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

1 Answer
Jun 4, 2018

Answer:

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]