The function is
.
All possible values of
, is domain of the function.
The domain of
is
.
Intercept:
\To find the
-intercept,substitute
in the function.
.
Apply zero product property.
\
and 
and 
and 
and 
The
-intercepts are
and
.
To find the
-intercept, substitute
in the function.

The
-intercept is
.
Find the extrema.
\The function is
.
Apply derivative on each side with respect to
.

Apply product rule of derivatives:
.
In this case
and
.


.
Find the critical numbers by equating
.


The critical numbers are
.
The domain of
is
.
Relative extrema points exist at critical numbers, Then intervals are
,
and
.
Perform first derivative test to identify the nature of the extrema.
\| Test itervals | \Test value | \Sign of ![]() | \
Conclusion | \
![]() | \
![]() | \
\
| \
Decreasing | \
![]() | \
![]() | \
\
| \
Increasing | \
![]() | \
![]() | \
\
| \
Decreasing | \
The function has relative maximum and minimum at critical number
.
Substitute
in
.



.
Substitute
in
.



.
The relative maximum is
.
The relative minimum is
.
Inflection points:
\
.
Again apply derivative on each side with respect to
.

Apply quotient rule of derivatives:
.

.
Find inflection points by equating
.

and
and
and
.
is not in the domain.
The possible inflection points occurs at
.
Substitute
in
.

.
The inflection point is
.
Consider the test intervals as
.
| \
Interval \ | \
Test Value | \Sign of ![]() | \
Concavity | \
![]() | \
![]() | \
\
| \
\
Up \ | \
![]() | \
![]() | \
\
| \
Down | \
Asymptotes :
\Vertical asymptote:
\Vertical asymptote exist when denominator is zero.
\Equate denominator to zero.
\The function is
.
There is no vertical asymptote.
\Horizontal asymptote:
\The line
is called a horizontal asymptote of the curve
if either
or 
.
.
There is no horizontal asymptote.
\Graph:
\Graph the function
.
Graph:
\Graph the function
.
.