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Math Help - Sketching graphs using partial fractions!

  1. #1
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    Sketching graphs using partial fractions!

    Sketch the graph of the following (showing turning points):

     <br />
y=\frac{2x+3}{x^2+2x+1}<br />

    My working out so far:

     y=\frac{2x+3}{x^2+2x+1}= \frac{2x+3}{(x+1)^2}

     \frac{2x+3}{(x+1)^{2}}\equiv \frac{a}{x+1} + \frac{b}{(x+1)^{2}}\equiv \frac{a(x+1)+b}{{(x+1)}^2}

     2x+3\equiv a(x+1)+b
    Letting x = -1
     b=1
    Letting x=0
     3=a +1
     a=2
    Vertical asymptotes: x=-1
    Horizontal asymptote:  y=0

     \frac{2}{x+1} + \frac{1}{(x+1)^2} = 0
     2(x+1)^2 +x+1=0
     2x^2+4x+2+x+1=0
     2x^2+5x+3=0
     2x^2+3x+2x+3=0
     x(2x+3)+1(2x+3)=0
     (x+1)(2x+3)=0
     x=-1
     x=\frac{-3}{2}

    I've drawn the corresponding graph, using addition-of-ordinates.
    But have I made any mistakes in my working out so far? And how do I find the turning points for this graph?

    Anyway help will be welcomed with open arms and be highly appreciated!
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  2. #2
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    Quote Originally Posted by Joker37 View Post
    Sketch the graph of the following (showing turning points):

     <br />
y=\frac{2x+3}{x^2+2x+1}<br />

    My working out so far:

     y=\frac{2x+3}{x^2+2x+1}= \frac{2x+3}{(x+1)^2}

     \frac{2x+3}{(x+1)^{2}}\equiv \frac{a}{x+1} + \frac{b}{(x+1)^{2}}\equiv \frac{a(x+1)+b}{{(x+1)}^2}

     2x+3\equiv a(x+1)+b
    Letting x = -1
     b=1
    Letting x=0
     3=a +1
     a=2
    Vertical asymptotes: x=-1
    Horizontal asymptote:  y=0

     \frac{2}{x+1} + \frac{1}{(x+1)^2} = 0
     2(x+1)^2 +x+1=0
    That's incorrect. Multiplying on both sides by (x+1)^2 and you get 2(x+ 1)+ 1= 0 so that 2x+ 3= 0 and then x= -3/2. The "-1" you get is an "extraneous" root.

     2x^2+4x+2+x+1=0
     2x^2+5x+3=0
     2x^2+3x+2x+3=0
     x(2x+3)+1(2x+3)=0
     (x+1)(2x+3)=0
     x=-1
     x=\frac{-3}{2}

    I've drawn the corresponding graph, using addition-of-ordinates.
    But have I made any mistakes in my working out so far? And how do I find the turning points for this graph?

    Anyway help will be welcomed with open arms and be highly appreciated!
    Follow Math Help Forum on Facebook and Google+

  3. #3
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    Hello, Joker37!

    I don't understand the method used in this problem.
    And evidently, neither do you.


    Sketch the graph of the following (showing turning points):

    . .  y\:=\:\dfrac{2x+3}{x^2+2x+1}

    You can find these two facts by "eyeballing" the original function.

    . . \begin{array}{ccc}\text{Vertical asymptotes:} & x\:=\:\text{-}1 \\<br />
\text{Horizontal asymptote:} & y\:=\:0\end{array}


    Similarly, you can also find these two facts:

    . . . . \begin{array}{ccc}x\text{-intercept:} & (\text{-}\frac{3}{2},\:0) \\ <br />
\,y\text{-intercept:} & (0,3) \end{array}


    To find turning points, we solve y' \,=\,0

    . . We have: . y' \:=\:\dfrac{\text{-}2(x+2)}{x+1} \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}2

    Hence, the turning point is: . (\text{-}2,\:\text{-}1)}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Using the process of Partial Fraction Decomposition,

    . . you found that: . \displaystyle  y \;=\;\frac{2}{x+1} + \frac{1}{(x+1)^2}

    But why?

    What possible purpose does it serve?


    PFD is normally used facilitate integration, which is not involved here.

