graphing rational functions example #6 end showend show slide #1 nextnext we want to graph this...

2

2

x +2x+1f(x)=

x +x -2

Graphing Rational FunctionsExample #6

End Show Slide #1 Next

We want to graph this rational function showing all relevant characteristics.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)


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First we must factor both numerator and denominator, but don’t reduce the fraction yet.

Both factor into 2 binomials.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1



Note the domain restrictions, where the denominator is 0.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1



Now reduce the fraction. In this case, it doesn't reduce.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1



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.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes.



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.Thus, there are no holes in this case.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11



Next look at the degrees of both the numerator and the denominator. Because both the denominator's and the numerator's degrees are the

same, 2, there will be a horizontal asymptote at y=(the ratio of the leading coefficients) and there is no oblique asymptote.

2

2

x +2x+1x

f(x)=+x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =1

1

2

2

2

2

=

x +2x+1=

x +2x+1x +x -

x +x -

12

2

x=-3



Next we need to find where the graph of f(x) would intersect the H.A. To do this we set the reduced form equal to the number from the H.A., and solve

for x.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =1

1

int.w/ H.A. -3at( ,1).

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3



Since the equation has a solution, the intersection will be the point with x-coordinate of the solution of the equation, and the y-coordinate will be the

number from the H.A.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11

int.w/ H.A.at(-3,1).

x - int.=-1

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3



We find the x-intercepts by solving when the function is 0, which would be when the numerator is 0. Thus, when x+1=0.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11


x - int.=-1 1 1

y- int.= =--2 2

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3



Now find the y-intercept by plugging in 0 for x.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11


x - int.=-1 1 1

y- int.= =--2 2

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3

2

2

x=2

2 +2(2)+1 9y= =

2 +2-2 49

pt: 2,4



Plot any additional points needed.Here I only plotted one more point at x=2 since a point hadn't been plotted

to the right of that V.A. You can always choose to plot more points than required to help you find the graph.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11


x - int.=-1 1 1

y- int.= =--2 2

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3

2

2

x=2

2 +2(2)+1 9y= =

2 +2-2 49

pt: 2,4



Finally draw in the curve.For the part to the right of the V.A., x=1, we use that it can't cross the H.A.

and it has to approach the V.A. and the H.A.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11


x - int.=-1 1 1

y- int.= =--2 2

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3

2

2

x=2

2 +2(2)+1 9y= =

2 +2-2 49

pt: 2,4



For the section between the V.A.'s, we use that it can't cross the H.A., it has to approach both V.A.'s and the multiplicity of the x-int. of -1 is 2, so

the graph bounces of the x-axis.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11


x - int.=-1 1 1

y- int.= =--2 2

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3

2

2

x=2

2 +2(2)+1 9y= =

2 +2-2 49

pt: 2,4



For -3<x<-2, we use that the graph has to approach the V.A. of x=-2, and the graph can't cross the x-axis in this interval.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11


x - int.=-1 1 1

y- int.= =--2 2

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3

2

2

x=2

2 +2(2)+1 9y= =

2 +2-2 49

pt: 2,4



For x<-3, we have to use that the graph has to approach the H.A., and to find out that the graph crosses the H.A., we can use either that the

intersection w/ the H.A. has a multiplicity of 1, or we could just plot another point at x=-4.

2

2

x +2x+1f(x)=

x +x -2

2(x+1)

=(x+2)(x -1)

; x -2,1

V.A.:x=-2,1

Noholes. 1

H.A.:y= =11


x - int.=-1 1 1

y- int.= =--2 2

2

2

2 2

x +2x+1=1

x +x -2x +2x+1= x +x -2

x=-3

2

2

x=2

2 +2(2)+1 9y= =

2 +2-2 49

pt: 2,4


Previous Slide #17 End Show

This finishes the graph.

graphing rational functions example #6 end showend show slide #1 nextnext we want to graph this...

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