What are the asymptote(s) and hole(s), if any, of # f(x) =(-2x^2-6x)/((x-3)(x+3)) #?

1 Answer
Feb 15, 2018

Answer:

Asymptotes at #x=3# and #y=-2#. A hole at #x=-3#

Explanation:

We have #(2x^2-6x)/((x-3)(x+3))#.

Which we can write as:

#(-2(x+3))/((x+3)(x-3))#

Which reduces to:

#-2/(x-3)#

You find the vertical asymptote of #m/n# when #n=0#.

So here,

#x-3=0#

#x=3# is the vertical asymptote.

For the horizontal asymptote, there exists three rules:

To find the horizontal asymptotes, we must look at the degree of the numerator (#n#) and the denominator (#m#).

If #n>m,# there is no horizontal asymptote

If #n=m#, we divide the leading coefficients,

If #n<##m#, the asymptote is at #y=0#.

Here, since the degree of the numerator is #2# and that of the denominator is #2# we divide the leading coefficients. As the coefficient of the numerator is #-2#, and that of the denominator is #1,# the horizontal asymptote is at #y=-2/1=-2#.

The hole is at #x=-3#.

This is because our denominator had #(x+3)(x-3)#. We have an asymptote at #3#, but even at #x=-3# there is no value of #y#.

A graph confirms this:

graph{(-2x^2-6x)/((x+3)(x-3)) [-12.29, 13.02, -7.44, 5.22]}