How do you find the vertical, horizontal and slant asymptotes of: #x/(1-x)^2#?
1 Answer
Explanation:
#"let " f(x)=x/(1-x)^2# 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 "(1-x)^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)/(1/x^2-(2x)/x^2+x^2/x^2)=(1/x)/(1/x^2-2/x+1)# as
#xto+-oo,f(x)to0/(0-0+1)#
#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)/((1-x)^2) [-10, 10, -5, 5]}
