Question

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

Answer 1

Vertical asymptotes are at `x = 2 and x = -2`
Horizontal asymptote is `y = 0` , slant asymptote is absent.

Explanation:

`f(x) = 1/(x^2-4) or f(x) = 1/((x+2)(x-2))`

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

the denominator is equal to zero. `x+2=0 :. x=-2` and

`x-2=0 :. x=2` .Thus, the graph will have vertical asymptotes

are at `x = 2 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]