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Math Help - question on quadratics

  1. #1
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    question on quadratics

    I've been trying to figure out how to do the last part of this question for a long time but I still can't..
    Can someone please help me..

    The equation of a curve is y=ax^2 -2bx +c ,where a,b and c are constants with a>0
    a) Find in terms of a,b and c the coordinates of the vertex of the curve
    b) Given that the vertex of the curve lies on the line y=x , find an expression for c in terms of a and b.
    Show that in this case, whatever the value of b, c is greater than or equal to -1/4a


    For a) I got ( b/a , c - b^2/a )
    For b) I got c= (b^2+b) /a

    Any help would be greatly appreciated...
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  2. #2
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    Re: question on quadratics

    Hey puresoul.

    Can you show us your working out? We can get a better idea of if the answer is right and if not what is going wrong.
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  3. #3
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    Re: question on quadratics

    Quote Originally Posted by chiro View Post
    Hey puresoul.

    Can you show us your working out? We can get a better idea of if the answer is right and if not what is going wrong.
    I thought of this...

    We know the vertex of the parabola and the y intercept and since a is greater than zero so c has to be greater than c - (b^2)/a
    and c= (b^2 +b) /a
    so (b^2 +b) /a >c - (b^2)/a
    (2b^2 + b )/a > c
    but that doesn't look anything like what they want ..

    And I don't know what else I could do
    Any hint that might help me?
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