# How do you find the asymptotes for f(x)=(2x+4)/(x^2-3x-4)?

##### 1 Answer
Sep 4, 2016

vertical asymptotes at x = -1 and x = 4
horizontal asymptote at y = 0

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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: ${x}^{2} - 3 x - 4 = 0 \Rightarrow \left(x + 1\right) \left(x - 4\right) = 0$

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

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 ${x}^{2}$

$f \left(x\right) = \frac{\frac{2 x}{x} ^ 2 + \frac{4}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{3 x}{x} ^ 2 - \frac{4}{x} ^ 2} = \frac{\frac{2}{x} + \frac{4}{x} ^ 2}{1 - \frac{3}{x} - \frac{4}{x} ^ 2}$

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

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(2x+4)/(x^2-3x-4) [-10, 10, -5, 5]}