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
.