope-intercept form to write an equation.
y = mx + b
2 = (6) + b Substitute for m, 6 for x, and 2 for y.
2 = 3 + b Simplify.
2 – 3 = 3 + b – 3 Subtract 3 from each side.
–1 = b Simplify.
The equation is y = x – 1.12
12
12
(continued)
Parallel and Perpendicular LinesParallel and Perpendicular LinesALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
6-5
Parallel and Perpendicular LinesParallel and Perpendicular Lines
The line in the graph represents the street in front of a new
house. The point is the front door. The sidewalk from the front door will
be perpendicular to the street. Write an equation representing the
sidewalk.
Step 1 Find the slope m of the street.
m = = = = – Points (0, 2) and (3, 0) are on the street.
y2 – y1
x2 – x1
0 – 23 – 0
–2 3
23
ALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
6-5
The equation for the
sidewalk is y = x – 3.32
Parallel and Perpendicular LinesParallel and Perpendicular Lines
(continued)
ALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
6-5
Step 2 Find the negative reciprocal of the slope.
The negative reciprocal of – is . So the slope
of the sidewalk is . The y-intercept is –3.
23
32
32
ALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
Parallel and Perpendicular LinesParallel and Perpendicular Lines
6-5
pages 314–317 Exercises
1.
2. –
3. 14. 0
5. –
6. 77. no, different slopes8. yes, same slopes and
different y-intercepts9. yes, same slopes and
different y-intercepts10. no, different slopes11. yes, same slopes and
different y-intercepts
12. yes, same slopes and different y-intercepts
13. y = 6x
14. y = –3x + 9
15. y = –2x – 1
16. y = – x – 20
17. y = 0.5x – 9
18. y = – x +
19. –
20.
21. –
22. 5
12
23
34
23.
24. undefined
25. y = – x
26. y = –x + 1027. y = 3x – 10
28. y = x +
29. y = – x + 24
30. y = – x + 2
31. y = x + 1
32. perpendicular33. parallel34. perpendicular
72
23
13
12
13
57
32
12
53
11 3
4512
54
ALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
Parallel and Perpendicular LinesParallel and Perpendicular Lines
6-5
35. neither36. parallel37. perpendicular38. parallel39. neither40. perpendicular41. neither42. Answers may vary.
Sample: same x and y coefficients
43. y = – x – ;
y = – x +
44. y = x +
y = –3x + 7
45. y = – x
y = 2x
46. y = x + ;
y = x –
47. y = 4; y = 248. y = x; y = –x + 1
49. about
50. Answers may vary. Sample: same slope
of –
51. Answers may vary.
Sample: • – = –1
45
19 5
45
35
13
43
52. a. The screen is not square.
b. The lines appear perpendicular.
53. Answers may vary. Sample: y = 4x + 1
54. No; the slopes are not equal.55. No; the slopes are not
neg. reciprocals.56. yes; same slopes and
different y-intercepts57. False; the product of two
positive numbers can’t be –1.
58. True; y = x + 2 and y = x + 3 are parallel.
12
25
35
25
21 5
54
12
54
12
/
ALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
Parallel and Perpendicular LinesParallel and Perpendicular Lines
6-5
59. False; all direct variations
go through the point (0, 0).
If they have the same slope,
they are the same line,
not parallel lines.
60. The slopes of AD and BC
are both undefined, so
they are parallel. The
slopes of AB and CD are
both so they are
parallel. The quadrilateral
is a parallelogram.
25
61. The slope of JK is .
The slope of KL is –2.
The slope of LM is .
The slope of JM is –4.
The quadrilateral is not
a parallelogram.
62. The slopes of PQ and
RS are both – .The
slopes of QR and SP
are both – .The
quadrilateral is
a parallelogram.
15
16
12
32
63.The slopes of AB and
CD are both .The
slopes of BC and AD
are both – . The
product is –1, so
the quadrilateral is a
rectangle.
