1. ## Graph Sketching

OK Here we go....

To Sketch the graph of

$
f(x)=\frac{4x-15}{x^2-9}
$

1. Find the domain.
The denominator is 0 when $
x= \pm3
$
so the domain of $
f
$
is all of $
R
$
except for $
\pm 3
$

2. Is the function odd or even.

$
f(2) = -\frac{7}{5} , f(-2)= \frac {-23}{-3}
$
as $
f(-2)
$
is not equal to $
\pm f(2)
$
$
f
$
is neither odd or even

3. The intercepts

The x intercept occurs when $
4x-15=0\\ x=\frac{15}{4}
$

The y intercept occurs at $
f(0) = \frac{5}{3}
$

4. Construct a sign table

$
f(x)=\frac{4x-15}{x^2-9}\\=\frac{4x-15}{(x-3)(x+3)}
$

Having constructed a sign table (don't know how to create it in LATEX) but gives me the following information to help in the sketching of the graph.

$
f
$
is positive on the interval $
(-3,0)\,(0,3)\,(3,\infty)
$
and negative on $
(-\infty,-3)
$

5. The Derivative of $
f(x)
$

$
f'(x)= \frac{(x^2-9)4-(4x-15)2x}{(x^2-9)^2}
$

$
f'(x)= \frac{(-4x+6)(x-6)}{(x^2-9)^2}
$

$
f(x)
$
has two fixed points at $
x=\frac{3}{2}
$
and $
x=6
$

As $
f(\frac{3}{2})=\frac{4}{3}
$
and $
f(6)=\frac{1}{3}
$
the fixed points are located at $(\frac{3}{2},\frac{4}{3})\,(6,\frac{1}{3})
$

After applying the first derivative test $(\frac{3}{2},\frac{4}{3})
$
is found to be a local minimum and $(6,\frac{1}{3})
$
is found to be a local maximum.

Also after constructing the sign table of $
f'(x)
$
the function is found to be decreasing on $(-\infty,-3)\,(-3,\frac{3}{2})\,(6,\infty)$ and increasing on $(\frac{3}{2},3)\,(3,6,)$

6. Asymptotes

As the denominator is 0 when $
x= \pm3
$
then the lines $
x=3
$
and $
x=-3
$
are vertical asymptotes

Now here is the question(s)

1 Is all the above correct ?
2 Does the graph have a horizontal asymptote ?
3 Is there enough information the sketch the graph? (I've had a go but I am unable to reproduce my result here.)

Thanks

2. Just taking a quick glance at your work it all seems good. To find the horizontal asymptotes, take $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$. You do have plenty of information to sketch the graph. You could take the second derivative though to test for concavity.

3. Hello, macca101!

You did a good job . . . one error.

$f(x)\:=\:\frac{4x-15}{x^2-9}$

Domain: all $x \neq \pm 3$

Intercepts: $\left(\frac{15}{4},0\right),\;\;\left(0,\frac{5}{3 }\right)$

Sign table
$f$ is positive on the intervals: $(-3,0),\;(0,3),\;(3,\infty)$ . . . no
and negative on $(-\infty,-3)$
$f$ is negative on $\left(3,\frac{15}{4}\right)$ . . . then positive on $\left(\frac{15}{4},\infty\right)$

Derivative of $f(x)$
$f(x)$ has fixed points at: $\left(\frac{3}{2}, \frac{4}{3}\right),\;\left(6, \frac{1}{3}\right)$

First derivative test
$\left(\frac{3}{2},\frac{4}{3}\right)$ is found to be a local minimum
and $\left(6,\frac{1}{3}\right)$ is found to be a local maximum.

Sign table of $f'(x)$
The function is found to be decreasing on $(-\infty,-3),\;\left(-3,\frac{3}{2}\right),\;(6,\infty)$
and increasing on $\left(\frac{3}{2},3\right),\;(3,6)$

Asymptotes
Vertical asymptotes are: $x = -3$ and $x = 3.$
For horizontal asymptotes, examine: . $\lim_{x\to\infty}f(x)$ and $\lim_{x\to-\infty}f(x)$

We have: . $\lim_{x\to\infty}\frac{4x - 15}{x^2 - 9} \;= \;$ $\lim_{x\to\infty}\frac{\frac{1}{x^2}(4x - 15)}{\frac{1}{x^2}(x^2 - 9)}\;=\;$ $\lim_{x\to\infty}\frac{\frac{4}{x} - \frac{15}{x^2}}{1 - \frac{9}{x^2}} \;= \;\frac{0 - 0}{1 - 0}\;=\;0$

Hence, the horizontal asymptote is: . $y = 0$, the x-axis.

There is plenty of information (which you found) for sketching the graph.
Code:
                    :*    |    *:
:     |     :
: *   |   * :
:  *  |     :
:    *|  *  :           *
:     |*    :       *      *
:     |     :    *               *
--------------:-----+-----:---*----------------------
*           :     |     :  *
*     :     |     :
*  :     |     : *
* :     |     :
:     |     :
*:     |     :*
I'll let you label the intercepts, asymptotes and critical points.

4. Thanks for spotting my deliberate mistake Soroban. My graph looks very similar to the one you have produced. Amazing

Thanks again