# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (1)/(x^2-4)?

Jun 4, 2018

Vertical asymptotes are at $x = 2 \mathmr{and} x = - 2$
Horizontal asymptote is $y = 0$ , slant asymptote is absent.

#### Explanation:

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

The vertical asymptotes will occur at those values of $x$ for which

the denominator is equal to zero. $x + 2 = 0 \therefore x = - 2$ and

$x - 2 = 0 \therefore x = 2$ .Thus, the graph will have vertical asymptotes

are at $x = 2 \mathmr{and} x = - 2$

The degree of numerator is $0$ and of denominator is $2$

Since the larger degree occurs in the denominator, the

graph will have a horizontal asymptote as $y = 0$ (i.e., the x-axis).

$$If the numerator's degree is greater (by a margin of 1), then


we have a slant asymptote . So here slant asymptote is absent.

graph{1/(x^2-4) [-11.25, 11.25, -5.625, 5.625]} [Ans]