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  4. #4
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    \frac{2x+3}{x^2+2x+1}=\frac{2(x+1)+1}{(x+1)^2}=\fr  ac{2}{x+1}+\frac{1}{(x+1)^2}

    Asymptotes:

    x\rightarrow\pm\infty\Rightarrow\ y\rightarrow\ 0

    x\rightarrow\ -1\Rightarrow\ y\rightarrow\pm\infty

    Intercepts:

    \frac{2}{x+1}+\frac{1}{(x+1)^2}=0 \Rightarrow\frac{2}{x+1}=-\frac{1}{(x+1)^2}\Rightarrow\frac{2(x+1)}{(x+1)^2}  =-\frac{1}{(x+1)^2}

    \Rightarrow\ x+1=-\frac{1}{2}\Rightarrow\ x=-\frac{3}{2}

    Remember also, the function is undefined for x=-1

    x=0\Rightarrow\ y=2+1=3


    Turning points:

    f(x)=\frac{2}{x+1}+\frac{1}{(x+1)^2}\Rightarrow\ f'(x)=-\frac{2}{(x+1)^2}-\frac{2}{(x+1)^3}=-\left[\frac{2(x+1)+2}{(x+1)^3}\right]=0\Rightarrow\ x=-2

    f(-2)=-2+1=-1
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  5. #5
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    Quote Originally Posted by Soroban View Post
    To find turning points, we solve y' \,=\,0

    . . We have: . y' \:=\:\dfrac{\text{-}2(x+2)}{x+1} \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}2
    How did you get y'?

    Quote Originally Posted by Archie Meade View Post
    Turning points:

    f(x)=\frac{2}{x+1}+\frac{1}{(x+1)^2}\Rightarrow\ f'(x)=-\frac{2}{(x+1)^2}-\frac{2}{(x+1)^3}
    How did you get f'(x)?
    Last edited by Joker37; January 22nd 2011 at 06:41 PM.
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  6. #6
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    Quote Originally Posted by Soroban View Post

    Using the process of Partial Fraction Decomposition,

    . . you found that: . \displaystyle y \;=\;\frac{2}{x+1} + \frac{1}{(x+1)^2}

    But why?

    What possible purpose does it serve?


    PFD is normally used facilitate integration, which is not involved here.
    Well, it gives you a better idea of the shape of the graph doesn't it? I still need to sketch the graph and this partial fraction is easier to analyze and graph than the more complex looking equation given originally.
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  7. #7
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    Quote Originally Posted by Joker37 View Post



    How did you get f'(x)?
    Using the Quotient Rule of differentiation on both partial fractions.
    Or write the fractions using negative exponents instead...


    f(x)=\frac{2}{x+1}+\frac{1}{(x+1)^2}=2(x+1)^{-1}+(x+1)^{-2}

    \Rightarrow\ f'(x)=-2(x+1)^{-2}-2(x+1)^{-3}=-\frac{2}{(x+1)^2}-\frac{2}{(x+1)^3}
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  8. #8
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    Quote Originally Posted by Archie Meade View Post
    \frac{2x+3}{x^2+2x+1}=\frac{2(x+1)+1}{(x+1)^2}=\fr  ac{2}{x+1}+\frac{1}{(x+1)^2}

    Asymptotes:

    x\rightarrow\pm\infty\Rightarrow\ y\rightarrow\ 0

    x\rightarrow\ -1\Rightarrow\ y\rightarrow\pm\infty

    Intercepts:

    \frac{2}{x+1}+\frac{1}{(x+1)^2}=0 \Rightarrow\frac{2}{x+1}=-\frac{1}{(x+1)^2}\Rightarrow\frac{2(x+1)}{(x+1)^2}  =-\frac{1}{(x+1)^2}

    \Rightarrow\ x+1=-\frac{1}{2}\Rightarrow\ x=-\frac{3}{2}

    Remember also, the function is undefined for x=-1

    x=0\Rightarrow\ y=2+1=3


    Turning points:

    f(x)=\frac{2}{x+1}+\frac{1}{(x+1)^2}\Rightarrow\ f'(x)=-\frac{2}{(x+1)^2}-\frac{2}{(x+1)^3}=-\left[\frac{2(x+1)+2}{(x+1)^3}\right]=0\Rightarrow\ x=-2

    f(-2)=-2+1=-1
    Was this all the complete working out that I had to show?
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  9. #9
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    Yes.

    The axis intercepts are useful to have.
    The major aspects are the asymptotes and the turning points.
    It's then possible to draw a rough freehand sketch of the graph.

    Accuracy is improved by calculating a few more points, though the axis intercepts help a lot there.
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