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

Sep 25, 2016

The vertical asymptotes are $x = 4$ and $x = - 1$. The horizontal asymptote is $y = 0$. There are no oblique asymptotes.

#### Explanation:

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

Factor the denominator.

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

To find the vertical asymptotes, set the denominator equal to zero and solve.
$\left(x - 4\right) \left(x + 1\right) = 0$
$x - 4 = 0 \textcolor{w h i t e}{a a a} x + 1 = 0$
$x = 4 \textcolor{w h i t e}{a a a} x = - 1 \textcolor{w h i t e}{a a a}$These are the VA's.

To find the horizontal asymptotes, compare the degree of the numerator to the degree of the denominator.

• If the deg of the num > deg of den, there is an oblique asymptote.

• If the deg of the num = deg of den, the HA is the leading coefficient of the numerator divided by the leading coefficient of the denominator.

• If the deg of the num < deg of den, the HA is $y = 0$.

In this example, the degree of the numerator is $\textcolor{red}{1}$ and the degree of the denominator is $\textcolor{b l u e}{2}$
$f \left(x\right) = \frac{2 {x}^{\textcolor{red}{1}} + 4}{{x}^{\textcolor{b l u e}{2}} - 3 x - 4}$

The degree of the numerator < degree of the denominator, so the HA is $y = 0$.

There is no oblique asymptote, because the degree of the numerator is not greater than the degree of the denominator.