<s>[INST] Parabolic range conditions proof 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 :-)
 [/INST] $$c=\frac{b^2+b}{a}=\frac{4b^2+4b}{4a}=\frac{(4b^2+4b+1)-1}{4a}=\frac{(2b+1)^2-1}{4a}\ge -\frac{1}{4a}$$
</s>