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I am having trouble solving this.

I know that the vertex is $\left(-\frac{b}{2a}, p(-\frac{b}{2a})\right)$, where $p(x) = ax^2+bx+c$, which is $(-\frac{b}{2a},c)$.

After that I am lost, how to show that $c \ge -\frac{1}{4a}$?

check around and I forgot that the vertex lines on the line y=x.

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    The vertex is on the main diagonal iff $p(-\frac b{2a})=-\frac b{2a}$. Show that this implies $4ac=$ some function of $b$.2012-09-09

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$(x-(-\frac{b}{2a}))^2=\frac{1}{a}(y-\frac{4ac-b^2}{4a})$

So, the vertex is $-\frac{b}{2a},\frac{4ac-b^2}{4a}$ which lies on $x=y$

So, $-\frac{b}{2a}=-\frac{b^2-4ac}{4a}\implies 4ac=b^2-2b=(b-1)^2-1≥-1$ for real $b$

$\implies 4ac≥-1\implies c≥-\frac{1}{4a}$ as $a>0$

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    Could you please go through http://en.wikipedia.org/wiki/Parabola#Equation_in_Cartesian_coordinates. From your question "check around and I forgot that the vertex lines on the line y=x.", the vertex point must satisfy the equation of line supplied.2012-09-11