Question

How do you find any asymptotes of `h(x)=(x-5)/(x^2+2x-4)`?

Answer 1

By (a) seeing if the denominator is zero for any value(s) of `x`, and by examining the behaviour for large positive and negative values of `x`.

Explanation:

The denominator is zero where `x=-1+-sqrt 5` so there `h(x)` has vertical asympototes at `x=-1+sqrt 5` and `x=-1=sqrt 5`.

For large `x` positive or negative `h(x)` approximates to `1/x` (`h(x)to 0` as `x to +- oo`), as you can see by ignoring the `-5` and the `2x-4` for large `x` magnitudes.

So the `x` axis is another asymptote.