Transcript
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Find Slope & Rate of Change
Lesson 2.2
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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
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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
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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.
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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
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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
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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)
for Examples 1 and 2
ANSWER 54
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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
–
for Examples 1 and 2
ANSWER12
–
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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
for Examples 1 and 2
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
–
for Examples 1 and 2
ANSWER12
–
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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
–
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Classify lines using slope
EXAMPLE 3
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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 =
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GUIDED PRACTICE for Example 3GUIDED PRACTICE
8. (7, 1), (7, – 1)
SOLUTION
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– 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
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GUIDED PRACTICE for Example 3GUIDED PRACTICE
10. (5, 6), (1, – 4)
SOLUTION
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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
–
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Classify parallel and perpendicular lines
EXAMPLE 4
m2 =3 – (– 1)
2 – (– 4) =4
6 =23
ANSWER
Because m1m2 = –2
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2= – 1, m1 and m2
are negative reciprocals of each other. So, the lines are perpendicular.
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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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GUIDED PRACTICE for Example 4GUIDED PRACTICE
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
Find the slopes of the two lines.a.
m1 =–4 – 8
2 – (– 2) = – 3
m2 = 2 – 1
– 2 – (– 5) =1
3
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GUIDED PRACTICE for Example 4GUIDED PRACTICE
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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GUIDED PRACTICE for Example 4GUIDED PRACTICE
12. Line 1: through (– 4, – 2) and (1, 7)Line 2: through (– 1, – 4) and (3, 5)
SOLUTION
Find the slopes of the two lines.a.
m1 =7 – (– 2)
1 – (– 4) =
m2 = 5 – (– 4) 3 – (– 1) =
9 4
9 5
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GUIDED PRACTICE for Example 4GUIDED PRACTICE
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.
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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=
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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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GUIDED PRACTICE for Example 5GUIDED 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.
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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
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