How do you graph #y=(x+2)/(x+3)# using asymptotes, intercepts, end behavior?

1 Answer
Aug 11, 2018

Answer:

Below

Explanation:

#y=(x+2)/(x+3)# can be rearranged to become #y=(x+3-1)/(x+3)# which when simplified becomes #y=1-1/(x+3)#

For horizontal asymptotes, imagine what the graph will look like when #x -> oo#. When #x ->oo#, then #x+3# will become a much bigger number than #1#. Now, if you divide a small number by a large number, then #1/(x+3) ->0# so the function becomes #y=1-0=1#. Therefore, #y=1# is an asymptote.

For vertical asymptotes, look at the denominator. It cannot equal to #0# as the graph will be undefined at that point. So we let the denominator equal to #1# so that we can find at what points the graph is undefined.

#x+3=0#
#x=-3# is an asymptote.

For intercepts,
When #x=0#, #y=2/3#
When #y=0#, #x=-2#

Now for the vertical and horizontal asymptotes, remember it only affects the endpoints of the graph. SO for the endpoints, it should be approaching the asymptotes either from above or below but it should never actually touch the graph. Anywhere else on the graph, it can definitely pass through the asymptotes.

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