# How do you find vertical, horizontal and oblique asymptotes for  (x^4 - 8x^2 +16) / x^2?

Jan 3, 2017

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

#### Explanation:

Let's rewrite the expression

$\frac{{x}^{4} - 8 {x}^{2} + 16}{x} ^ 2$

Let $f \left(x\right) = \frac{{x}^{4} - 8 {x}^{2} + 16}{x} ^ 2$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{0\right\}$

As you canot divide by $0$, $x \ne 0$

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} {x}^{4} / {x}^{2} = {\lim}_{x \to \pm \infty} {x}^{2} = + \infty$

graph{(y-(x^4-8x^2+16)/(x^2))(y-1000x)=0 [-10, 10, -5, 5]}