How do you find the vertical, horizontal or slant asymptotes for  g(x) = (X+6) /( x^2 - 36)?

Dec 27, 2016

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

Explanation:

Factorise and simplify.

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

Since (x + 6) has been removed, x = -6 is an excluded value.

The denominator of g(x) cannot be zero as this would make g(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

${\lim}_{x \to \pm \infty} , g \left(x\right) \to c \text{ ( a constant)}$

divide terms on numerator/denominator by x

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

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

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

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-6) [-10, 10, -5, 5]}