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

1 Answer
Nov 8, 2016

Answer:

vertical asymptote at x = 1
oblique asymptote at y = x + 1

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #x-1=0rArrx=1" is the asymptote"#

Horizontal asymptotes occur when the degree of the numerator ≤ degree of the denominator. This is not the case here ( numerator-degree 2 , denominator-degree 1 ) Hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. Hence there is an oblique asymptote.

Polynomial division gives.

#y=x+1+1/(x-1)#

as #xto+-oo,ytox+1+0#

#rArry=x+1" is the asymptote"#
graph{(x^2)/(x-1) [-10, 10, -5, 5]}