Rate of Change and Slope

transcript

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.

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 6-1

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

–2–8

–3–5

ALGEBRA 1 LESSON 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 Fee100 $30150200250

$42$54$66

Rate of Change and SlopeALGEBRA 1 LESSON 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 SlopeALGEBRA 1 LESSON 6-1

change in cost Cost depends on the change in number of miles number of miles.rate of change =

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)

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 – 0400 20 20

Use two points.

Divide the vertical change by the horizontal change.

Simplify.

The slope of the line is – .32

slope =riserun

Find the slope of each line.

= – 32

= 3 –2

4 – 1 0 – 2

The slope of the line is 2.

slope =riserunb. Find the slope of the line.

(continued)

= –2–1

–1 – 1 –2 – (–2)

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

2 – 2 1 – (–4)

Substitute (1, 2) for (x2, y2) and (–4, 2)

for (x1, y1).

Simplify.

slope =y2 – y1

x2 – x1

The slope of the horizontal line is 0.

Find the slope of each line.

Division by zero is undefined.The slope of the vertical line in undefined.

(continued)

Find the slope of the line.

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.

8. –3

41 22 3

13. –

14. –1

15. –1

18. –

19. –

5 95 3 6

21. –5

23. undefined

25. undefined

27. in./month

28. $5/person

29. 30 mi/hr

30.2 5

41. a.

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

40. a. C

b. C greatest;

A least; the slope

33. –20

34. undefined

35. –336.

ALGEBRA 1 LESSON 6-1Rate of Change and Slope

1 61 4

46. a.

b. c.d. equal

47. a. Answers may vary. Sample: (0, 0), (1,

) b. Answers may vary.

Sample: (0, 0), (1, – )

48. 049. –650. 651. 452. 12

53. 354–60. 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

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

74. [2] = ; the slope is .

= 12.5%; the grade

is 12.5%.

[1] finds grade only

62. Friend found instead

64. –

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

2d – bc – 2a

runrise

75. A76. c = 3.5n77. p = 4.95q – 23278. 0

81. 182. –383. –4

84.85. 786. 587. –10

riserun

1. Find the rate of change for the data in the table.

5 $75678

$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).

5. State whether the slope is zero or undefined.a. b.

$15 per ticket400 calories per hour 4

1. Find the rate of change for the data in the table.

5 $75678

$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).

undefined 0

5. State whether the slope is zero or undefined.a. b.

(For help, go to Lessons 1-6 and 2-6.)

ALGEBRA 1 LESSON 6-2Slope-Intercept Form

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

Slope-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

Solutions

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.

Use the slope-intercept form.

Substitute for m and 4 for b.

y = mx + b

y = x + 425

Write an equation of the line with slope and y-intercept 4.

y = mx + b

y = – x + 1Substitute – for m and 1 for b.

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

Step 2 The slope is . Use slope to plot a second point.

Step 1 The y-intercept is –2.So plot a point at (0,–2).

Step 3 Draw a line through the two points.

Graph y = x – 2.13

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.

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.

(continued)

Step 3 Draw a line through the points.

15 100

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 +

22. y = – x + 1

23. y = x + 2

24. y = 2x – 2

25. y = x +

26. y = – x + 2.8

27. y = x –

y = x + 4

y = x – 1

y = –5x + 231.

y = 2x + 5

y = x + 433.

y = –x + 234.

y = 4x – 3

y = – x

y = x – 3

y = – x + 2

y = – x + 4

y = –0.5x + 240. a.

t = 14c – 4

b. $80

y = –2x52.

y = – x – 2

y = 5x – 6

41. –3; 2

42. – ; 0

43. 9; 44. 3; –9

45. ; 3

46. 9; –1547. c; d48. 2 – a; a49. –3; –2n50.

y = 7 – 3x

y = – x –

y = 6x – 856. The slope was used for

the y-int., and the y-int. was used for the slope.

