Graph Sketching

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June 24th 2006, 05:51 AM
macca101

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
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