# How do you find the vertical, horizontal and slant asymptotes of: (8x^2-x+2)/(4x^2-16)?

Nov 7, 2016

Vertical asymptotes: $x = - 2$ and $x = 2$
Horizontal asymptote: $y = 2$
No slant asymptotes, as slant asymptotes only occur when the degree on the numerator is greater than the degree on the denominator.

#### Explanation:

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

Vertical Asymptotes:
First, factor as much as possible. The numerator cannot be factored. The denominator can be factored into $4 \left(x + 2\right) \left(x - 2\right)$
$f \left(x\right) = \frac{8 {x}^{2} - x + 2}{\left(4\right) \left(x + 2\right) \left(x - 2\right)}$

Vertical asymptotes occur when the denominator is equal to zero. In this case, that is when $x = - 2$ and when $x = 2$.

Horizontal Asymptote:
Horizontal asymptotes are the value that the function approaches when the x value approaches $\pm \infty$.
$\lim x \to \pm \infty \left[\frac{8 {x}^{2} - x + 2}{4 {x}^{2} - 16}\right] = \frac{8}{4} = 2$