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example 1 find a positive slope let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1...

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  • EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute
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EXAMPLE 1 Find a positive slope (x 1 , y 1 ) = (–4, 2) = (x 2 , y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute. Write formula for slope. Find the slope of the line shown.

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Page 1: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 1 Find a positive slope

Let (x1, y1) = (–4, 2) = (x2, y2) = (2, 6).

m =y2 – y1

x2 – x1

6 – 22 – (–4)

=

=46

23= Simplify.

Substitute.

Write formula for slope.

Find the slope of the line shown.

Page 2: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 2 Write an equation from a graphGUIDED PRACTICE for Example 1

Find the slope of the line that passes through the points.

ANSWER 3

1. (5, 2) and (4, –1)

Page 3: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

Write an equation from a graphGUIDED PRACTICE for Example 1

2. (–2, 3) and (4, 6)

1 2ANSWER

Find the slope of the line that passes through the points.

Page 4: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

Write an equation from a graphGUIDED PRACTICE for Example 1

3. ( , 5) and ( , –3)92

12

ANSWER 2

Find the slope of the line that passes through the points.

Page 5: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 2 Find a negative slope

Find the slope of the line shown.

m =y2 – y1

x2 – x1

Let (x1, y1) = (3, 5) and (x2, y2) = (6, –1).

–1 – 56 – 3

=

– 63= = –2

Write formula for slope.

Substitute.

Simplify.

Page 6: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 3 Find the slope of a horizontal line

Find the slope of the line shown.

Let (x1, y1) = (–2, 4) and (x2, y2) = (4, 4).

m =y2 – y1

x2 – x1

4 – 44 – (–2)

=

06= = 0

Write formula for slope.

Substitute.

Simplify.

Page 7: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 4 Find the slope of a vertical line

Find the slope of the line shown.

Let (x1, y1) = (3, 5) and (x2, y2) = (3, 1).

m =y2 – y1

x2 – x1Write formula for slope.

1 – 53 – 3

= Substitute.

Division by zero is undefined.

ANSWER

Because division by zero is undefined, the slope of a vertical line is undefined.

– 40=

Page 8: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 2 Write an equation from a graphGUIDED PRACTICE for Examples 2, 3 and 4

Find the slope of the line that passes through the points.

4. (5, 2) and (5, –2)

ANSWER undefined

Page 9: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

Write an equation from a graphGUIDED PRACTICE for Examples 2, 3 and 4

5. (0, 4) and (–3, 4)

Find the slope of the line that passes through the points.

0ANSWER

Page 10: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

Write an equation from a graphGUIDED PRACTICE for Examples 2, 3 and 4

6. (0, 6) and (5, –4)

Find the slope of the line that passes through the points.

ANSWER –2

Page 11: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 5 Find a rate of change

INTERNET CAFEThe table shows the cost of using a computer at an Internet cafe for a given amount of time. Find the rate of change in cost with respect to time.

Time (hours) 2 4 6

Cost (dollars) 7 14 21

Page 12: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 5 Find a rate of change

Rate of change =change in costchange in time

14 – 74 – 2

=72= 3.5=

ANSWER

The rate of change in cost is $3.50 per hour.

SOLUTION

Page 13: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

Time(minute) 30 60 90

Distance (miles)

1.5 3 4.5

GUIDED PRACTICE for Example 5

The table shows the distance a person walks for exercise. Find the rate of change in distance with respect to time.

7. EXERCISE

ANSWER 0.05 mi/min

Page 14: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 6 Use a graph to find and compare rates of change

COMMUNITY THEATER

A community theater performed a play each Saturday evening for 10 consecutive weeks. The graph shows the attendance for the performances in weeks 1, 4, 6, and 10. Describe the rates of change in attendance with respect to time.

Page 15: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

SOLUTION

EXAMPLE 6 Use a graph to find and compare rates of change

Find the rates of change using the slope formula.

Weeks 1–4: 232 – 1244 – 1 =

1083

= 36 people per week

Weeks 4–6: 204 – 2326 – 4 =

–28 2

= –14 people per week

Weeks 6–10: 72 – 20410 – 6 =

–1324 = –33 people per week

ANSWER

Attendance increased during the early weeks of performing the play. Then attendance decreased, slowly at first, then more rapidly.

Page 16: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 7 Interpret a graph

COMMUTING TO SCHOOL

A student commutes from home to school by walking and by riding a bus. Describe the student’s commute in words.

Page 17: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 7 Interpret a graph

The first segment of the graph is not very steep, so the student is not traveling very far with respect to time. The student must be walking. The second segment has a zero slope, so the student must not be moving. He or she is waiting for the bus. The last segment is steep, so the student is traveling far with respect to time. The student must be riding the bus.

SOLUTION

Page 18: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 7 Interpret a graphGUIDED PRACTICE for Examples 6 and 7

WHAT IF? How would the answer to Example 6 change if you knew that attendance was 70 people in week 12?

8.

Sample answer: The attendance did not decrease as rapidly between weeks 10 and 12.

ANSWER

Page 19: EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute

EXAMPLE 7 Interpret a graphGUIDED PRACTICE for Examples 6 and 7

WHAT IF? Using the graph in Example 7, draw a graph that represents the student’s commute from school to home.

9.

ANSWER


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