How do you find vertical, horizontal and oblique asymptotes for #f(x) = (4x)/(x^2-1)#?

1 Answer
Mar 18, 2016

Answer:

vertical asymptotes x = ± 1
horizontal asymptote y = 0

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s equate the denominator to zero.

solve : # x^2 - 1 = 0 → (x-1)(x+1) = 0 #

# rArr x = ± 1" are the asymptotes " #

Horizontal asymptotes occur as # lim_(x→±∞) f(x) → 0 #

If the degree of the numerator is less than the degree of the denominator , as is the case here then equation is y = 0 .

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator.This is not the case here and so there is no oblique asymptote.

Here is the graph of the function.
graph{4x/(x^2-1) [-10, 10, -5, 5]}