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This problem is getting the better of me, since I have no idea where to start: The equation of a curve is $y=ax^2-2bx+c$, where a, b and c are constants with $a>0$. 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\ge -\frac{1}{4a} $.

My working so far: Vertex:$\left(\frac ba,c-\frac {b^2}a\right)$ (Completion of square) $$ y=x $$$$ c- \frac {b^2}a = \frac ba $$Thus,$$ c=\frac{b^2+b}a $$When $a>0$ The discriminant of the original equation comes to $-4b$

Thanks in advance :-)

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$$c=\frac{b^2+b}{a}=\frac{4b^2+4b}{4a}=\frac{(4b^2+4b+1)-1}{4a}=\frac{(2b+1)^2-1}{4a}\ge -\frac{1}{4a}$$

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    $\begingroup$ Thanks :-) I didn't think rewriting would make such a big difference $\endgroup$
    – Wilhelm Erasmus
    Jul 8, 2015 at 19:49

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