Completing the square and the vertex

A few words this lesson leans on:

• A quadratic is an expression like $2x^2 - 8x + 5$: the highest power of $x$ is 2.
• A coefficient is the number multiplying a letter. In $2x^2 - 8x + 5$ the coefficient of $x^2$ is 2 and the coefficient of $x$ is $-8$. The lone number, 5, is the constant.
• The graph of a quadratic is a curve called a parabola. If the $x^2$ coefficient is positive it is shaped like a valley, ∪. If it is negative it is shaped like a hill, ∩.
• The vertex is the turning point of that curve: the bottom of the valley or the top of the hill.

Why rewrite a quadratic at all? From $x^2 - 6x + 11$ you cannot see where the bottom of the valley is. Rewrite the same thing as $(x - 3)^2 + 2$ and it jumps out: the bracket is squared, so it is never negative, and it is smallest (zero) when $x = 3$. So the lowest point of the graph is at $x = 3$, height 2. Completing the square is the tool that does this rewrite.

Expand a perfect square and look at the pattern:

$(x + 3)^2 = (x + 3)(x + 3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9$

The $6x$ is two lots of $3x$, so the number in the bracket (3) is half of the coefficient of $x$ (6). The $+9$ is that half, squared.

So run it backwards. To "complete" $x^2 + 6x$:

1. Halve the coefficient of $x$ and put it in a bracket: $(x + 3)^2$. This bracket gives you the $x^2 + 6x$ you wanted...
2. ...but it also brings an unwanted $+9$. Subtract it to cancel it out: $(x + 3)^2 - 9$.

Check: $(x + 3)^2 - 9 = x^2 + 6x + 9 - 9 = x^2 + 6x$ ✓. Nothing has changed except the way it is written.

In general, for any number $b$: $x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2$.