Answer:
See below
Step-by-step explanation:
I assume the function is 
A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, 
is the only vertical asymptote.
B) Set the first derivative equal to 0 and solve:
Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:
Therefore, the function increases on the interval 
and decreases on the interval 
.
C) Since we determined that the slope is 0 when from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, 
, meaning there's an extreme at the point 
, but is it a maximum or minimum? To answer that, we will plug in into the second derivative which is 
. If 
, then it's a minimum. If 
, then it's a maximum. If 
, the test fails. So, 
, which means is a local maximum.
D) Now set the second derivative equal to 0 and solve:
We then test where 
is negative or positive by plugging in test values. I will use -1 and 3 to test this:
, so the function is concave down on the interval 
, so the function is concave up on the interval 
The inflection point is where concavity changes, which can be determined by plugging in into the original function, which would be 
, or 
.
E) See attached graph
