4.1 slopes of lines from graphs

13
Chapter 4 – Linear Functions Answer Key CK-12 Algebra I Honors Concepts 1 4.1 Slopes of Lines from Graphs Answers 1. Find two points on the line. Calculate . 2. The steepness of the line and whether the line goes up or down from left to right. 3. Positive. 4. From left to right, if the line goes up then the slope is positive, if the line goes down the slope is negative. 5. 0 6. Undefined 7. 2 8. 1 3 9. 3 4 10. -3 11. βˆ’ 1 4 12. -2 13. -1 14. 3 15. 1 3 16. 3 2

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Page 1: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 1

4.1 Slopes of Lines from Graphs

Answers

1. Find two points on the line. Calculate π‘Ÿπ‘–π‘ π‘’

π‘Ÿπ‘’π‘›.

2. The steepness of the line and whether the line goes up or down from left to right.

3. Positive.

4. From left to right, if the line goes up then the slope is positive, if the line goes down the slope is

negative.

5. 0

6. Undefined

7. 2

8. 1

3

9. 3

4

10. -3

11. βˆ’1

4

12. -2

13. -1

14. 3

15. 1

3

16. 3

2

Page 2: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 2

4.2 Slopes of Lines from Two Points

Answers

1. βˆ’2

3

2. 26

5

3. βˆ’3

10

4. 1

4

5. βˆ’2

5

6. βˆ’1

4

7. βˆ’12

5

8. undefined

9. βˆ’19

23

10.

200 400 600-20

20

40

60

80

100

120

140

160

180

x

y

11. The slope of the line is 1

7, which means that it costs $1 for each additional 7 miles driven.

Page 3: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 3

12.

-4 -2 2 4 6 8 10 12 14 16 18-750

750

1500

2250

3000

3750

x

y

13. The slope of the line is 180, which means that each additional computer costs $180.

14.

-1 1 2 3 4

-1

1

2

3

4

x

y

15. The slope of the line is 1.3 which means that each additional quart of milk costs $1.30.

Page 4: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 4

16.

-1 1 2 3 4 5 6 7 8

10

20

30

40

50

60

x

y

17. The slope of the line is 7.5, which means that each additional hour of tutoring costs $7.50.

Page 5: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 5

4.3 Equations of Lines from Two Points

Answers

1. 𝑦 = 2π‘₯ βˆ’ 5

2. 𝑦 =1

3π‘₯ + 6

3. 𝑦 =3

4π‘₯ βˆ’ 8

4. 𝑦 = βˆ’3π‘₯ + 6

5. 𝑦 = βˆ’1

4π‘₯ βˆ’ 1

6. π‘₯ = βˆ’3

7. 𝑦 = βˆ’5

8. 𝑦 = βˆ’π‘₯ βˆ’ 4

9. 𝑦 =2

5π‘₯ +

28

5

10. 𝑦 = βˆ’2π‘₯ + 5

11. 𝑦 = 6π‘₯ βˆ’ 17

12. a) Two points are (320, 124) and (600, 164). The independent variable is the miles traveled and the

dependent variable is the cost.

b) 𝑦 =1

7π‘₯ + 78.3.

1

7 means every 7 miles costs $1 and 78.3 is the base cost of having the car for the

month.

13. a) Two points are (10, 1950) and (15, 2850). The independent variable is the computers sold and

the dependent variable is the profit.

b) 𝑦 = 180π‘₯ + 150. $180 is the cost per computer and $150 is the base cost before any computers

are sold.

14. a) Two points are (1, 1.65) and (2, 2.95). The independent variable is the number of quarts and the

dependent variable is the cost.

b) 𝑦 = 1.3π‘₯ + .35. The 1.3 means each additional quart costs $1.30 and the $0.35 is the base cost

(perhaps for packaging).

15. a) Two points are (3, 25) and (7, 55). The independent variable is the number of hours spent

tutoring and the dependent variable is the money earned.

b) 𝑦 = 7.5π‘₯ + 2.5. This means that $7.50 was the money earned per hour and $2.50 was the base

charge for tutoring.

Page 6: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 6

4.4 Graphs of Lines from Equations

Answers

1. Slope is 5

8 , y-intercept is (0, 3).

2. Slope is βˆ’4

5, y-intercept is (0,

3

5).

3. Slope is 4

3, y-intercept is (0, 7).

4. Slope is 0, y-intercept is (0, βˆ’7).

5. Slope is 8

9, y-intercept is (0, 3).

