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

Jul 3, 2016

vertical asymptote x = -2
horizontal asymptote y = 0

#### Explanation:

The first step is to factorise and simplify the function.

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

The denominator of this function cannot be zero. This is undefined.Setting the denominator equal to zero and solving for x gives us the value that x cannot be and if the numerator is non-zero for this x then it is a vertical asymptote.

solve: x + 2 = 0 → x = - 2 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

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

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

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator.This is not the case here (numerator-degree 0 , denominator-degree 1) Hence there are no slant asymptotes.
graph{1/(x+2) [-10, 10, -5, 5]}