Minimum Value of Real Quadratic Function

Theorem

$\map Q x = a x^2 + b x + c$

$\map Q x$ achieves a minimum at $x = -\dfrac b {2 a}$, at which point $\map Q x = c - \dfrac {b^2} {4 a}$.

$\ds \map Q x$

$=$

$\ds a x^2 + b x + c$

$\ds $

$=$

$\ds \dfrac {4 \paren {a x}^2 + 4 a b x + 4 a c} {4 a}$

$\ds $

$=$

$\ds \dfrac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}$

As $\paren {2 a x + b}^2 > 0$, it follows that:

$\ds a x^2 + b x + c$

$\ge$

$\ds \dfrac {-\paren {b^2 - 4 a c} } {4 a}$

$\ds $

$=$

$\ds c - \dfrac {b^2} {4 a}$

Equality occurs when $2 a x + b = 0$, that is:

$x = -\dfrac b {2 a}$

$\blacksquare$