Graphing Rational Functions Example #2 END SHOWEND SHOW Slide #1 NextNext We want to graph this...

2

2

3x -2x -1f(x)=

2x - x -1

Graphing Rational FunctionsExample #2

END SHOW Slide #1 Next

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

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1)=

(2x+1)(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

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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



Note the domain restrictions, where the denominator is 0.

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)



Now reduce the fraction. In this case, we cancel the common factor of(x-1) in both the numerator and the denominator.

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2



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

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3



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 is a hole at 1. From the reduced form, y=(3·1+1)/(2·1+1)=4/3.

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

( x +1)3=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

H.A.: y = 32



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

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

H.A.: y = 32

(3x +1)(2x+1)

=32

6x +2=6x +3 2=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

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

3H.A.: y =

2

(3x +1) 3=

(2x+1) 2

6x +2=6x +3 2=3

No I nt. w/ H.A.



In this case, solving the equation, led to a statement that is always false. Thus, there are no values of x where the 2 graph intersect. Hence, there

are no intersections between the graph of f(x) and the H.A.

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

3H.A.: y =

2

(3x +1) 3=

(2x+1) 2

6x +2=6x +3 2=3

No I nt. w/ H.A. 1

x - int=-3



We find the x-intercepts by solving when the function is 0, which would be when the numerator is 0. Thus, when 3x+1=0.

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

3H.A.: y =

2

(3x +1) 3=

(2x+1) 2

6x +2=6x +3 2=3

No I nt. w/ H.A. 1

x - int=-3

0+0-1y - int= =1

0-1



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

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

3H.A.: y =

2

(3x +1) 3=

(2x+1) 2

6x +2=6x +3 2=3

No I nt. w/ H.A. 1

x - int=-3

0+0-1y - int= =1

0-1

x=-1

-3+1 -2

y= = =2-2+1 -2



Plot any additional points needed.Here I only plotted one more point at x=-1 since a point hadn't been plotted

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

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

3H.A.: y =

2

(3x +1) 3=

(2x+1) 2

6x +2=6x +3 2=3

No I nt. w/ H.A. 1

x - int=-3

0+0-1y - int= =1

0-1

x=-1

-3+1 -2

y= = =2-2+1 -2



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

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

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

3H.A.: y =

2

(3x +1) 3=

(2x+1) 2

6x +2=6x +3 2=3

No I nt. w/ H.A. 1

x - int=-3

0+0-1y - int= =1

0-1

x=-1

-3+1 -2

y= = =2-2+1 -2



For the part to the left of the V.A., we use that it can't cross the H.A. and it has to approach the V.A. and the H.A.

2

2

3x -2x -1f(x)=

2x - x -1

(3x +1)(x -1) 1

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

(3x +1)=

(2x+1)

1V.A.: x = -

2

4Hole at 1,

3

3H.A.: y =

2

(3x +1) 3=

(2x+1) 2

6x +2=6x +3 2=3

No I nt. w/ H.A. 1

x - int=-3

0+0-1y - int= =1

0-1

x=-1

-3+1 -2

y= = =2-2+1 -2


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This finishes the graph.

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