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

Jun 2, 2016

vertical asymptote x = -2
horizontal asymptote y = 0

#### Explanation:

The first step here is to factorise and simplify f(x).

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

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

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

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