# How do you identify all asymptotes or holes for f(x)=1/(-2x-2)?

Dec 26, 2016

$\text{vertical asymptote at } x = - 1$
$\text{horizontal asymptote at } y = 0$

#### 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: $- 2 x - 2 = 0 \Rightarrow x = - 1 \text{ is the asymptote}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

divide terms on numerator/denominator by x

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

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Holes occur when there are duplicate factors on the numerator/denominator. This is not the case here hence there are no holes.
graph{1/(-2x-2) [-10, 10, -5, 5]}