# How do you find vertical, horizontal and oblique asymptotes for (x+6 )/ (x^2 - 36)?

Aug 26, 2016

vertical asymptote at x = 6
horizontal asymptote at y = 0

#### Explanation:

The first step is to factorise and simplify the function.

$f \left(x\right) = \frac{\cancel{x + 6}}{\cancel{\left(x + 6\right)} \left(x - 6\right)} = \frac{1}{x - 6}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: $x - 6 = 0 \Rightarrow x = 6 \text{ is the asymptote}$

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

$f \left(x\right) = \frac{\frac{1}{x}}{\frac{x}{x} - \frac{6}{x}} = \frac{\frac{1}{x}}{1 - \frac{6}{x}}$

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 the denominator. This is not the case here ( numerator-degree 0, denominator-degree 1) Hence there are no oblique asymptotes.
graph{1/(x-6) [-10, 10, -5, 5]}