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Math Help - Graph Sketching

  1. #1
    Junior Member
    Joined
    Feb 2006
    Posts
    29

    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
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  2. #2
    MHF Contributor
    Joined
    Oct 2005
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    Earth
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    1,599
    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.
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  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Thanks
    697
    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.
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  4. #4
    Junior Member
    Joined
    Feb 2006
    Posts
    29
    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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