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Math Help - graph of function with asymtotes

  1. #1
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    graph of function with asymtotes

    Sketch the graph for each of the following functions , showing clearly the asymtotes .

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

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

    For (1) I know that x=1 and x=2 are asymtotes as they make the function undefined . And that 's the only information i can gather . I'm not sure bout the rest ..

    Thanks for your help.
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  2. #2
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    Hello, thereddevils!

    Sketch the graph for each of the following functions, showing clearly the asymptotes.

    (1)\;\;y=\:\:\frac{x+1}{(x-1)(x-2)}

    I know that x = 1, x = 2 are asymptotes, as they make the function undefined.
    You're right . . . Those are the vertical asymptotes.
    We must also look for horizontal asymptotes.

    They occur when y \:=\:\lim_{x\to\infty}f(x) has a finite value.

    We have: . y \:=\:\lim_{x\to\infty}\frac{x+1}{x^2-3x+2}

    Divide top and bottom by x^2\!:\;\;\frac{\dfrac{x}{x^2} + \dfrac{1}{x^2}}{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2} + \dfrac{2}{x^2}}

    So we have: . y \:=\:\lim_{x\to\infty}\left(\frac{\frac{1}{x} + \frac{1}{x^2}}{1 - \frac{3}{x} + \frac{2}{x^2}}\right) \:=\:\frac{0+0}{1-0+0} \:=\:0

    Therefore, y = 0 (the x-axis) is a horizontal asymptote.


    Code:
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                          |   :           :    * 
                          | * :           :           *
      --------------------+---:-----------:--------------------
        *                 *   :           :
             *           *|   :     *     :
                *      *  |   :   *   *   :
                   *      |   :  *     *  :
                          |   :           :
                          |   : *       * :
                          |   :           :
                          |   :           :
                          |   :*         *:
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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, thereddevils!

    You're right . . . Those are the vertical asymptotes.
    We must also look for horizontal asymptotes.

    They occur when y \:=\:\lim_{x\to\infty}f(x) has a finite value.

    We have: . y \:=\:\lim_{x\to\infty}\frac{x+1}{x^2-3x+2}

    Divide top and bottom by x^2\!:\;\;\frac{\dfrac{x}{x^2} + \dfrac{1}{x^2}}{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2} + \dfrac{2}{x^2}}

    So we have: . y \:=\:\lim_{x\to\infty}\left(\frac{\frac{1}{x} + \frac{1}{x^2}}{1 - \frac{3}{x} + \frac{2}{x^2}}\right) \:=\:\frac{0+0}{1-0+0} \:=\:0

    Therefore, y = 0 (the x-axis) is a horizontal asymptote.

    Code:
                          |   :           :*
                          |   :           :
                          |  *:           : *
                          |   :           :  *
                          |   :           :    * 
                          | * :           :           *
      --------------------+---:-----------:--------------------
        *                 *   :           :
             *           *|   :     *     :
                *      *  |   :   *   *   :
                   *      |   :  *     *  :
                          |   :           :
                          |   : *       * :
                          |   :           :
                          |   :           :
                          |   :*         *:
                          |   :           :

    Thanks a lot Soroban . So I will have enough information to sketch by just finding the vertical and horizontal asymtotes ? I don see how you get the three curves there .
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  4. #4
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    Quote Originally Posted by thereddevils View Post
    Sketch the graph for each of the following functions , showing clearly the asymtotes .


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

    This is my attempt for #2

    The vertical asymtotes are x=0,-1 and the horizontal asymtotes is y=0 .

    I tried to find the max and min points using the algebraic method . This is where i get stucked .

    yx^2+(y-1)x-x+1=0

     (y-1)^2-4y\geq0

    y\geq5.828 and y\leq0.1716

    I know that the graph doesn't exist between 0 and 0.1716 and hence it exists between 0.1716 and 5.828 .

    I am not sure whether my working is correct . Can someone pls correct me ? Really Thanks .
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