graphing rational functions example #8 previouspreviousslide #1 nextnext we want to graph this...

41- xf(x)= 2x+1

x

y

Graphing Rational FunctionsExample #8

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

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

x

y



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

Numerator: Factor out the negative, then difference of squares twice.Denominator: It's prime.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

x

y



Note the domain restrictions, where the denominator is 0.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

x

y



Now reduce the fraction. In this case, there are no common factors. So it doesn't reduce.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2

x

y



Any places where the reduced form is undefined, the denominator is 0, forms a vertical asymptote. Remember to give the V. A. and the full

equation of the line and to graph it as a dashed line.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2

NoHoles

x

y



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.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2

NoHoles NoH.A.or O.A.

x

y



Next look at the degrees of both the numerator and the denominator. Because the denominator's degree, 1, is less than the numerator's, 4,by

more than 1, there is neither a horizontal asymptote nor an oblique asymptote.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2


3Opt.:End behavior like y=-0.5x

x

y



Optional step: Even though there isn't a H.A. or an O.A. we can find out the end behavior of the graph. By dividing the leading terms, -x4 and 2x, we get -0.5x3. So the end behavior of the graph of f(x) will be like that of y= -

0.5x3, a vertical S that opens down on the right and up on the left.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2


3Opt.:End behavior like y=-0.5x x - int.=-1,1

x

y



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 and x-1=0. Note that x2+1=0

would lead to imaginary solutions which is why we ignored it.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2


3Opt.:End behavior like y=-0.5x x - int.=-1,1 y- int.=1

x

y



Now find the y-intercept by plugging in 0 for x, but in this case that would lead to a 0 in the denominator. Thus, there can't be a y-intercept.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2



x

y



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.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2



x

y



Finally draw in the curve.For x>1, we can use the right end behavior to know the graph has to curve down to the right of x=1. You could also plot more points to determine this.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2



x

y



For -1/2<x<1, we can use the y-intercept, that the graph has to approach the V.A, and that there are no x-intercepts for -1/2<x<0.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2



x

y



For x<-1, we can use the left end behavior to know the graph has to curve up to the left of x=-1. You could also plot more points to determine this.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2



x

y



For -1<x<-1/2, we can use that the graph has to approach the V.A, and that there are no more x-intercepts for -1<x <-1/2.

41- xf(x)= 2x+1

2-(x +1)(x+1)(x -1)= 2x+1

1; x - 2

1V.A.: x=- 2



x

y



This finishes the graph.

graphing rational functions example #8 previouspreviousslide #1 nextnext we want to graph this...

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