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Math Help - Constructing the equation of an Irreducible Quadratic from it's graph.

  1. #1
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    Constructing the equation of an Irreducible Quadratic from it's graph.

    I have a Quadratic with Minima (1,3) and y-intercept of 5. When I did this with quadratics with rational zeros I just reconstructed the polynomials off of those Zeros and then added the terms necessary to get the correct maxima/minima. Since I can not, or do not know, how to do this with an Irreducible Quadratic I'm running into trouble.

    I have a Quadratic that satisfies the (1,3) and the y-intercept 5 requirements but the facing is incorrect. (-2x^2)+5 = f(#9)

    Also, some what related, I assume that I could find my complex zeros by inspection (as this is the procedure that I am using to find rational zeros) if the drawing was set up correctly disregarding the fact that I can not visually discern between values that are near each other. Another way of putting this is to say; I assume that the complex plane can be "drawn". I'm I thinking of this correctly?
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  2. #2
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    In turning point form you have \displaystyle y = a(x-k)^2+h where \displaystyle (k,h) is the turning point.

    You should be able to find \displaystyle  a easily enough after substituting in these values.
    Last edited by pickslides; May 24th 2011 at 02:14 PM.
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  3. #3
    Super Member bigwave's Avatar
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    did you get (x-1)^2=(4)\left\(\frac{1}{8 }\right\) (y-3)

    or y=2x^2-4x+5

    for your quadratic
    Last edited by bigwave; May 24th 2011 at 02:13 PM. Reason: latex
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