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Math Help - Trouble finding b and c. I know the vertex.

  1. #1
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    Trouble finding b and c. I know the vertex.

    Find b and c so that has vertex .
    I really dont even know where to start
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by dtownangel View Post
    Find b and c so that has vertex .
    I really dont even know where to start
    the vertex of quadratics function given by \frac{-b}{2a}

    here we have a=-5 and b=b and c=c so

    you want the vertex have the x-coordinate equal 5 so

    \frac{-b}{2(-5)} = 5 \rightarrow b=50

    we have now y=-5x^2+50x+c when x=5 y=-2

    -2=-5(25)+50(5)+c find c value


    for more information see this link


    Vertex of a quadratic curve
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  3. #3
    MHF Contributor red_dog's Avatar
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    If y=ax^2+bx+c then the vertex has the coordinates \left(-\frac{b}{2a},-\frac{\Delta}{4a}\right), where \Delta=b^2-4ac.

    In this case a=-5, \ \Delta=b^2+20c.

    Then you have to solve the system

    \left\{\begin{array}{ll}\displaystyle\frac{b}{10}=  5\\\displaystyle\frac{b^2+20c}{20}=-2\end{array}\right.
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