Transcript Ch.2:4

Section 2.4
More on Slope
Parallel and Perpendicular Lines
y=2x+7
Parallel Lines
Find a line parallel to -x+6y=8 and passing through (-2,3).
-x+6y=8
6y=x+8
1
4
y= x 
6
3
1
y-3= ( x  2)
6
1
1
y-3= x 
6
3
1
1
y= x   3
6
3
1
10
y= x 
6
3
Solve for y
1
Use the slope
6
Substitute into the Point-Slope form
Example
Write the equation in slope intercept form for a line
that is parallel to 3x-4y=12 and passing through
(5,2).
Perpendicular Lines
Find a line perpendicular to -x+6y=8 and passing through (-2,3).
-x+6y=8
Solve for y
6y=x+8
1
4
y= x 
6
3
y-3= - 6( x  2)
y-3=  6 x  12
y= - 6x  12  3
y=-6x  9
1
6
Substitute into the Point-slope form
Use the negative reciprocal slope of
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.
y
change in y
m
or
x
change in x
or
y2  y1
x2  x1
Interpreting a real life situation
The 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 2007
Percent Adults
(0,24.7)
y










(10,19.5)




change in y 19.5  24.7
m

change in x
10  0
-5.2
=
10
=-.52


b. What does this slope
represent?










x









X years after 1997



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
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.
y

( x2 , f ( x2 ))


f ( x2 )  f ( x1)
x2  x1

Secant line


( x1, f ( x1))



x












The slope of this line between the
points (1,3.83) and (5,7.83) is
f ( x2 )  f ( x1)
x2  x1
y


7.83  3.83 4
  1
5 1
4
(5,7.83)




(1,3.83)



x












The slope of this line between the
points (1,3.83) and (4,7.34) is
y
f ( x2 )  f ( x1)
x2  x1



(4,7.34)
7.34  3.83 3.51
 1.17

3
4 1



(1,3.83)



x





Continuation of
same problem







The slope of this line between the
points (1,3.83) and (3,6.5) is
y
f ( x2 )  f ( x1)
x2  x1



6.5  3.83 2.67
 1.34

3 1
2
(3,6.5)



(1,3.83)



x






Continuation
of same problem






Let’s look at the different slopes
from the point (1, 3.83).
y
x

y
3 6.5
4 7.34
5 7.83












Continuation
of same problem




1.34
1.17
1
•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.
x

Slope of the
secant line



Find the average rate of change of
f(x)= x3
y
a. When x1  2 and x 2  3

f (x1 )  8, f (x 2 )  27
c. When x1  2 and x 2  2
f (x1 )  8, f ( x2 )  8
8  (8) 16

4
2  (2) 4
27  8 19

 19
3 2
1
b. When x1  3 and x 2  4
x
f (x1 )  27, f (x 2 )  64


64  27 37

 37
43
1

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
The average rate of change from x1  x to x 2  x  h is
f ( x2 )  f ( x1 ) f ( x  h)  f ( x) f ( x  h)  f ( x)


x2  x1
xhx
h
The last expression is the difference quotient. The
difference quotient gives the average rate of change
of a function from x to x+h. In the difference quotient,
h 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.
Average rate of change and the
difference quotient
f ( x2 )  f ( x1)
x2  x1
y
(( x  h), f ( x  h))

f ( x  h)

Suppose x1=x and
x2=x+h, then

f ( x  h)  f ( x )
( x  h)  x


f ( x)
( x, f ( x))

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.
x

x




h



( x  h)





Average Rate of Change Application
Example
• 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?
concentration




(1, 3.96)




(4, .72)













hours
Average Rate of Change Application
Example
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?
concentration
hours cncntr
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





y








x












hours
Find the equation of the line that is parallel
to 3x-5y=8 and passes through the point (4,-5).
(a)
(b)
(c)
(d)
3
7
y= x 
5
5
3
y= x  9
5
5
7
y=- x 
3
5
3
37
y= x 
5
5
Find the equation of the line that is perpendicular
to 8x-3y=6 and passes through (-1,3).
(a)
(b)
(c)
(d)
3
y= x  4
5
5
5
y=- x 
3
3
3
y=- x  5
5
5
y=- x  5
3
Find the average rate of change of the function
f(x)=x 2 - 2x from x1 =0 to x 2 =3.
(a) 1
(b) 0
(c) 3
(d) 1