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Math Help - Graphing a Quadratic Function Given in General Form

  1. #1
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    Graphing a Quadratic Function Given in General Form

    If I understand correctly, the formula is:

    -\frac{b}{2a}, \frac{4ac-b^2}{4a}

    Here's what I have:

    33x^2 -2x +15

    First:

    -\frac{2}{2(33)} = \frac{1}{33}

    Easy enough. Now though...

    \frac{4(33)(15)}\frac{4(33)}

    Doing it that way comes up with wildly incorrect answers. I'd type it out but it's just all wrong. The answer in the book is...

    \frac{494}{33}

    However, doing...

    15-\frac{-2^2}{4(33)} does get me

    \frac{494}{33}

    So that's really C-\frac{(-b)^2}{4(a)}

    So my question is: Why? I'm missing something in the translation and I'm hoping to get some kind of answer. Thanks for any and all help.
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  2. #2
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    Re: Graphing a Quadratic Function Given in General Form

    Quote Originally Posted by Trypanosoma View Post
    If I understand correctly, the formula is:

    -\frac{b}{2a}, \frac{4ac-b^2}{4a}
    Being psychic, I know that you meant the formula for the vertex coordinates of the parabola ax^2+bx+c .

    Quote Originally Posted by Trypanosoma View Post
    33x^2 -2x +15

    First:

    -\frac{2}{2(33)} = \frac{1}{33}
    Should be -\frac{-2}{2(33)}

    Quote Originally Posted by Trypanosoma View Post
    \frac{4(33)(15)}\frac{4(33)}
    Should be \frac{4(33)(15)-4}\frac{4(33)}

    Quote Originally Posted by Trypanosoma View Post
    Doing it that way comes up with wildly incorrect answers.
    Not wildly incorrect; the difference between this and the correct answer is only 1/33.

    Quote Originally Posted by Trypanosoma View Post
    So that's really C-\frac{(-b)^2}{4(a)}

    So my question is: Why?
    The y-coordinate of the vertex is \frac{4ac-b^2}{4a}=c-\frac{b^2}{4a}.
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