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