Transcript Graphing Rational Functions Example #3 f(x)= x3 +4x2 x2 +2x+1 We want to graph this rational function showing all relevant characteristics. END SHOW Slide #1 Next Graphing Rational Functions Example.

Graphing Rational Functions
Example #3
f(x)=
x3 +4x2
x2 +2x+1
We want to graph this rational function showing all relevant characteristics.
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Slide #1
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
=
(x+1)2
f(x)=
First we must factor both numerator and denominator, but don’t reduce the
fraction yet.
Numerator: Factor out a GCF of x2.
Denominator: Factor as a perfect square trinomial.
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Slide #2
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
f(x)=
Note the domain restrictions, where the denominator is 0.
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Slide #3
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
f(x)=
Now reduce the fraction. In this case, there isn't a common factor. Thus, it
doesn't reduce.
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Slide #4
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
f(x)=
Any places where the reduced form is undefined, the denominator is 0,
forms a vertical asymptote. Remember to give the V. A. as the full equation
of the line and to graph it as a dashed line.
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Slide #5
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
f(x)=
Any values of x that are not in the domain of the function but are not a V.A.
form holes in the graph. In other words, any factor that reduced completely
out of the denominator would create a hole in the graph where it is 0.
Since this example didn't reduce, it has no holes.
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Slide #6
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
f(x)=
Next look at the degrees of both the numerator and the denominator.
Because the denominator's degree,2, is exactly 1 less than the
numerator's degree,3, there will be an oblique asymptote, but no horizontal
asymptote.
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Slide #7
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
f(x)=
x +2
x +2x +1 x +4x +0x + 0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
- 5x -2
To find the O.A. we must divide out the rational expression. In this case,
since the fraction didn't reduce we will use the original form.
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Slide #8
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
f(x)=
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
- 5x -2
The O.A. will be y=(what is on top of the division).
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Slide #9
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
f(x)=
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 + 4x +2)
- 5x -2
-5x -2 = 0
-5x =2
2
x=5
Now we need to find the intersections between the graph of f(x) and the
O.A. Usually the easiest way to do this is to set the remainder from the
division equal to 0 and solve for x.
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Slide #10
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
 2 8
Int. w/ O.A. at  - , 
 5 5
f(x)=
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
-5x -2
2
-5x -2 = 0
x=5
-5x =2
8
 2
y =  -  +2 =
2
5
x= 5
5
Next we need to find the y coordinate of the intersection by plugging the x
we just found into the equation from the O.A.
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Slide #11
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
 2 8
Int. w/ O.A. at  - , 
 5 5
x -int =0,-4
f(x)=
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
-5x -2
2
-5x -2 = 0
x=5
-5x =2
8
 2
y =  -  +2 =
2
5
x= 5
5
We find the x-intercepts by solving when the function is 0, which would be
when the numerator is 0. Thus, when x2=0 and x+4=0.
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Slide #12
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
 2 8
Int. w/ O.A. at  - , 
 5 5
x -int =0,-4
f(x)=
y -int =
02 (0+4)
=0
(0+1)2
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
-5x -2
2
-5x -2 = 0
x=5
-5x =2
8
 2
y =  -  +2 =
2
5
x= 5
5
Now find the y-intercept by plugging in 0 for x.
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Slide #13
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
 2 8
Int. w/ O.A. at  - , 
 5 5
x -int =0,-4
f(x)=
y -int =
02 (0+4)
=0
(0+1)2
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
-5x -2
2
-5x -2 = 0
x=5
-5x =2
8
 2
y =  -  +2 =
2
5
x= 5
5
Plot any additional points needed.
In this case we don't need any other points to determine the graph.
Though, you can always plot more points if you want to.
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Slide #14
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
 2 8
Int. w/ O.A. at  - , 
 5 5
x -int =0,-4
f(x)=
y -int =
02 (0+4)
=0
(0+1)2
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
-5x -2
2
-5x -2 = 0
x=5
-5x =2
8
 2
y =  -  +2 =
2
5
x= 5
5
Finally draw in the curve.
Let's start on the interval for x<-1, the graph has to pass through the point
(-4,0) and approach both asymptotes.
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Slide #15
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
 2 8
Int. w/ O.A. at  - , 
 5 5
x -int =0,-4
f(x)=
y -int =
02 (0+4)
=0
(0+1)2
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
-5x -2
2
-5x -2 = 0
x=5
-5x =2
8
 2
y =  -  +2 =
2
5
x= 5
5
For -1<x<0, the graph to go through the points (0,0) and (-2/5,8/5) and
approach the V.A. The graph has to approach the V.A. going up since it
can cross the x-axis between x=-1 and x=-2/5.
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Slide #16
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
 2 8
Int. w/ O.A. at  - , 
 5 5
x -int =0,-4
f(x)=
y -int =
02 (0+4)
=0
(0+1)2
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
-5x -2
2
-5x -2 = 0
x=5
-5x =2
8
 2
y =  -  +2 =
2
5
x= 5
5
For x>0, the multiplicity of the x-intercept of 0 is 2. Which since it is even
the graph must bounce off the x-axis. Then the graph must approach the
O.A.
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Slide #17
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Graphing Rational Functions
Example #3
x3 +4x2
x2 +2x+1
x2 (x+4)
;x  -1
=
(x+1)2
V.A.: x = -1
No Holes
O.A.: y = x+2
 2 8
Int. w/ O.A. at  - , 
 5 5
x -int =0,-4
f(x)=
y -int =
02 (0+4)
=0
(0+1)2
x +2
x +2x +1 x +4x +0x +0
2
3
2
-(x 3 +2x2 +x)
2x2 - x +0
-(2x2 +4x +2)
-5x -2
2
-5x -2 = 0
x=5
-5x =2
8
 2
y =  -  +2 =
2
5
x= 5
5
This finishes the graph.
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Slide #18
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