The function is
.
Find the horizontal asymptote.
\
Therfore the horizontal asymptote is
.
Find the vertical asymptote.
\To find the vertical asymptote, equate denominator of the function to zero.
\So
.
Here the roots are imaginary.
\Therefore there is no vertical asymptote.
\The function is
.



Find the critical points.
\Thus, the critical points exist when
.




The critical points are
and
.
The test intervals are
,
and
.
| Interval | \Test Value | \ \
Sign of | \
Conclusion | \
![]() | \
![]() | \
\
| \
Decreasing | \
| \
| \
![]() | \
\
| \
Increasing | \
![]() | \
![]() | \
\
| \
Decreasing | \
The function is increasing on the interval
.
The function is decreasing on the intervals
and
.
Find the local maximum and local minimum.
\The function
has a local minimum at
, because
changes its sign from negative to positive.
Substitute
in
.

Local minimum is
.
The function
has a local maximum at
, because
changes its sign from positive to negative.
Substitute
in
.

Local maximum is
.
.
Apply derivative on each side with respect to
.



.
Find the inflection points by equate the second derivative to zero.
\The function
does not exist when
.
The denominator does not have real roots.
\Equate the numerator of
to zero.



and 
and
.
Substitute the values of
in original function.
.
.
.
Inflection points are
,
and
.
The test intervals are
,
,
and
.
| Interval | \Test Value | \Sign of ![]() | \
Conclusion | \
![]() | \
![]() | \
\
| \
Down | \
| \
| \
![]() | \
\
| \
Up | \
| \
| \
![]() | \
\
| \
Down | \
![]() | \
![]() | \
\
| \
Up | \
The graph is concave up on the intervals
and
.
The graph is concave down on the intervals
and
.
The inflection points are
,
and
.
Graph :
\Graph the function
:
Horizontal asymptote
.
The function is increasing on
and decreasing on
and
.
The graph is concave up on
and
and concave down on
and
.
Graph of the function
is
.