How do you find the vertical, horizontal or slant asymptotes for #f(x)=x/(x-1)^2#?

1 Answer
Aug 9, 2017

Answer:

#"vertical asymptote at "x=1#
#"horizontal asymptote at "y=0#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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)^2=0rArrx=1" is the asymptote"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc"( a constant)"#

Divide terms on numerator/denominator by the highest power of x, that is #x^2#

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

as #xto+-oo,f(x)to0/(1-0+0)#

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{x/(x-1)^2 [-10, 10, -5, 5]}