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

1 Answer
Jan 18, 2018

Answer:

Vertical asymptote at #x=1#

Oblique asymptote at #y = x+1#

Explanation:

A vertical asymptote occurs where the denominator is equal to #0#.

So we have:

#x-1=0->x=1#

So a vertical asymptote will occur where #x=1#.

The degree of the numerator is greater than the degree of the denominator so the function will not have horizontal asymptotes but will have oblique ones. To find them: we must split the fraction up like so:

#x^2/(x-1)=x+x/(x-1)=x+1+1/(x-1)#

So for very large values of #x# the fraction term will become insignificantly small and the function will approach the oblique asymptote: #y = x+1#.