# How do you find vertical, horizontal and oblique asymptotes for ( x²-x-12 )/(x+5)?

Oct 21, 2016

The vertical asymptote is $x = - 5$
The oblique asymptote is $y = x - 6$

#### Explanation:

The domain of the function is $\mathbb{R} - \left(5\right)$
as we cannot divide by $0$
So the vertical asymptote is $x = - 5$
The degree of the polynomial of the numerator $>$ degree of the denominator, so we have an oblique asymptote
doing a long division give

$\frac{{x}^{2} - x - 12}{x + 5} = x - 6 + \frac{18}{x + 5}$

So $y = x - 6$ is an oblique asymptote

Verification

$x - 6 + \frac{18}{x + 5} = \frac{\left(x - 6\right) \left(x + 5\right) + 18}{x + 5} = \frac{{x}^{2} - x - 30 + 18}{x + 5}$

$= \frac{{x}^{2} - x - 12}{x + 5}$