Question

How do you find the asymptotes for `(x+3)/(x^2-9)`?

Answer 1

To find the asymptotes, you need to find out where the denominator approaches zero for the vertical asymptotes and then establish what happens as `x -> +-oo`

Explanation:

The denominator in this case equals the difference of two squares, `x^2` and `3^2`, so
`(x+3)/(x^2-9) = (x+3)/((x-3)(x+3))`
The `(x+3)` terms cancel out so there is a removable asymptote at `x=-3`
There is a vertical asymptote at `x=3`

As `x->+-oo`, `(x+3) ->x` and `(x^2 - 9) -> x^2`

`(x+3)/(x^2-9) -> x/x^2 = 1/x` which gives the slant asymptotes.