# How do you find the vertical, horizontal or slant asymptotes for (x-2)/(x^2-4)?

Apr 24, 2016

vertical asymptote x = -2
horizontal asymptote y = 0

#### Explanation:

The first step we have to take here is to factorise and simplify the function.

rArr (x-2)/(x^2-4) = (x-2)/((x-2)(x+2) ) =cancel((x-2))/(cancel((x-2)) (x+2)

$= \frac{1}{x + 2}$

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

solve : x + 2 = 0 → x = -2 is the asymptote

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

divide terms on numerator/denominator by x

$\Rightarrow \frac{\frac{1}{x}}{\frac{x}{x} + \frac{2}{x}} = \frac{\frac{1}{x}}{1 + \frac{2}{x}}$

as $x \to \pm \infty , y \to \frac{0}{1 + 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(x-2)/(x^2-4) [-10, 10, -5, 5]}