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

Nov 26, 2016

vertical asymptote at x = 0
slant 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.

solve: $- 4 x = 0 \Rightarrow x = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the numerator. This is the case here ( numerator-degree 2, denominator-degree 1 ) Hence there is a slant asymptote.

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

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

as $x \to \pm \infty , f \left(x\right) \to - \frac{1}{4} x - \frac{1}{2} + 0$

$\Rightarrow y = - \frac{1}{4} x - \frac{1}{2} \text{ is the asymptote}$

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here hence there are no holes.
graph{(x^2+2x-8)/(-4x) [-10, 10, -5, 5]}