# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)=(x-4)/(x^2-3x-4)?

Jun 2, 2016

vertical asymptote x = -1
horizontal asymptote y = 0

#### Explanation:

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

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

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 + 1 = 0 → x = -1 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

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

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

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

Oblique 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 oblique asymptotes.
graph{1/(x+1) [-10, 10, -5, 5]}