How do you identify all asymptotes for #f(x)=1/(x-1)#?

1 Answer
Apr 2, 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=0rArrx=1" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#f(x)=(1/x)/(x/x-1/x)=(1/x)/(1-1/x)#

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

#rArry=0" is the asymptote"#
graph{1/(x-1) [-10, 10, -5, 5]}