Completing the Square

Expand any squared binomial and look at the pattern: ${(x+p)}^2=x^2+2px+p^2$. The middle coefficient is $2p$ and the constant is $p^2$, so the constant is always the square of half the $x$-coefficient. For example $x^2+6x+9$ factors as ${(x+3)}^2$ because $9={\left(\frac{6}{2}\right)}^2$. The picture explains why. Build $x^2+6x$ from an $x$-by-$x$ square plus two $3$-by-$x$ strips placed along two sides. The shape is almost a big square of side $x+3$, but one corner is missing: a $3$-by-$3$ piece with area $9$. "Completing the square" literally means filling in that missing corner. Since we cannot change the value of the expression, we add the $9$ and immediately subtract it again, which is the same as adding zero.

1. Take the coefficient of $x$, halve it, and square the result: ${\left(\frac{b}{2}\right)}^2$. For $y=x^2-8x+10$: half of $-8$ is $-4$, and ${(-4)}^2=16$.
2. Add and subtract that number right after the $x$-term: $y=x^2-8x+16-16+10$.
3. Group the first three terms as a perfect square binomial, using half of $b$ (with its sign) inside the bracket: $y={(x-4)}^2-16+10$.
4. Simplify the leftover constants: $y={(x-4)}^2-6$, so the vertex is $(4,-6)$.

Two details make this reliable. First, the number inside the bracket is always half of $b$ with the same sign as $b$: $x^2-8x$ becomes ${(x-4)}^2$, and $x^2+8x$ becomes ${(x+4)}^2$. Second, odd values of $b$ simply produce fractions, and the method still works. For $y=x^2+5x+1$, half of $5$ is $\frac{5}{2}$ and its square is $\frac{25}{4}$, so $y={\left(x+\frac{5}{2}\right)}^2-\frac{25}{4}+1={\left(x+\frac{5}{2}\right)}^2-\frac{21}{4}\text{.}$ The vertex is $\left(-\frac{5}{2},-\frac{21}{4}\right)$, or $(-2.5,-5.25)$. Keep fractions exact until the last line; rounding early spoils the vertex.

The perfect-square pattern only works when $x^2$ has coefficient $1$. So when $a\ne 1$, first factor $a$ out of the $x^2$ and $x$ terms only, leaving the constant term outside and untouched. Complete the square inside the bracket as before, then distribute $a$ back through before simplifying. The coefficient inside may become a fraction, which is fine: $y=2x^2+6x+1$ becomes $2(x^2+3x)+1$; half of $3$ is $\frac{3}{2}$, squared is $\frac{9}{4}$, so $y=2\left({\left(x+\frac{3}{2}\right)}^2-\frac{9}{4}\right)+1=2{\left(x+\frac{3}{2}\right)}^2-\frac{9}{2}+1=2{\left(x+\frac{3}{2}\right)}^2-\frac{7}{2}\text{.}$ When $a$ is negative, factor out the negative too: $-3x^2+12x$ becomes $-3(x^2-4x)$, and every sign inside the bracket flips.

Once in vertex form $y=a{(x-h)}^2+k$, the vertex is $(h,k)$. If $a>0$, $k$ is the minimum value of the function; if $a<0$, $k$ is the maximum value — because a squared term is always $\ge 0$, so $a{(x-h)}^2$ only ever adds to $k$ (when $a>0$) or subtracts from it (when $a<0$), and it equals zero only at $x=h$. This is why completing the square is the standard tool for optimization. The completed form also tells you about the zeros. Solving $a{(x-h)}^2+k=0$ gives ${(x-h)}^2=-\frac{k}{a}$, which has two solutions when $-\frac{k}{a}>0$, one when $k=0$, and none when $-\frac{k}{a}<0$ (a square cannot be negative). For $y={(x-4)}^2-6$: ${(x-4)}^2=6$, so $x=4\pm \sqrt{6}$.

1.3 - Completing the Square - Functions 11