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

May 18, 2016

vertical asymptote x = 2
horizontal asymptote y = 0

#### Explanation:

The first step here is to factor and simplify the function.

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

and cancelling gives

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

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 ${\lim}_{x \to \pm \infty} , y \to 0$

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 (numerator-degree 0 , denominator-degree 1). Hence there are no slant asymptotes.
graph{1/(x-2) [-10, 10, -5, 5]}