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Math Help - Quadratics

  1. #1
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    Quadratics

    Can you please help me with the last part of this question...

    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
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  2. #2
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    Re: Quadratics

    So what did you get for parts a) and b)?
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  3. #3
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    Re: Quadratics

    Quote Originally Posted by Prove It View Post
    So what did you get for parts a) and b)?
    For a) I completed the square and got the coordinated as
    ( b/a , c-(b^2/a) )
    b) c-(b^2/a)=b/a
    c=b(b-1)/a
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  4. #4
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    Re: Quadratics

    I am confused in the last part..
    I don't get it why whatever the value of b, c should always be greater than or equal to -1/4a
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  5. #5
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    Re: Quadratics

    I had an idea of making b=o and finding the value of c but that won't work i'll get 0
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