How do you graph #f(x)=2/(x-3)+1# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Aug 1, 2018

Below

Explanation:

#f(x)=1+2/(x-3)#

For vertical asymptote, look at the denominator. It cannot equal to #0# as the graph will be undefined at that point. Hence, you let denominator equal to #0# to find at what point the graph cannot equal to #0#.

#x-3=0#
#x=-3#

For horizontal asymptote, imagine what happens to the graph when #x ->oo#. As #x ->oo#, #2/(x-3) ->0# so #f(x)=1+0=1# ie #y=1#

For intercepts,
When #y=0#, #x=1#
When #x=0#, #y=1/3#

Below is what the graph looks like

graph{1+2/(x-3) [-10, 10, -5, 5]}