Question

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

Answer 1

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]}