# How do you find vertical, horizontal and oblique asymptotes for (2x^2 - x - 38) / (x^2 - 4)?

Jul 16, 2017

$\text{vertical asymptotes at } x = \pm 2$
$\text{horizontal asymptote at } y = 2$

#### 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.

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

$\Rightarrow x = - 2 \text{ and "x=2" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

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

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

$f \left(x\right) = \frac{\frac{2 {x}^{2}}{x} ^ 2 - \frac{x}{x} ^ 2 - \frac{38}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = \frac{2 - \frac{1}{x} - \frac{38}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{2 - 0 - 0}{1 - 0}$

$\Rightarrow y = 2 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both degree 2) hence there are no oblique asymptotes.
graph{(2x^2-x-38)/(x^2-4) [-20, 20, -10, 10]}