# How do you find vertical, horizontal and oblique asymptotes for (2x+4)/(x^2-3x-4)?

Apr 7, 2016

vertical asymptotes x = -1 , x = 4
horizontal asymptote y = 0

#### Explanation:

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero.

solve:  x^2 - 3x - 4 = 0 → (x -4)(x + 1) = 0

$\Rightarrow x = - 1 , x = 4 \text{ are the asymptotes}$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

When the degree of the numerator < degree of the denominator , as is the case here then the equation is always
y = 0.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.

Here is the graph of the function.
graph{(2x+4)/(x^2-3x-4) [-10, 10, -5, 5]}