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