Lesson 02.2

1

Find Slope & Rate of Change

Lesson 2.2

2

Definition of Slope

• Slope (m) of a nonvertical line is the ratio of the vertical change (the rise) to the horizontal change (the run)

• Algebraic definition

run

rise

12

12

xx

yym

3

Find slope in real lifeEXAMPLE 1

Skateboarding

A skateboard ramp has a rise of 15 inches and a run of 54 inches. What is its slope?

SOLUTION

slope = riserun =

ANSWER

The slope of the ramp is 518

.

518

=1554

4

Standardized Test PracticeEXAMPLE 2

SOLUTION

Let (x1, y1) = (–1, 3) and (x2, y2) = (2, –1).

m =y2 – y1

x2 – x1=

– 1 – 32 – (–1) =

43

ANSWER

The correct answer is A.

5

for Examples 1 and 2GUIDED PRACTICE

1. What If ? In Example 1, suppose that the rise of the ramp is changed to 12 inches without changing the run. What is the slope of the ramp?

SOLUTION

slope = riserun =

ANSWER

The slope of the ramp is 2 9

.

1254

= 2 9

6

GUIDED PRACTICE

SOLUTION

Let (x1, y1) = (–4, 9) and (x2, y2) = (–8, 3).

m =y2 – y1

x2 – x1=

ANSWER

The correct answer is D.

3 – (9)– 8 – (–4) =

32

2. What is the slope of the line passing through the points (– 4, 9) and (– 8, 3) ?

for Examples 1 and 2

7

GUIDED PRACTICE

Let (x1, y1) = (0, 3) and (x2, y2) = (4, 8).

m =y2 – y1

x2 – x1=

8 – 3 4 – 0 =

54

Find the slope of the line passing through the given points.

SOLUTION

3. (0, 3), (4, 8)


ANSWER 54

8

GUIDED PRACTICE

Let (x1, y1) = (– 5, 1) and (x2, y2) = (5, – 4)

m =y2 – y1

x2 – x1=

4. (– 5, 1), (5, – 4)

SOLUTION

– 4 – 1 5 – (–5) =

12

–


ANSWER12

–

9

GUIDED PRACTICE

Let (x1, y1) = (– 3, – 2) and (x2, y2) = (6, 1).

m =y2 – y1

x2 – x1=

1 –( – 2) 6 – (–3) =

13

5. (– 3, – 2), (6, 1)

SOLUTION


ANSWER 13

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GUIDED PRACTICE

Let (x1, y1) = (7, 3) and (x2, y2) = (– 1, 7).

m =y2 – y1

x2 – x1=

6. (7, 3), (– 1, 7)

SOLUTION

7 – 3 – 1 – 7 =

12

–


ANSWER12

–

11

Classification of Lines by Slope

• Positive slope rises from left to right• Negative slope falls from left to right• Zero slope is a horizontal line• Undefined slope is a vertical line

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Classify lines using slope

EXAMPLE 3

Without graphing, tell whether the line through the given points rises, falls, is horizontal, or is vertical.

b. (– 6, 0), (2, –4)

d. (4, 6), (4, –1)c. (–1, 3), (5, 8)

SOLUTION

a. (– 5, 1), (3, 1)

1 – 13– (–5) =m =a. Because m = 0, the line is

horizontal.

– 4 – 02– (–6) =m =b. Because m < 0, the line

falls.

08 = 0

– 48 =

12

–

13

Classify lines using slope

EXAMPLE 3

56

8 – 35– (–1) =m =c. Because m > 0, the line rises.

– 7 0

– 1 – 6 4 – 4 =m =d. Because m is undefined, the

line is vertical.

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GUIDED PRACTICE for Example 3GUIDED PRACTICE

Without graphing, tell whether the line through the given points rises, falls, is horizontal, or is vertical.

7. (– 4, 3), (2, – 6)

SOLUTION

96

– Because m < 0, the line falls.

– 6 – 3 2 – (–4) =m =

15


8. (7, 1), (7, – 1)

SOLUTION

20

– Because m is undefined, the line is vertical.

– 1 – 1 7 – 7 =m =

9. (3, – 2), (5, – 2)

SOLUTION

Because m = 0, line is horizontal.

=– 2 – (– 2) 5 – 3

m =02 = 0

16


10. (5, 6), (1, – 4)

SOLUTION

52

Because m > 0 the line rises.

