# How do you identify all asymptotes or holes for f(x)=(x^2-2x-8)/(-4x)?

May 20, 2017

$\text{vertical asymptote at } x = 0$
$\text{oblique asymptote } y = - \frac{1}{4} x + \frac{1}{2}$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

$\text{solve " -4x=0rArrx=0" is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is an oblique asymptote.

$\text{dividing numerator by denominator gives}$

$f \left(x\right) = - \frac{1}{4} x + \frac{1}{2} - \frac{8}{- 4 x} = - \frac{1}{4} x + \frac{1}{2} + \frac{2}{x}$

as $x \to \pm \infty , f \left(x\right) \to - \frac{1}{4} x + \frac{1}{2}$

$\Rightarrow y = - \frac{1}{4} x + \frac{1}{2} \text{ is the asymptote}$
graph{(x^2-2x-8)/(-4x) [-10, 10, -5, 5]}