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

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.

b. Rectangle; check students’ work.

c. y = x OR y = – x;

explanations may vary.82. a. r = –15d + 265

r = –15d + 265c. 18 days

76. a. ;

b. 2; –2c. same slopes

77. Check students’ work.

78. –

79. –5

81. a.

x = –2; x = 2y = 3; y = –3

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

5 – 31 – 0

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

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

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 Form

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

Standard FormALGEBRA 1 LESSON 6-3

Solutions7. 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

2x+ 4 525

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

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

The y-intercept is .

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 Form

Draw a line through the points.

a. Graph y = 4 b. Graph x = –3.

Standard 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.

The equation in standard form is –2x + 3y = 18.

–2x + 3y = 18 Subtract 2x from each side.

y = x + 623

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

Define: Let = the hours mowing lawns.

Let = the hours delivering newspapers.

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.

pages 301–303 Exercises1. 18; 9

2. 3; –9

3. –6; 30

4. ; –3

5. – ; –

6. –8; 12

7. 6; 4

8. ; –2

9. –5; 4

19. horizontal20. vertical21. horizontal22. vertical23.

ALGEBRA 1 LESSON 6-3Standard Form

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.

47. a. 3x + 7y = 28b. 7 oz

48. 4.29x + 3.99y = 3049.

51. y = – x – 3

52. y = x –

y = x + 1045

46.y = x + 36

53. y = – x – 8

54. y = x –

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 3x58. y = 259. y = –260. x = 161. 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. H68. [2] y = x + 1;

x – 4y = –4[1] correct slope-

intercept equation69. [4] a. 48x + 56y = 2008

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.

[3] correct equation and graph

[2] wrong equation, but a corresponding correct graph

[1] Correct equation only70. yes71. no72. no

74.75. 476. 2.2577. –0.578. –21

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 Form

x = 4, y = 12

horizontal vertical

–5x + 2y = –6 or 5x – 2y = 6

x = – , y = –394

Point-Slope Form and Writing Linear Equations

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

–2–8–14

x y–3 –5–113

–4–3–2

x y10 47.55

–1–6–11

Point-Slope Form and Writing Linear EquationsALGEBRA 1 LESSON 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

–5 + 4–3 + 1

4 – (–2) 2 – 5

4 – (–1 ) 10 – 7.5

–5 – (–4)–3 – (–1)

4 + 12.5

–1–2

6 –3

Solutions

The equation of a line that passes

through (1, 2) with slope .

Graph the equation y – 2 = (x – 1).

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.

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)

The slope is – .13

Step 1 Find the slope.

= my2 – y1

x2 – x1

4 – 3 –1 – 2 = – 1

Write equations for the line in point-slope form and in slope-intercept form.

Step 3 Rewrite the equationfrom Step 2 in slope–intercept form.

y – 4 = – (x + 1)

y – 4 = – x –

y = – x + 3

(continued)

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

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 62–1–3

4–2–6

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)

The relationship is not linear.

Is the relationship shown by the data linear? If so, model the data with an equation.

1 ( ) 12 ( ) 11 ( ) 1

–2 –2–112

–101

Step 1 Find the rate of change for consecutive ordered pairs.

= 1–

ALGEBRA 1 LESSON 6-4Point-Slope Form and Writing Linear Equations

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

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);

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 +

521.96.8

1.96.8

7.16.8

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);

43. y – 6 = – (x + 6);

x + 3y = 12

44. y – 3 = – (x – 2);

2x + 3y = 1345. y + 3 = – (x – 5);

7x + 2y = 29

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. 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

68. 469. 670.

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

–10 –705

–3–15

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.

y – 4 = – (x – 0), or y = – x + 423

y + 5 = – (x – 3); y = – x –

25yes; y + 3 = (x – 0)

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

Parallel and Perpendicular Lines

Parallel and Perpendicular LinesALGEBRA 1 LESSON 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

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.

2y = –4x + 6 Subtract 4x from each side.

y = –2x + 3 Simplify.

