Section 1.5More on Slope

Parallel and Perpendicular Lines

y=2x+7

Find a line parallel to -x+6y=8 and passing through (-2,3).

-x+6y=8 Solve for y

6y=x+8

1 4 1 y= Use the slope

6 3 61

y-3= ( 2) Substitute into the Point-Slope form6

y

x

x

1 1-3=

6 31 1

y= 36 31 10

y=6 3

x

x

x

Parallel Lines

Example

Write the equation in slope intercept form for a line that is parallel to 3x-4y=12 and passing through (5,2).

Find a line perpendicular to -x+6y=8 and passing through (-2,3).

-x+6y=8 Solve for y

6y=x+8

1 4 1 y= Use the negative reciprocal slope of

6 3 6y-3= - 6( 2) S

x

x

ubstitute into the Point-slope form

y-3= 6 12

y= - 6 12 3

y=-6 9

x

x

x

Perpendicular Lines

Example

Write the equation in slope intercept form for a line perpendicular to 3x-4y=12 and passing through (5,2).

Slope as Rate of Change

Definition- Slope is defined as the ratio of a change in y to a corresponding change in x.

change in y or

change in x

ym

x

or 2 1

2 1

y y

x x

Interpreting a real life situationThe line graphs the percent of US adults who smoke cigarettes x years after 1997.

a. Find the slope of the line segment from 1997 to 2007Percent Adults

X years after 1997

x

y(0,24.7)

(10,19.5)

change in y

change in xm

-5.2 =

10

change in y

change in

24.719.5

10 0 xm

=-.52

b. What does this slope represent?

The percent of US adult cigarette smokers is decreasing by .52 percent each year. The change is consistent each year.

The Average Rate of Change of a Function

• If the graph of a function is not a straight line, the average rate of change between any two points is the slope of the line containing the two points. This line is called a secant line.

x

y

The average rate of change of a function.

x

y

Secant line

1 1( , ( ))x f x

2 2( , ( ))x f x

2

1

1

2( ) ( )f xf

x x

x

The slope of this line between the points (1,3.83) and (5,7.83) is

x

y

3.837.83

15

2

1

1

2( ) ( )f xf

x x

x

4

4 1

(1,3.83)

(5,7.83)

The slope of this line between the points (1,3.83) and (4,7.34) is

x

y

2

1

1

2( ) ( )f xf

x x

x

3.837.34

14

1.17

3.51

3

(1,3.83)

(4,7.34)

Continuation of same problem

The slope of this line between the points (1,3.83) and (3,6.5) is

2

1

1

2( ) ( )f xf

x x

x

x

y

3.836.5

13

2.67

2 1.34

(1,3.83)

(3,6.5)

Continuation of same problem

x

y

Let’s look at the different slopes from the point (1, 3.83).

x

y

x ySlope of the secant line

3 6.5 1.34

4 7.34 1.17

5 7.83 1

Continuation of same problem

•Notice how the slope changes depending upon the point that you choose because this function is a curve, not a line. So the average rate of change varies depending upon which points you may choose.

Find the average rate of change of f(x)= x3

1 2

1 2

. When x 2 and x 3

(x ) 8, (x ) 27

27 8 1919

3 2 1

a

f f

1 2

1 2

. When x 3 and x 4

(x ) 27, (x ) 64

64 27 3737

4 3 1

b

f f

1 2

1 2

. When x 2 and x 2

(x ) 8, ( ) 8

8 ( 8) 164

2 ( 2) 4

c

f f x

x

y

Example

Find the average rate of change of f(x)=3x-1 from x1 =0 to x2=1

x1 =1 to x2=2

x1 =2 to x2=3

1 2

2 1

2 1

The average rate of change from x to x is

( ) ( ) ( ) ( ) ( ) ( )

The last expression is the difference quotient. The

difference quotient gives the average rate of chang

x x h

f x f x f x h f x f x h f x

x x x h x h

e

of a function from x to x+ . In the difference quotient,

is thought of as a number very close to 0. In this way,

the average rate of change can be found for a very short

interval.

h

h

Average rate of change and the difference quotient

x

y

2

1

1

2( ) ( )f xf

x x

x

(( )

( h)

)f xx

x

f h

x

( )f x h

( )f x

( h)x xh

( , ( ))x f x

( h), ( )( )x f x h

Do you recognize the difference quotient that we studied in section 2.2?

You will study more about the difference quotient in future math classes.

Suppose x1=x and x2=x+h, then

Average Rate of Change Application

• When a person receives a drug injection, the concentration of the drug in the blood is a function of the hours elapsed after the injection. X represents the hours after the injection and f(x) represents the drug’s concentration in milligrams per 100 milliliters.

• a. Find the average rate of change of the drug’s concentration between the 1st and 4th hours.

• b. What does this value mean in terms of the drug’s concentration?

(1, 3.96)

(4, .72)

hours

concentration

Example

0.00 0.00 0.50 6.14 1.00 8.10 1.50 8.54 2.00 8.43 2.50 8.13 3.00 7.77 3.50 7.41 4.00 7.05 4.50 6.71 5.00 6.38

Average Rate of Change Application

x

y

Sometimes it takes a while for a drug to diffuse sufficiently to affect the desired organ. The curve below is a close approximation of the concentration in that organ compared to the time after the drug was taken. a. What is the average rate of change from the time the drug was taken until the first hour? b. What was the average rate of change from the second hour until the fourth.? c. What is the interpretation of each answer?

hours cncntr

hours

concentration

Example

(a)

(b)

(c)

(d)

Find the equation of the line that is parallel

to 3x-5y=8 and passes through the point (4,-5).

3 7y=

5 53

y= 955 7

y=-3 5

3 37y=

5 5

x

x

x

x

(a)

(b)

(c)

(d)

Find the equation of the line that is perpendicular

to 8x-3y=6 and passes through (-1,3).

3y= 4

55 5

y=-3 33

y=- 555

y=- 53

x

x

x

x

(a)

(b)

(c)

(d)

21 2

Find the average rate of change of the function

f(x)=x - 2x from x =0 to x =3.

1

0

3

1