64.The slopes of KL and
MN are both – .The
slopes of LM and KN
are both 5. The
product is not –1, so
the quadrilateral is not
a rectangle.
25
52
16
ALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
Parallel and Perpendicular LinesParallel and Perpendicular Lines
6-5
The diagonal AC has
a slope of . The
diagonals are
perpendicular.
ABCD is a rhombus.
67. RP has a slope of .
RQ has a slope
of – . RP is the neg.
reciprocal of RQ, so
PQR is a right
triangle.
68. parallel
69. perpendicular
65. The slopes of PQ and RS
are both . The slopes
of PS and QR are both –2.
The product is –1, so
the quadrilateral is a
rectangle.
66. BC and AD both have
a slope of zero. BC and
AD are parallel. AB and
CD both have a slope
of . AB and CD are
parallel. The diagonal BD
has a slope of –2.
12
43
23
32
12
70. y = – x – ;
y = x + 7
71. y = x – 3;
y = –2x + 7
72. –1.5; 24
73. D
74. F
75. C
38
17 8
3812
ALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
Parallel and Perpendicular LinesParallel and Perpendicular Lines
6-5
76. [2] Find slope of
2x + y = 3:
y = –2x + 3, therefore
slope is –2.
Find x: = –2,
= –2,
–4 = –2 + 2x,
x = –1 (OR equivalent
explanation)[1] correct value of x but
no work OR minor computation error in work
77. C
78. A
79. B
80. y = 3x + 4
81. y = –4x – 8
82. y + 3 = (x – 5)
83. y + 9 = – (x + 1)
84. y – 4 = – (x + 6)
85. y – 11 = (x – 7)
86. 7; 11; 15
87. –9; –17; –25
88. yes
2 – 61 – x
– 4 1 – x
89. yes
90. no
91. yes
34
2335
12
1. Find the slope of a line parallel to 3x – 2y = 1.
2. Find the slope of a line perpendicular to 4x + 5y = 7.
Tell whether the lines for each pair of equations are parallel, perpendicular, or neither.
3. y = 3x – 1, y = – x + 2
4. y = 2x + 5, 2x + y = –4
5. y = – x – 1, 2x + 3y = 6
6. Write an equation of the line that contains (2, 1) and is
perpendicular to y = – x + 3.
32
54
perpendicular
neither
parallel
y = 2x – 3
Parallel and Perpendicular LinesParallel and Perpendicular LinesALGEBRA 1 LESSON 6-5ALGEBRA 1 LESSON 6-5
6-5
13
23
12
1 2
2
3
4
–3
8
9
5 –25
yx
1.
1 21
2
3
4
15
12
9
5 7
yx
2.
Use the data in each table to draw a scatter plot.
Scatter Plots and Equations of LinesScatter Plots and Equations of Lines
(For help, go to Lessons 1-9.)
ALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
6-6
1. Graph the points: (1, 2) (2, –3) (3, 8) (4, 9)
(5, –25)
Scatter Plots and Equations of LinesScatter Plots and Equations of Lines
2. Graph the points: (1, 21) (2, 15) (3, 12) (4, 9) (5, 7)
ALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
Solutions
6-6
Make a scatter plot to represent the data. Draw a trend line
and write an equation for the trend line. Use the equation to predict the
time needed to travel 32 miles on a bicycle.
Miles
Time (min)
5
27
10
46
14
71
18
78
22
107
Speed on a Bicycle Trip
Scatter Plots and Equations of LinesScatter Plots and Equations of Lines
Step 1 Draw a scatter plot. Use a straight edge to draw atrend line. Estimate twopoints on the line.
ALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
6-6
y – y1 = m(x – x1) Use point-slope for m.
y – 27 = 5(x – 5) Substitute 5 for m and (5, 27) for (x1, y1).Step 3 Predict the time needed to travel 32 miles.
y – 27 = 5(32 – 5) Substitute 32 for x.
y – 27 = 5(27) Simplify within the parentheses.
y – 27 =135 Multiply.
y = 162 Add 27 to each side.
