# Graph Sketching

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

To Sketch the graph of

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

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

2. Is the function odd or even.

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

3. The intercepts

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

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

4. Construct a sign table

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

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.

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

5. The Derivative of $\displaystyle f(x)$

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

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

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

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

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

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

6. Asymptotes

As the denominator is 0 when $\displaystyle x= \pm3$ then the lines $\displaystyle x=3$ and $\displaystyle x=-3$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
• Jun 24th 2006, 06:23 AM
Jameson
Just taking a quick glance at your work it all seems good. To find the horizontal asymptotes, take $\displaystyle \lim_{x \to \infty}f(x)$ and $\displaystyle \lim_{x \to -\infty}f(x)$. You do have plenty of information to sketch the graph. You could take the second derivative though to test for concavity.
• Jun 24th 2006, 09:36 AM
Soroban
Hello, macca101!

You did a good job . . . one error.

Quote:

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

Domain: all $\displaystyle x \neq \pm 3$

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

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

Quote:

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

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

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

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

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

Hence, the horizontal asymptote is: .$\displaystyle y = 0$, 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.
• Jun 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