The document discusses slope and rate of change. It defines slope as the ratio of rise over run, and gives examples of calculating slope from points on a line. It also discusses using slope to find the average rate of change between quantities and solving multi-step problems involving rate of change. Lines can be classified by their slopes as horizontal, vertical, increasing, or decreasing. Parallel and perpendicular lines have specific relationships between their slopes.

1. Find Slope & Rate of Change Lesson 2.2

3. Find slope in real life EXAMPLE 1 Skateboarding SOLUTION A skateboard ramp has a rise of 15 inches and a run of 54 inches. What is its slope ? slope = rise run = ANSWER The slope of the ramp is 5 18 . 5 18 = 15 54

4. Standardized Test Practice EXAMPLE 2 SOLUTION Let ( x 1 , y 1 ) = ( –1 , 3 ) and ( x 2 , y 2 ) = ( 2 , –1 ). m = y 2 – y 1 x 2 – x 1 = – 1 – 3 2 – ( –1 ) = 4 3 ANSWER The correct answer is A.

5. for Examples 1 and 2 GUIDED 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 = rise run = ANSWER The slope of the ramp is 2 9 . 12 54 = 2 9

6. GUIDED PRACTICE SOLUTION Let ( x 1 , y 1 ) = (–4, 9) and ( x 2 , y 2 ) = (–8, 3). for Examples 1 and 2 m = y 2 – y 1 x 2 – x 1 = ANSWER The correct answer is D. 3 – (9) – 8 – (–4) = 3 2 2. What is the slope of the line passing through the points (– 4, 9) and (– 8, 3) ?

7. GUIDED PRACTICE Let ( x 1 , y 1 ) = (0, 3) and ( x 2 , y 2 ) = (4, 8). Find the slope of the line passing through the given points . SOLUTION 3. (0, 3), (4, 8) for Examples 1 and 2 m = y 2 – y 1 x 2 – x 1 = 8 – 3 4 – 0 = 5 4 ANSWER 5 4

8. GUIDED PRACTICE Let ( x 1 , y 1 ) = (– 5, 1) and ( x 2 , y 2 ) = (5, – 4) 4. (– 5, 1), (5, – 4) SOLUTION for Examples 1 and 2 m = y 2 – y 1 x 2 – x 1 = – 4 – 1 5 – (–5) = 1 2 – ANSWER 1 2 –

9. GUIDED PRACTICE Let ( x 1 , y 1 ) = (– 3, – 2) and ( x 2 , y 2 ) = ( 6, 1). 5. (– 3, – 2), (6, 1) SOLUTION for Examples 1 and 2 m = y 2 – y 1 x 2 – x 1 = 1 –( – 2) 6 – (–3) = 1 3 ANSWER 1 3

10. GUIDED PRACTICE Let ( x 1 , y 1 ) = (7, 3) and ( x 2 , y 2 ) = (– 1 , 7). 6. (7, 3), (– 1, 7) SOLUTION for Examples 1 and 2 m = y 2 – y 1 x 2 – x 1 = 7 – 3 – 1 – 7 = 1 2 – ANSWER 1 2 –

12. Classify lines using slope EXAMPLE 3 Without graphing, tell whether the line through the given points rises, falls, is horizontal, or is vertical. SOLUTION Because m = 0 , the line is horizontal. Because m < 0, the line falls. b. (– 6, 0), (2, –4) d. (4, 6), (4, –1) c. (–1, 3), (5, 8) a. (– 5, 1), (3, 1) 1 – 1 3– ( –5) = m = a. – 4 – 0 2– ( –6) = m = b. 0 8 = 0 – 4 8 = 1 2 –

13. Classify lines using slope EXAMPLE 3 Because m > 0, the line rises. Because m is undefined, the line is vertical. 5 6 8 – 3 5– ( –1) = m = c. – 7 0 – 1 – 6 4 – 4 = m = d.

