# How do you find the asymptotes for (8x^2-x+2)/(4x^2-16)?

Apr 21, 2016

vertical asymptotes x = ± 2
horizontal asymptote y = 2

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve:  4x^2-16 = 0 → 4(x^2-4) = 0 → 4(x-2)(x+2) = 0

$\Rightarrow x = - 2 , x = 2 \text{ are the asymptotes }$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

divide all terms on numerator/denominator by ${x}^{2}$

$\Rightarrow \frac{\frac{8 {x}^{2}}{x} ^ 2 - \frac{x}{x} ^ 2 + \frac{2}{x} ^ 2}{\frac{4 {x}^{2}}{x} ^ 2 - \frac{16}{x} ^ 2} = \frac{8 - \frac{1}{x} + \frac{2}{x} ^ 2}{4 - \frac{16}{x} ^ 2}$

as$x \to \pm \infty , \frac{1}{x} , \frac{2}{x} ^ 2 \text{ and } \frac{16}{x} ^ 2 \to 0$

and $y \to \frac{8 - 0 + 0}{4 - 0} = \frac{8}{4} = 2$

$\Rightarrow y = 2 \text{ is the asymptote }$
graph{(8x^2-x+2)/(4x^2-16) [-10, 10, -5, 5]}