The time needed to travel 32 miles is about 162 minutes.
(continued)
Scatter Plots and Equations of LinesScatter Plots and Equations of LinesALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
6-6
Step 2 Write an equation of the trend line.
m = = = 5y2 – y1
x2 – x1
120 – 27 25 – 5
9320
Scatter Plots and Equations of LinesScatter Plots and Equations of LinesALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
6-6
1996 5086.6
1997
1998
1999
4930.0
4619.3
4266.8
Year No. of Crimes
1995 5275.9
[Source: Crime in the United States,1999, FBI, Uniform Crime Reports]
Use a graphing calculator to find the equation of the line of
best fit for the data below. What is the correlation coefficient? U.S. Crime Rate
(per 100,000 inhabitants)
The equation for the line of best fit is y = –248.55x + 28,945.07 for
values a and b rounded to the nearest hundredth. The value of the
correlation coefficient is –0.9854908654.
Scatter Plots and Equations of LinesScatter Plots and Equations of Lines
(continued)
Step 1 Use the EDIT feature of the screen on your graphing calculator. Let 95 correspond to 1995. Enter the data foryears and then enter data for crimes.
Step 2 Use the CALC feature in the screen.
Find the equation for the line of best fit.
ALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
6-6
pages 320–324 Exercises
1–6. Trend lines may vary. Samples given.
1. y – 52.5 = 2(x – 91)
2. y – 100 =15.71(x – 5)
3. y – 16.4 = 0.64(x – 69.9)
4. y – 85 = – 15.25(x – 1)5.
y – 5= 0.66(x – 240)
6.
y = 1.6x – 80; 40 in.
7. y = –1.06x + 92.31; –0.9701709306
8. y = 10.60x – 772.66; 0.990733298
9. y = –2.29x + 613.93; –0.8108238756
10. y = 2.64x + 70.51; 0.9900170523
11. y = 1.35x – 31.42; 0.999808967
ALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
Scatter Plots and Equations of LinesScatter Plots and Equations of Lines
6-6
12. a.
b. Answers may vary. y = 3.25x – 1c. Answers may vary.
Sample: The slope is the approximate ratio of the circumference to the diameter.
d. 14 cm13. a.
b. Answers may vary. Sample: y = 0.939x + 13,800
c. 143,800,000d. Answers may vary.
Sample: No, the year is too far in the future.
14. a. Check students’ work.b. 1
15. Answers may vary. Sample: pos. slope; as temp. increases, more students are absent.
16. a. y = 0.61x + 35.31b. Answers may vary.
Sample: small set of data with weak correlation
17. y = 0.37x – 28.66; $12.04 billion
ALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
Scatter Plots and Equations of LinesScatter Plots and Equations of Lines
6-6
18. a. y = –16.70x + 297.57b. –0.6681863355c. No, the correlation
coefficient is not close to 1 or –1, so the equation does not closely model the data.
19. a. (2, 3) and (6, 6); y = 0.75x + 1.5
b. y = 0.75x + 1.2120. a.
y = 4.82x – 29.65
b. 404 ftc. The speed is much faster
than those speeds used to find the equation of a trend line.
21. B22. H23. [4] a.
b–c. Answers may vary. Samples:b. Let 1960 = 60. Two points
on line are (62, 18) and (90, 30).y = 0.429x – 8.6
c. y = 0.429(105) – 8.6; 36.4 million people
ALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
Scatter Plots and Equations of LinesScatter Plots and Equations of Lines
6-6
ALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
Scatter Plots and Equations of LinesScatter Plots and Equations of Lines
6-6
[3] appropriate methods but one computational error
[2] incorrect points used correctly OR correct points used incorrectly; function written appropriately, given previous results.
