# How do you find the vertical, horizontal and slant asymptotes of: g(x) = (x²-x-12) /( x+5)?

Nov 28, 2016

vertical asymptote at x = - 5
slant asymptote is y = x - 6

#### Explanation:

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: $x + 5 = 0 \Rightarrow x = - 5 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$g \left(x\right) = \frac{{x}^{2} / {x}^{2} - \frac{x}{x} ^ 2 - \frac{12}{x} ^ 2}{\frac{x}{x} ^ 2 + \frac{5}{x} ^ 2} = \frac{1 - \frac{1}{x} - \frac{12}{x} ^ 2}{\frac{1}{x} + \frac{5}{x} ^ 2}$

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

This is undefined hence there are no horizontal asymptotes.

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is a slant asymptote.

$\text{Using "color(blue)"polynomial division}$

$g \left(x\right) = x - 6 + \frac{18}{x + 5}$

as $x \to \pm \infty , g \left(x\right) \to x - 6 + 0$

$\Rightarrow y = x - 6 \text{ is the asymptote}$
graph{(x^2-x-12)/(x+5) [-10, 10, -5, 5]}