6.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

7.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

Page 7: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 7

8.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

9.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

10.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

Page 8: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 8

11.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

12. Slope is undefined.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

13. Slope is 0.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

Page 9: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 9

14. Slope is 0.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

15. Slope is undefined.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y

Page 10: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 10

4.5 Equations of Lines from Graphs

Answers

1. 𝑦 = βˆ’3π‘₯ + 5

2. 𝑦 = 2π‘₯ βˆ’ 3

3. 𝑦 = βˆ’4

3π‘₯ + 5

4. 𝑦 =5

4π‘₯ βˆ’ 3

5. 𝑦 = βˆ’3

2π‘₯ +

5

2

6. 𝑦 =1

2π‘₯ βˆ’

1

2

7. 𝑦 =1

4π‘₯ +

3

4

8. 𝑦 = βˆ’2

3π‘₯ +

1

3

9. 5π‘₯ βˆ’ 𝑦 βˆ’ 4 = 0

10. π‘₯ + 2𝑦 βˆ’ 4 = 0

11. 5π‘₯ βˆ’ 6𝑦 βˆ’ 4 = 0

12. 5π‘₯ βˆ’ 6𝑦 βˆ’ 4 = 0

13. You can’t if you cannot easily identify two points on the graph.

14. A vertical line will have the equation π‘₯ = __ and a horizontal line will have the equation 𝑦 = ___.

Where the ___ is the x or y intercept.

15. π‘₯ βˆ’ 4𝑦 βˆ’ 20 = 0

16. 2π‘₯ βˆ’ 3𝑦 + 3 = 0

17. 7π‘₯ βˆ’ 21𝑦 βˆ’ 9 = 0

Page 11: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 11

4.6 Equations of Parallel and Perpendicular Lines

Answers

1. Parallel

2. Perpendicular

3. Parallel

4. Neither

5. Perpendicular

6. Parallel

7. 𝑦 = 5π‘₯ βˆ’ 31

8. 𝑦 =1

4π‘₯ + 7.25

9. 𝑦 = βˆ’3

2π‘₯ βˆ’ 6.5

10. 𝑦 = βˆ’1

2π‘₯ βˆ’ 6

11. 𝑦 = 2π‘₯

12. 𝑦 = βˆ’1

5π‘₯ + 4.8

13. 𝑦 = βˆ’5

2π‘₯ + 4

14. 𝑦 =1

7π‘₯ βˆ’

11

7

15. 𝑦 = βˆ’7π‘₯ + 27

Page 12: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 12

4.7 Applications of Linear Functions

Answers

1. The dependent variable is the profit and the independent variable is the number of candles sold.

2. (4, 30) and (12, 70)

3.

2 4 6 8 10 12 14

10

20

30

40

50

60

70

80

90

Number of Candles Sold

Profit

4. 𝑦 = 5π‘₯ + 10

5. The slope is 5. This means each candle sold creates a profit of $5.

6. The profit-intercept is 10. This means each player will have at least a $10 profit even if he/she

doesn't sell any candles.

7. The maximum profit is 𝑦 = 5(24) + 10 = $130.

8. Domain: {π‘₯|π‘₯ β‰₯ 0, π‘₯ ∈ π‘Š} Range: {𝑦|𝑦 = 5π‘₯ + 50, π‘₯ ∈ π‘Š}

9. 16 candles.

10. This data is discrete because you cannot sell portions of candles.

11. The dependent variable is distance from home and the independent variable is time spent driving.

Page 13: 4.1 Slopes of Lines from Graphs

Chapter 4 – Linear Functions Answer Key

CK-12 Algebra I Honors Concepts 13

12. (5, 112) and (7, 15)

13.

1 2 3 4 5 6 7

50

100

150

200

250

300

350

Time Spent Driving (Hours)

Distance from Home (km)

14. 𝑦 = βˆ’48.5π‘₯ + 354.5

15. The slope is -48.5 which means his speed in kilometers per hour was 48.5.

16. The distance intercept is 354.5 kilometers which means he started 354.5 kilometers from home.

17. It took about 7.3 hours.

18. Domain: {π‘₯|0 ≀ π‘₯ ≀ 7.31, π‘₯ ∈ 𝑅} Range: {𝑦|0 ≀ 𝑦 ≀ 354.5, 𝑦 ∈ 𝑅}

19. He was 160.5 miles from home

20. 3 hours