[1] correct function, without work shown
24. y + 3 = 5(x – 2)
25. y – 5 = –x
26. y – 4 = – (x + 1)
27. y + 4 = – (x – 3)
28. y + 1 = –2(x + 2)
29. y – 2 = (x + 1)
30. x > 1
31. x < 5
32. x > –1
33. x –5
34. x >
35. x < 3
23
32
12
23
34
<–
Number of Households in the U.S.
[Source: U.S. Census Bureau, Current Population Reports.From Statistical Abstract of the United States, 2000]
1980 80.8
1985
1990
1995
86.8
93.3
99.0
Year Households(millions)
1975 71.1
1. Graph the data in a scatter plot.
Draw a trend line.
2. Write an equation for the trend
line.
3. Predict the number of households
in the U.S. in 2005. (Use 105 for x.)
4. Use a calculator to find the line of
best fit for the data.
5. What is the correlation coefficient?
y – 86.8 = 1.3(x – 85)
about 112.8 million households
y = 1.366x – 29.91
0.9943767027
Scatter Plots and Equations of LinesScatter Plots and Equations of LinesALGEBRA 1 LESSON 6-6ALGEBRA 1 LESSON 6-6
6-6
Simplify each expression.
1. |2 – 7| 2. |7 – 12|
3. |38 – 56| 4. |–24 + 12|
Model each rule using a table of values.
5. y = 6 – x 6. y = |x| + 1
7. y = |x + 1|
Graphing Absolute Value EquationsGraphing Absolute Value Equations
(For help, go to Lessons 1-5 and 5-3.)
ALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
6-7
1. |2 – 7| = |–5| = 5 2. |7 – 12| = |–5| = 5
3. |38 – 56| = |–18| = 18 4. |–24 + 12| = |–12| = 12
5. 6. 7.
0 6
1
2
3
5
4
3
x y
0 1
1
–1
–2
2
0
1
x y
0 1
1
–1
–2
2
2
3
x y
Graphing Absolute Value EquationsGraphing Absolute Value EquationsALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
Solutions
6-7
The graphs are the same shape. The y-intercept of the first graph
is 0. The y-intercept of the second graph is 1.
Describe how the graphs of y = |x| and y = |x| + 1 are the
same and how they are different.
Graphing Absolute Value EquationsGraphing Absolute Value EquationsALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
6-7
Start with a graph of y = |x|.
Translate the graph down 3 units.
Graphing Absolute Value EquationsGraphing Absolute Value Equations
Graph y = |x| – 3 by translating y = |x|.
ALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
6-7
The equation is y = |x| + 13.
Write an equation for each translation of y = |x|.
a. 9 units down b. 13 units up
The equation is y = |x| – 9.
Graphing Absolute Value EquationsGraphing Absolute Value EquationsALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
6-7
Graph each equation by translating y = |x|.
a. y = |x + 2.5|
Graphing Absolute Value EquationsGraphing Absolute Value Equations
Start with a graph of y = |x|.
Translate the graph left 2.5 units.
ALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
b. y = |x – 2.5|
Start with a graph of y = |x|.
Translate the graph right 2.5 units.
6-7
The equation is y = |x – 7|.The equation is y = |x + 10|.
b. 7 units right
Write an equation for each translation of y = |x|.
a. 10 units left
Graphing Absolute Value EquationsGraphing Absolute Value EquationsALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
6-7
pages 327–329 Exercises
1. Answers may vary. Sample: same shape, shifted 3 units up
2. Answers may vary. Sample: same shape, shifted 3 units down
3. Answers may vary. Sample: same shape, shifted 7 units down
4.
5.
6.
7.
8.
9.
10. y = |x| + 9
11. y = |x| – 6
12. y = |x| + 0.25
13. y = |x| +
14. y = |x| + 5.90
15. y = |x| – 1
ALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
Graphing Absolute Value EquationsGraphing Absolute Value Equations
6-7
52
16.
