\
The function is .
Find the horizontal asymptote.
\Therefore the horizontal asymptote is at .
To find the vertical asymptote, equate denominator of the function to zero.
\So
Here the roots of the function is imaginary, so there is no vertical asymptote.
\\
The function is .
Apply derivative on each side with respect to .
Find the critical points.
\Thus, the critical points exist when .
The critical point is .
The test intervals are and
.
Interval | \Test Value | \Sign of ![]() | \
Conclusion | \
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\
| \
Decreasing | \
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\
| \
Increasing | \
The function is increasing on the interval .
The function is decreasing on the interval .
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 .
\
.
Again apply derivative on each side with respect to .
Find the inflection points.
\Equate to zero.
The inflection point is at and
.
The test intervals are ,
and
.
\
\
Interval \ | \
Test Value | \Sign of ![]() | \
Concavity | \
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![]() | \
Down | \
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\
Up \ | \
![]() | \
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![]() | \
Down | \
The graph is concave up on the interval .
The graph is concave down on the intervals and
.
The inflection points are and
.
\
Graph :
\Graph the function :
\
The horizontal asymptote is at .
The function is increasing on and decreasing on
.
The graph is concave up on and concave down on
and
.
Graph of the function is
.