How do you find vertical, horizontal and oblique asymptotes for #y=3/(x-2)+1#?

1 Answer
Nov 9, 2016

Answer:

The vertical asymptote is #x=2#
The horizontal asymptote is #y=1#
There is no oblique asymptote.

Explanation:

As we cannot divide by #0#, the vertical asymptote is #x=2#
And #lim_(x->+-oo)y=1#
So #y=1#is a horizontal asymptote.
There is no oblique asymptote as the degree of the numerator #=# the degree of the denominator
graph{(y-1-3/(x-2))(y-1)=0 [-14.24, 14.24, -7.12, 7.12]}