14. GUIDED PRACTICE for Example 3 GUIDED PRACTICE Without graphing, tell whether the line through the given points rises , falls , is horizontal , or is vertical . 7. (– 4, 3), (2, – 6) SOLUTION Because m < 0, the line falls. 9 6 – – 6 – 3 2 – ( –4) = m =

15. GUIDED PRACTICE for Example 3 GUIDED PRACTICE 8. (7, 1), (7, – 1) SOLUTION Because m is undefined, the line is vertical. 9. (3, – 2), (5, – 2) SOLUTION Because m = 0, line is horizontal. 2 0 – – 1 – 1 7 – 7 = m = = – 2 – (– 2) 5 – 3 m = 0 2 = 0

16. GUIDED PRACTICE for Example 3 GUIDED PRACTICE 10. (5, 6), (1, – 4) SOLUTION Because m > 0 the line rises. 5 2 – 4 – 6 1 – 5 = m =

18. Classify parallel and perpendicular lines EXAMPLE 4 Tell whether the lines are parallel, perpendicular, or neither. SOLUTION 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) Find the slopes of the two lines. a . m 1 = – 1 – 2 0 – (– 2) = – 3 2 = 3 2 –

19. Classify parallel and perpendicular lines EXAMPLE 4 m 2 = 3 – (– 1) 2 – (– 4) = 4 6 = 2 3 ANSWER Because m 1 m 2 = – 2 3 3 2 = – 1, m 1 and m 2 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 . m 1 = – 3 – 2 4 – 1 = – 5 3 = 5 3 – m 2 = – 2 – 3 – 1 – (– 4) = – 5 3 = 5 3 – ANSWER Because m 1 = m 2 (and the lines are different), you can conclude that the lines are parallel.

21. GUIDED PRACTICE for Example 4 GUIDED PRACTICE Tell whether the lines are parallel , perpendicular , or neither . SOLUTION – 3 11. Line 1 : through ( – 2, 8) and (2, – 4) Line 2 : through ( – 5, 1) and ( – 2, 2) Find the slopes of the two lines. a . m 1 = – 4 – 8 2 – (– 2) = m 2 = 2 – 1 – 2 – (– 5) = 1 3

22. GUIDED PRACTICE for Example 4 GUIDED PRACTICE ANSWER Because m 1 m 2 = – 3 = – 1, m 1 and m 2 are negative reciprocals of each other. So, the lines are perpendicular. 2 3

23. GUIDED PRACTICE for Example 4 GUIDED PRACTICE SOLUTION 12. Line 1 : through ( – 4, – 2) and (1, 7 ) Line 2 : through ( – 1, – 4) and ( 3 , 5) Find the slopes of the two lines. a . m 1 = 7 – (– 2) 1 – (– 4) = m 2 = 5 – (– 4) 3 – (– 1) = 9 4 9 5

24. GUIDED PRACTICE for Example 4 GUIDED PRACTICE ANSWER Because m 1 = m 2 and m 1 and m 2 are not reciprocals of each other. So, the lines are neither

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.

27. Solve a multi-step problem EXAMPLE 5 SOLUTION STEP 1 Find the average rate of change. = 0.1 inch per year 141 in. – 137 in. 2005 – 1965 = 4 in. 40 years = Average rate of change Change in diameter Change in time =

28. Solve a multi-step problem EXAMPLE 5 STEP 2 Predict 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 = 60 Increase 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.

29. GUIDED PRACTICE for Example 5 GUIDED PRACTICE 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 .

30. SOLUTION STEP 1 Find the average rate of change. = 0.075 inch per year for Example 5 GUIDED PRACTICE 251 in. – 248 in. 2005 – 1965 = 3 in. 40 years = Average rate of change Change in diameter Change in time =

31. STEP 2 Predict 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 for Example 5 GUIDED PRACTICE ANSWER In 2105, the diameter of the sequoia will be about 251 + 7.5 = 258.5 inches.