– 4 – 6 1 – 5 =m =

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Parallel and Perpendicular Lines

• Parallel lines– Two lines are parallel if and only if they have the

same slope

• Perpendicular lines– Two lines are perpendicular if and only if the

product of their slopes is -1• Also known as negative reciprocals

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Classify parallel and perpendicular linesEXAMPLE 4

Tell whether the lines are parallel, perpendicular, orneither.

Line 1: through (– 2, 2) and (0, – 1)a.Line 2: through (– 4, – 1) and (2, 3)

Line 1: through (1, 2) and (4, – 3)b.Line 2: through (– 4, 3) and (– 1, – 2)

SOLUTION

Find the slopes of the two lines.a.

m1 =–1 – 2

0 – (– 2) =– 3

2 =32

–

19

Classify parallel and perpendicular lines

EXAMPLE 4

m2 =3 – (– 1)

2 – (– 4) =4

6 =23

ANSWER

Because m1m2 = –2

33

2= – 1, m1 and m2

are negative reciprocals of each other. So, the lines are perpendicular.

20

Classify parallel and perpendicular lines

EXAMPLE 4

Find the slopes of the two lines.b.

m1 =–3 – 2

4 – 1 =– 5

3 =53

–

m2 =– 2 – 3

– 1 – (– 4) =– 5

3 =53

–

ANSWER

Because m1 = m2 (and the lines are different), you can conclude that the lines are parallel.

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Tell whether the lines are parallel, perpendicular, or neither.

11. Line 1: through (– 2, 8) and (2, – 4)Line 2: through (– 5, 1) and (– 2, 2)

SOLUTION


m1 =–4 – 8

2 – (– 2) = – 3

m2 = 2 – 1

– 2 – (– 5) =1

3

22


ANSWER

Because m1m2 = – 3 2

3= – 1, m1 and m2

are negative reciprocals of each other.So, the lines are perpendicular.

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12. Line 1: through (– 4, – 2) and (1, 7)Line 2: through (– 1, – 4) and (3, 5)

SOLUTION


m1 =7 – (– 2)

1 – (– 4) =

m2 = 5 – (– 4) 3 – (– 1) =

9 4

9 5

24


ANSWER

Because m1 = m2 and m1 and m2 are not reciprocals of each other. So, the lines are neither

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Rate of Change

• Slopes can be used to represent an average rate of change, or how much one quantity changes, on average, relative to the change in another quantity.

26

Solve a multi-step problem

EXAMPLE 5

Forestry

Use the diagram, which illustrates the growth of a giant sequoia, to find the average rate of change in the diameter of the sequoia over time. Then predict the sequoia’s diameter in 2065.

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Solve a multi-step problemEXAMPLE 5

SOLUTION

STEP 1

Find the average rate of change.

141 in. – 137 in.2005 – 1965=

4 in. 40 years=

= 0.1 inch per year

Average rate of changeChange in diameter

Change in time=

28

Solve a multi-step problem

EXAMPLE 5

STEP 2Predict the diameter of the sequoia in 2065.

Find the number of years from 2005 to 2065. Multiply this number by the average rate of change to find the total increase in diameter during the period 2005–2065.

Number of years = 2065 – 2005 = 60Increase in diameter =(60 years) (0.1 inch/year) = 6 inches

ANSWER

In 2065, the diameter of the sequoia will be about 141 + 6 = 147 inches.

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13. What If ? In Example 5, suppose that the diameter of the sequoia is 248 inches in 1965 and 251 inches in 2005. Find the average rate of change in the diameter, and use it to predict the diameter in 2105.

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SOLUTION

STEP 1

Find the average rate of change.

251 in. – 248 in.2005 – 1965=

3 in. 40 years=

= 0.075 inch per year

Average rate of changeChange in diameter

Change in time=

for Example 5GUIDED PRACTICE

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STEP 2Predict the diameter of the sequoia in 2105.

Find the number of years from 2005 to 2105. Multiply this number by the average rate of change to find the total increase in diameter during the period 2005–2105.

Number of years = 2105 – 2005 = 100

Increase in diameter =(100 years) (0.075 inch/year) = 7.5 inches

ANSWER

In 2105, the diameter of the sequoia will be about 251 + 7.5 = 258.5 inches.

for Example 5GUIDED PRACTICE

Lesson 02.2

Technology

Transcript of Lesson 02.2