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Math Help - manipulating the graph of a rational function

  1. #1
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    manipulating the graph of a rational function

    Hello-
    I have a problem that says "Find the equation given the following graph"

    Below is an example of what my graph looks like, however this is not the exact one. Mine has Vertical asymptotes at (-3,0) and (4,0). X intercept is (2,0) and the Y intercept is (0,-2)

    http://upload.wikimedia.org/wikipedi...ree2byXedi.gif

    Based off this information and the graph i've come up with the following equation (which is wrong)
    after solving for A in the equation y=a(x-2)^3/((x+3)(x-4)^2)
    Y=4(x-2)^3/((x+3)(x-4)^2)

    and my graph on my calculator the right hand portion of the graph needs to be flipped over and on the other side of the X axis... is there a trick to doing this?

    Thanks
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  2. #2
    MHF Contributor
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    Hello xsavethesporksx
    Quote Originally Posted by xsavethesporksx View Post
    Hello-
    I have a problem that says "Find the equation given the following graph"

    Below is an example of what my graph looks like, however this is not the exact one. Mine has Vertical asymptotes at (-3,0) and (4,0). X intercept is (2,0) and the Y intercept is (0,-2)

    http://upload.wikimedia.org/wikipedi...ree2byXedi.gif

    Based off this information and the graph i've come up with the following equation (which is wrong)
    after solving for A in the equation y=a(x-2)^3/((x+3)(x-4)^2)
    Y=4(x-2)^3/((x+3)(x-4)^2)

    and my graph on my calculator the right hand portion of the graph needs to be flipped over and on the other side of the X axis... is there a trick to doing this?

    Thanks
    I'm not entirely sure which question you are asking here, but if you want an equation that produces a graph like the one you attached, it's this (assuming the marks on the axes indicate unit values):

    1. It has asymptotes x = \pm2,\;y=1
      .
    2. Intercepts are approximately (0, \tfrac12), (-\tfrac12,0) and (\tfrac72,0)

    From 1. we can deduce that the equation is of the form:
    y = 1+\frac{ax+b}{(x-2)(x+2)} = 1 +\frac{ax+b}{x^2-4}
    and from 2. we get:
    \tfrac12 = 1+\frac{b}{-4}

    \Rightarrow b = 2
    and
    0 = 1+\frac{-\tfrac12a+2}{\tfrac14-4}

    \Rightarrow ...

    \Rightarrow a\approx -3
    So the equation is
    y=1+\frac{2-3x}{x^2-4}
    =\frac{x^2-3x-2}{x^2-4}
    Grandad
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