# How do you find the asymptotes for h(x)=(x^2-4)/ x?

Nov 15, 2016

The vertical asymptote is $x = 0$
The slant asymptote is $y = x$
No horizontal asymptote

#### Explanation:

As you cannot divide by $0$,

$x \ne 0$

So $x = 0$ is a vertical asymptote.

The degree of the numerator is $>$ the degree of the denominator, so we expect a slant asymptote.

Let's simplify the $h \left(x\right)$

$h \left(x\right) = \frac{{x}^{2} - 4}{x} = x - \frac{4}{x}$

Therefore, $y = x$ is a slant asymptote.

${\lim}_{x \to \pm \infty} h \left(x\right) = {\lim}_{x \to \pm \infty} x = \pm \infty$

graph{(y-(x^2-4)/x)(y-x)=0 [-11.25, 11.25, -5.63, 5.62]}