Divide each side by 2.2y –4x + 6 2 2=

Step 1 Identify the slope

of the given line.

y = x – 4

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.

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

y – 3 = x + (2) Use the Distributive Property.

Step 1 Identify the slope of the given line.

y = – 2 x + 7

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.

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

(continued)

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

The equation for the

sidewalk is y = x – 3.32

(continued)

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.

ALGEBRA 1 LESSON 6-5Parallel and Perpendicular Lines

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. –

21. –

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

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

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.

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.

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.

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.

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. parallel69. perpendicular

65. The slopes of PQ and RSare 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.

70. y = – x – ;

y = x + 7

71. y = x – 3;

y = –2x + 7

72. –1.5; 24

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

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

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.

perpendicular

neither

parallel

y = 2x – 3

1 2234

–389

5 –25

1 21234

Use the data in each table to draw a scatter plot.

Scatter Plots and Equations of Lines

(For help, go to Lessons 1-9.)

1. Graph the points: (1, 2) (2, –3) (3, 8) (4, 9)

(5, –25)

2. Graph the points: (1, 21) (2, 15) (3, 12) (4, 9) (5, 7)

Solutions

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.

Time (min)

Speed on a Bicycle Trip

Step 1 Draw a scatter plot. Use a straight edge to draw atrend line. Estimate twopoints on the line.

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 LinesALGEBRA 1 LESSON 6-6

Step 2 Write an equation of the trend line.

m = = = 5y2 – y1

x2 – x1

120 – 27 25 – 5

1996 5086.6

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.

(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.

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)

y = 1.6x – 80; 40 in.7. y = –1.06x + 92.31; –0.9701709306 8. y = 10.60x – 772.66; 0.9907332989. y = –2.29x + 613.93; –0.810823875610. y = 2.64x + 70.51; 0.990017052311. y = 1.35x – 31.42; 0.999808967

ALGEBRA 1 LESSON 6-6Scatter Plots and Equations of Lines

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

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

[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

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

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

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 Equations

1. |2 – 7| = |–5| = 5 2. |7 – 12| = |–5| = 5

3. |38 – 56| = |–18| = 18 4. |–24 + 12| = |–12| = 12

5. 6. 7.

0 6123

x y0 11

–1–2

x y0 11

–1–2

Graphing Absolute Value EquationsALGEBRA 1 LESSON 6-7

Solutions

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.

Start with a graph of y = |x|.

Translate the graph down 3 units.

Graph y = |x| – 3 by translating y = |x|.

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.

Graph each equation by translating y = |x|.

a. y = |x + 2.5|

Translate the graph left 2.5 units.

b. y = |x – 2.5|

Translate the graph right 2.5 units.

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

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

10. y = |x| + 911. y = |x| – 612. y = |x| + 0.2513. y = |x| + 14. y = |x| + 5.9015. y = |x| – 1

ALGEBRA 1 LESSON 6-7Graphing Absolute Value Equations

22. y = |x + 9|

23. y = |x – 9|

24. y = |x – |

25. y = |x + |

26. y = |x + 0.5|

27. y = |x – 8.2|

32. y = –|x| + 2

33. y = –|x + 2.25|

34. y = –|x| –

35. y = –|x – 4|

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.

x y–1 20 01 2

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 – 413,00050. y = 4000x – 313,000

51. 12 5 2 6

52. 4 1 1 4

53. 2.5 14–2.0 10.7

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|

ALGEBRA 1 CHAPTER 6Linear Equations and Their Graphs

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. –

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

24. y – 7 = –2x25. D26. y = 5x – 1127. y = 628. y = x – 229. y = 230. Answers may vary.

Sample: y = 0.5x + 231. a. y = 5x – 30

b. intercepts: 6; –30

32. y = |x| – 2

ALGEBRA 1 CHAPTER 6Linear Equations and Their Graphs

33. y = |x – |

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

Rate of Change and Slope

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