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

Dec 11, 2016

#### Answer:

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 $x = - 1 \pm \sqrt{5}$ so there $h \left(x\right)$ has vertical asympototes at $x = - 1 + \sqrt{5}$ and $x = - 1 = \sqrt{5}$.

For large $x$ positive or negative $h \left(x\right)$ approximates to $\frac{1}{x}$ ($h \left(x\right) \to 0$ as $x \to \pm \infty$), as you can see by ignoring the $- 5$ and the $2 x - 4$ for large $x$ magnitudes.

So the $x$ axis is another asymptote.