# Graph Sketching

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• June 24th 2006, 05:51 AM
macca101
Graph Sketching
OK Here we go....

To Sketch the graph of

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

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

2. Is the function odd or even.

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

3. The intercepts

The x intercept occurs when $<br /> 4x-15=0\\ x=\frac{15}{4}<br />$

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

4. Construct a sign table

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

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.

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

5. The Derivative of $<br /> f(x)<br />$

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

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

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

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

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

Also after constructing the sign table of $<br /> f'(x)<br />$ 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 $<br /> x= \pm3<br />$ then the lines $<br /> x=3<br />$ and $<br /> x=-3<br />$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
• June 24th 2006, 06:23 AM
Jameson
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.
• June 24th 2006, 09:36 AM
Soroban
Hello, macca101!

You did a good job . . . one error.

Quote:

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

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 $\left(\frac{15}{4},\infty\right)$

Quote:

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.
• June 24th 2006, 10:12 AM
macca101
Thanks for spotting my deliberate ;) mistake Soroban. My graph looks very similar to the one you have produced. Amazing

Thanks again
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