# How do you find the Vertical, Horizontal, and Oblique Asymptote given y = (2x+4)/( x^2-3x-4)?

Jun 28, 2016

so we have vertical asymptotes at $x = - 1 , 4$

for horizontal and slope asmptotes, ${\lim}_{x \setminus \to \pm \infty} y = 0$

#### Explanation:

for vertical aympptotes, we look at when the demoninator is zero

so
${x}^{2} - 3 x - 4 = \left(x + 1\right) \left(x - 4\right) \implies x = - 1 , 4$

to check for possible indeterminates we note that

$y \left(- 1\right) = \frac{2}{0}$ = ndef
and
$y \left(4\right) = \frac{12}{0}$ = ndef

so we have vertical asymptotes at $x = - 1 , 4$

for horizontal and slope we look at the behaviour of the function as $x \setminus \to \pm \infty$

so we re-write
${\lim}_{x \to \pm \infty} \frac{2 x + 4}{{x}^{2} - 3 x - 4}$

as

${\lim}_{x \to \pm \infty} \frac{\frac{2}{x} + \frac{4}{x} ^ 2}{1 - \frac{3}{x} - \frac{4}{x} ^ 2} = 0$