Say we have an equation:$y=$ ${x^2} + 2x + 1$

Completing the square we get:

$\eqalign{ & y={x^2} + 2x + 1 \cr & = {(x + 1)^2} - {(1)^2} + 1 \cr & = {(x + 1)^2} \cr} $

The minimum point of this parabola is (-1,0)

What I would like to know is how/why does putting a quadratic equation in completing the square form give you the minimum point of a parabola? What is it about this form that corresponds to give you the minimum point? I hope i've made myself clear, if not please ask me to make myself clearer.

Thank you!

  • 2
    $\begingroup$ No matter what $x$ is, $(x + 1)^2 \geq 0$. The smallest it can ever get is $0$, which happens when $x = -1$. $\endgroup$ – littleO Oct 3 '13 at 20:29

The expression $(x+a)^2+b$ attains its minimum at $x=-a$ because $(x+a)^2\ge 0$ for all real values of $x$ with equality if and only if $x=-a$.


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