\

The function \"\".

\

Find the horizontal asymptote.

\

\"\"

\

Therefore the horizontal asymptote is at \"\".

\

Find the vertical asymptote.

\

To find the vertical asymptote, equate denominator of the function to zero.

\

So \"\".

\

Therefore the vertical asymptote is at \"\".

\

\

The function is \"\".

\

Apply derivative on each side with respect to \"\".

\

\"\"

\

Find the critical points.

\

Here \"\" is never zero, but it is undefined at \"\".

\

The critical point is \"\".

\

The test intervals are \"\" and \"\". 

\ \
\ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
Interval Test Value Sign of \"\"Conclusion
\"\"\"\" \

\"\"

\
Decreasing
\"\" \"\" \

\"\"

\
Decreasing
\

The function is decreasing on the interval \"\".

\

So there is neither local maximum nor local minimum, since the function is decreasing on \"\".

\

\

Concavity :

\

\"\".

\

Again apply derivative on each side with respect to \"\".

\

\"\"

\

\"\" is never zero, but it is undefined at \"\".

\

There is no inflection points.

\

Consider the test interval are \"\" and \"\". 

\

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\

Interval

\
Test Value Sign of \"\"Concavity
\"\" \"\" \

\"\"

\
Down
\"\" \"\" \

\"\"

\
\

Up

\
\

The graph is concave up on the interval \"\".

\

The graph is concave down on the interval \"\".

\

\

Graph :

\

Graph the function \"\" :

\

\"\"

\

\

The horizontal asymptote is \"\".

\

The function is decreasing on \"\".

\

The graph is concave up on \"\" and concave down on \"\".

\

Graph of the function \"\"  is

\

\"\".