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Transcript
Graphing Rational Functions
Steps
Step #1
Factor both numerator and denominator, but don’t reduce the fraction yet.
Example
f(x)=
End Show
3x2 +7x+2 (3x+1)(x+2)
=
x2 -4
(x -2)(x+2)
Slide #1
Next
Graphing Rational Functions
Steps
Step #2
Note the domain restrictions. For rational functions, x can not be any
number that makes the denominator 0.
Example
f(x)=
Previous
(5x+1)(x+5)
2
; x  , -5
(3x -2)(x+5)
3
Slide #2
Next
Graphing Rational Functions
Steps
Step #3
Reduce the fraction.
Example
f(x)=
Previous
(2x+1)(x+1) (2x+1)
=
(5x -2)(x+1) (5x -2)
Slide #3
Next
Graphing Rational Functions
Steps
Step #4
Find the vertical asymptotes (V.A.). The V.A. will be where the
denominator of the reduced form is 0. Remember to give the V.A. as the
full equation of the line.
Notice that you are using the reduced form from step #3 not the factored
form from step #1. Thus, the numbers for the V.A. will be among the
numbers in the domain restriction in step #2, but they won't always be all
the numbers.
Example
(2x +1)(x +1)
2
f(x)=
; x  ,-1
(5x -2)(x +1)
5
(2x +1)
=
(5x -2)
2
V.A.: x =
5
Previous
Slide #4
Next
Graphing Rational Functions
Steps
Step #5
If there are any numbers that are in the domain restriction but are not in
the V.A., there will be a hole in the graph at those numbers.
To find the y-coordinate of the hole plug the number into the reduced
form.
Example
(3x -1)(x +2)
1
; x  - , -2
(2x +1)(x +2)
2
(3x -1)
=
(2x +1)
1
V.A.: x = 2
f(x)=
Previous
Slide #5
x = -2
[3(-2)-1] 7
y=

[2(-2)+1] 3
7

Hole at  -2, 
3

Next
Graphing Rational Functions
Steps
Step #6
Next find the horizontal asymptote (H.A.) or the oblique asymptote
(O.A.) and the intersections with the H.A. or O.A.
Previous
Slide #6
Next
Graphing Rational Functions
Steps
Step #6.1
First does the numerator or the denominator have the larger degree.
Choose the NEXT under the case you want to see.
Case A
Denominator has
larger degree
Case B
Both degrees are
the same
Case C or D
Numerator has
larger degree
Example
Example
Example
f(x)=
(3x -1)
(2x+1)(x+1)
NEXT
Previous
f(x)=
(3x2 -1)
(2x+1)(x+1)
NEXT
Slide #7
f(x)=
(3x5 -1)
(2x+1)(x+1)
NEXT
Graphing Rational Functions
Steps
Step #6.2 Case A
When the denominator has the larger degree, y=0 is always the H.A.
and there is no O.A.
Example
(3x -1)
(2x +1)(x +1)
H.A.: y =0
f(x)=
Previous
Slide #8
Next
Graphing Rational Functions
Steps
Step #6.3 Case A
Since the H.A. is y=0 and y=0 is the x-axis, the intersections with the
H.A. are the x-intercepts which will be done in a later step.
Previous
Slide #9
Next
Graphing Rational Functions
Steps
Step #6.2 Case B
When the degrees are the same the H.A. is y=( the ratio of the leading
coefficients).
Example
3x2 -1
2x2 +3x +1
3
H.A.: y =
2
f(x)=
Previous
Slide #10
Next
Graphing Rational Functions
Steps
Step #6.3 Case B
To find the intersections with the H.A. set the reduced form equal to the
H.A. and solve for x. The y-coordinate will be the same number as the H.A.
Example
3x2 -1
2x2 + 3x +1
3
H.A.: y =
2
f(x)=
Previous
3x2 -1
3
=
2
2x + 3x +1 2
6x2 -2 =6x2 +9x + 3
-5= 9x
5
 =x
9
Slide #11
 5 3
Intersection w/H. A. at   , 
 9 2
Next
Graphing Rational Functions
Steps
Step #6.1.1 Case C or D
Is the degree of the numerator exactly 1 more than the degree of the
denominator?
Case C
Yes, exactly one
more.
Case D
No, 2 or more.
Example
Example
f(x)=
2x2 -1
x +1
f(x)=
NEXT
Previous
x5 +1
1- x
NEXT
Slide #12
Graphing Rational Functions
Steps
Step #6.2 Case C
Divide out the fraction to get the form
mx +b+
r(x)
q(x)
The O.A. is y=mx+b.
Example
f(x)=
2x2 -1
x +1
1
=2x -2+
x +1
2x - 2
x +1 2x + 0x - 1
2
-(2x2 +2x)
O.A.: y =2x -2
-2x - 1
-(-2x -2)
1
Previous
Slide #13
Next
Graphing Rational Functions
Steps
Step #6.3 Case C
To find the intersection with the O.A., set the function equal to the O.A.
and solve for x. This is equivalent to setting the remainder to 0.
Example
f(x)=
2x2 -1
x +1
=2x -2+
Previous
1
x +1
2x2 - 1
= 2x -2
x +1
1
2x -2 +
= 2x -2
x +1
1
=0
x +1
1=0
Slide #14
Since 1  0, there are
no intersections w/ the O.A.
Next
Graphing Rational Functions
Steps
Step #6.2 Case D
In this case there is neither a H.A. nor an O.A. All that you are required
to know is that the end behavior of the graph is to either turn up or turn
down at each end.
However, there is an optional step from College Algebra that can help
determine the end behavior.
Skip Optional
Steps
Previous
Go to Optional
Steps
Slide #15
Graphing Rational Functions
Steps
Step #6.3 Case D
Optional Step: Take the non-factored form and drop off all but the
leading terms in the numerator and denominator. Then reduce to get a
monomial. Note in the example below I didn't use an = when I dropped off
the terms since the 2 expressions are not equal.
The end behavior of the f(x) is the same as the end behavior of the
monomial.
Example
x5 +1
f(x)=
1- x
x5

= -x 4
-x
Previous
Since - x4 has a negative coefficient and an even degree,
the end behavior is that both ends turn down.
Slide #16
Next
Graphing Rational Functions
Steps
Step #7
Find x-intercept by finding where the numerator is 0 which would be
where each factor of the numerator is 0.
Example
(x+3)(2x -1)(x2 +1)
f(x)=
x -1
Previous Slide
Step #6
x +3 = 0
x = -3
2x -1= 0
2x =1
1
x=
2
Slide #17
x2 +1=0
x2 = -1
No real Soln.
x -int.= -3,
1
2
Next
Graphing Rational Functions
Steps
Step #8
Find the y-intercept by plugging in 0 for x.
Example
(x +4)(5x -1)
f(x)=
x +2
Previous
y -int.=
(0+4)[5(0)-1] (4)(-1)
=
= -2
0+2
2
Slide #18
Next
Graphing Rational Functions
Steps
Step #9
If you haven't done so already plot what you know about the graph from
previous steps.
Then plot any additional points, by choosing values of x where either
1) there is a section of the graph with no points plotted already.
or 2) you are unsure about where the graph is.
Previous
Slide #19
Next
Graphing Rational Functions
Steps
Step #10
Finally, draw the graph using the information from previous steps.
Previous
Slide #20
Restart
End Show