17.
18.
19.
28.
29.
30.
31.
20.
21.
22. y = |x + 9|
23. y = |x – 9|
24. y = |x – |
25. y = |x + |
26. y = |x + 0.5|
27. y = |x – 8.2|
5232
ALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
Graphing Absolute Value EquationsGraphing Absolute Value Equations
6-7
32. y = –|x| + 2
33. y = –|x + 2.25|
34. y = –|x| –
35. y = –|x – 4|
36. B
37.
38.
39.
40.
41. a.
b. (2, 3)
c. Answers may vary. Sample: The x-coordinate is the horizontal translation, and the y-coordinate is the vertical translation.
d. Use (a, b) for the vertex. Graph y = x and y = –x above the vertex.
42.a.
b.
32
ALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
Graphing Absolute Value EquationsGraphing Absolute Value Equations
6-7
x y–1 20 01 2
c.
d.
43. a. y = |x|; y = 1
b. –1 and 1
c. y = – x or
y = – x + 244. B45. H
46. A47. G48. [4] a. and b.
c. Part (a) graph is shifted 8 units up to get part (b) graph.
[3] one incorrect graph but correct answer to part (c) based on graphs drawn OR incorrect answer based on correct graphs
[2] both graphs incorrect but correct answer to part (c) based on graphs drawn
[1] both graphs incorrect and no answer given for part (c)
49. y = 5000x – 413,000
50. y = 4000x – 313,000
51. 12 5 2 6
52. 4 1 1 4
53. 2.5 14–2.0 10.7
ALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
Graphing Absolute Value EquationsGraphing Absolute Value Equations
1212
6-7
1. Describe how the graphs of y = |x| and y = |x| – 9 are the same and how they are different.
2. Graph y = |x| – 2 by translating y = |x|.
3. Write an equation for the translation of y = |x|, 1.5 units down.
4. Graph y = |x – 5|.
5. Write an equation for the translation of y = |x|, 7 units left.
The graphs are the same shape. The y-intercept of the first graph is 0.The y-intercept of the second graph is –9.
y = |x| – 1.5
y = |x + 7|
Graphing Absolute Value EquationsGraphing Absolute Value EquationsALGEBRA 1 LESSON 6-7ALGEBRA 1 LESSON 6-7
6-7
ALGEBRA 1 CHAPTER 6ALGEBRA 1 CHAPTER 6
Linear Equations and Their GraphsLinear Equations and Their Graphs
6-A
1. False; a rate of change could also be negative or 0.
2. False; a vertical line has an undefined rate of change.
3. –5
4. –
5.
6.
7.
8.
9. y = – x +
10. y = x + 6
11. y = – x + 25
12. y = x –
13. –8; –6
14. ; –4
15. –12; 6
16. 1; 1
17. y + 7 = (x + 2)
18. y + 8 = 3(x – 4)
19. y – 3 = – x
20. y = –5(x – 9)
21–24. Samples given.
21. y – 9 = (x – 4)
22. y = (x + 1)
23. y + 8 = 0
14
87
37
13
54
92
13 2
52
43
83
12
52
24. y – 7 = –2x
25. D
26. y = 5x – 11
27. y = 6
28. y = x – 2
29. y = 2
30. Answers may vary. Sample: y = 0.5x + 2
31. a. y = 5x – 30b. intercepts: 6; –30
32. y = |x| – 2
ALGEBRA 1 CHAPTER 6ALGEBRA 1 CHAPTER 6
Linear Equations and Their GraphsLinear Equations and Their Graphs
6-A
12
33. y = |x – |
34.
35.
36. a. y = 0.0436x + 15.34 (for 1967 = 67)
b. For sample in (a): 20,100 municipalities
37. a. y = –0.197x + 31.95 (for 1967 = 67)
b. For sample in (a): 10,300 school districts
34