# How do you find the vertical, horizontal and slant asymptotes of: f(x)=(2x+4)/(x^2-3x-4)?

Aug 3, 2016

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

#### Explanation:

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: ${x}^{2} - 3 x - 4 = 0 \Rightarrow \left(x - 4\right) \left(x + 1\right) = 0$

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

Horizontal asymptotes occur as

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

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

$\frac{\frac{2 x}{x} ^ 2 + \frac{4}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{3 x}{x} ^ 2 - \frac{4}{x} ^ 2} = \frac{\frac{2}{x} + \frac{4}{x} ^ 2}{1 - \frac{3}{x} - \frac{4}{x} ^ 2}$

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

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1, denominator-degree 2)Hence there are no slant asymptotes.
graph{(2x+4)/(x^2-3x-4) [-10, 10, -5, 5]}