Minimum Value of Real Quadratic Function

Theorem
Let $a \in \R_{>0}$ be a (strictly) positive real number.

Consider the quadratic function:


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

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

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

Equality occurs when $2 a x + b = 0$, that is:
 * $x = -\dfrac b {2 a}$