Question

How do you find the vertical, horizontal or slant asymptotes for `(x+3)/(x^2-9)`?

Answer 1

This function has a horizontal asymptote `y=0`, a vertical asymptote `x=3` and a removable singularity at `x=-3`.

Explanation:

`f(x) = (x+3)/(x^2-9) = color(red)(cancel(color(black)(x+3)))/((x-3)color(red)(cancel(color(black)((x+3))))) = 1/(x-3)`

excluding `x=-3`

As `x->+-oo`, `1/(x-3)->0`. So there is a horizontal asymptote `y=0`

As `x->3^+`, `1/(x-3)->+oo`

As `x->3^-`, `1/(x-3)->-oo`

`f(3)` is undefined since division by `0` is undefined.

So `f(x)` has a vertical asymptote at `x=3`

`f(-3)` is undefined, since both numerator `(x+3) = 0` and denominator `(x^2-9) = 0`. Note however, that both left and right limits exist at `x=-3` and are both equal to `-1/6`. So there is a removable singularity at `x=-3` (removable by redefining `f(-3) = -1/6`).