Given f(x) = Ax^{2 }+ Bx +C, Domain = [a,b]
This function is polynomial function,so it is continuous & differentiable in its domain [a,b].
Hence LMVT(Lagrange's Mean Value Theorem) is applicable here.
Let there exists a 'ζ' such that a < ζ < b, So from LMVT ,we have,
f(b) - f(a) /(b-a) = f'(ζ) => [(Ab^{2 }+Bb +C) - (Aa^{2 }+Ba +C) ]/(b-a) = 2Aζ + B
=> [A(b^{2} - a^{2}) + B(b-a)]/(b-a) = 2Aζ + B
Since b & a are not equal so cancelling (b-a) from Numerator & denominator,we get,
=>A(b+a) + B = 2Aζ + B => ζ = (b+a)/2
Hence, ζ = (b+a)/2