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Math Help - Solve quadratic eqiation using 3 points?

  1. #1
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    Solve quadratic equation using 3 points?

    The points are (0, 189), (-116, 0), (116, 0), in general form.

    I tried, but my answer didn't go with the parabola graph, help?
    Last edited by dumbblonde; May 15th 2013 at 01:29 AM.
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  2. #2
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    Re: Solve quadratic eqiation using 3 points?

    The question is not clear. Do you want to form quadratic equation in general form. In that case
    let the quadratic equation be y = ax^2+ bx + c. Now the points ( 0,189 ) lies on it so we get c = 189
    now we have y = ax^2 + bx + 189; the points ( 116,0) and ( -116,0) lie on it. thus we get
    ( 116)^2 a + 116 b + 189 = 0 and
    (116)^2 - 116 b + 189 = 0
    Solve these two equations for a and b.
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  3. #3
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    Re: Solve quadratic equation using 3 points?

    Hello, dumbblonde!

    \text{Find the equation of the parabola containing: }(0, 189),\:(\text{-}116, 0),\:(116, 0)

    Code:
                    |
                    *189
                    |
                    |
                    |
                    |
          - - * - - + - - * - -
            -116    |    116
                    |
    The parabola opens downward
    . . and has x-intercepts -116 and +116.

    The equation has the form: . y \:=\:a(x-116)(x+116) + b

    From (\pm116,0), we have: . a(0)+b \:=\:0 \quad\Rightarrow\quad b  \,=\,0

    From (0,189), we have: . a(\text{-}116)(116) \:=\:189 \quad\Rightarrow\quad a \,=\,-\tfrac{189}{116^2}


    Therefore: . y \;=\;-\tfrac{189}{13,456}(x-116)(x+116)
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