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Math Help - Finding the turning point for the curve f(x)2x^2-6x-8

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    Finding the turning point for the curve f(x)2x^2-6x-8

    I have found out the coordinates are (-1,0)(4,0)(0,-8) but I can't find the turning point

    Thanks!
    Last edited by agehayoshina; December 15th 2012 at 11:50 AM.
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    Re: Need help finding the turning point for the curve f(x)2x^2-6x-8

    the vertex, or 'turning point", of f(x) = ax^2 + bx + c is at x = \frac{-b}{2a}
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    Re: Need help finding the turning point for the curve f(x)2x^2-6x-8

    Hmm this is till kind of confusing. If I plug the numbers in i get -6/2(2) =-1.5. That's x. How do I find y?


    Thanks!
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    Re: Need help finding the turning point for the curve f(x)2x^2-6x-8

    Quote Originally Posted by agehayoshina View Post
    Hmm this is till kind of confusing. If I plug the numbers in i get -6/2(2) =-1.5. That's x. How do I find y?
    you have miscalculated the x-value of the vertex.

    f(x) = 2x^2 - 6x - 8

    x = \frac{-b}{2a} = \frac{-(-6)}{2(2)} = \frac{3}{2}

    the y-value of the vertex is \color{red}y = f\left(\frac{3}{2}\right)
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