# How do you find the vertical, horizontal or slant asymptotes for (x-4)/(x^2-3x-4)?

Nov 6, 2016

The vertical asymptote is $x = - 1$
The horizontal asymptote is $y = 0$
There is no slant asymptote

#### Explanation:

We can factorise the denominator ${x}^{2} - 3 x - 4 = \left(x + 1\right) \left(x - 4\right)$
$\therefore \frac{x - 4}{{x}^{2} - 3 x - 4} = \frac{\cancel{x - 4}}{\left(x + 1\right) \cancel{x - 4}} = \frac{1}{x + 1}$

So the vertical asymptote is $x = - 1$ as we cannot divide by $0$
Limit $\frac{1}{x + 1} = {0}^{-}$
$x \to - \infty$

Limit $\frac{1}{x + 1} = {0}^{+}$
$x \to + \infty$

So the horizontal asymptote is $y = 0$
And the intercept with the y axis is $\left(0 , 1\right)$
As the degree of the numerator is $<$ the degree of the denominator, there is no slant asymptote
graph{1/(x+1) [-12.66, 12.65, -6.33, 6.33]}