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

1 Answer
Jul 30, 2016

Answer:

vertical asymptote x = 1
horizontal asymptote y = 0

Explanation:

The denominator of f(x) cannot be zero as this would be 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^2-2x+1=0rArr(x-1)^2=0rArrx=1#

#rArrx=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#

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

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

#rArry=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2 ) Hence there are no oblique asymptotes.
graph{(1)/(x^2-2x+1) [-10, 10, -5, 5]}