How do you identify all asymptotes or holes for #f(x)=(3x-6)/(x-1)#?

1 Answer
Oct 18, 2016

Answer:

A vertical asymptote is #x=1#
A horizontal asymptote is #y=3#

Explanation:

As you cannot divide by 0
so #x-1!=0#
Also you calculate the limit when #x->oo#
#f(x)->3#
So #y=3# is a horizontal asymptote
See graph below
#f(x)=3 -3/(x-1)#
graph{(3x-6)/(x-1) [-10, 10, -5, 5]}