\(f\) is differentiable across its domain, so the MVT says there is some value of \(c\) in the open interval \((a,b)\) such that
\(f'(c)=\dfrac{f(b)-f(a)}{b-a}\)
You have
\(f(x)=Ax^2+Bx+C\implies f'(x)=2Ax+B\)
so the equation above becomes
\(2Ac+B=\dfrac{(Ab^2+Bb+C)-(Aa^2+Ba+C)}{b-a}\)
Solve for \(c\).
\(2Ac+B=\dfrac{A(b^2-a^2)+B(b-a)}{b-a}\)
\(2Ac+B=A(b+a)+B\)
\(2Ac=A(b+a)\)
\(c=\dfrac{b+a}2\)
so \(c\) is indeed the average of the endpoints, i.e. the midpoint.
