Completing The Square

Occasionally you can be asked to turn an equation such as $y=ax^2+bx+c$ into the "standard form" or "vertex form" that looks like $y=a(x-h)^2+k$.

### Solution #1

If this happens on Paper 2, where you can use your calculator, then you're in luck. Your calculator can find the minimum (or maximum) of the quadratic thus giving you the coordinates of the vertex which can then simply be substituted in to standard/vertex form. Where h and k are the x and y coordinates of the vertex, respectively. The a value stays the same.

### Solution #2

A second option is to use the equation for the axis of symmetry $x=\frac{-b}{2a}$. This gives you the x-coordinate of the vertex which can then be substituted into the original equation to find the value of the y-coordinate of the vertex. With both of these in hand you can easily rewrite the equation in standard/vertex form. The a value stays the same.

### Solution #3 - Completing the Square

A third options exists. I present it only because there could be a problem where one of the above approaches does not work, although I can't think of one. The third option is to "complete the square." This terminology comes from the fact that quadratic equations can (sometimes) be represented by geometrical diagrams. For example the expression $x^2+4x+2$ can be represented geometrically as:

Note that this quadratic makes a shape that is almost a square… If we add the dashed parts below then we would have a complete square.

This addition to the diagram represents an addition to the algebraic expression of $+2$. But we can not simply add to an expression… We can only add zero (or multiply by 1)! So the expression becomes:

(1)
\begin{equation} x^2+4x+2 +(2) - (2) \end{equation}

You might say that nothing has changed which is sort of true. But if we leave the minus 2 by itself and regroup the other bits (this makes some sense as we only wanted to deal with the +2 anyways…) we get an expression that looks like

(2)
\begin{equation} (x^2+4x+4)-2 \end{equation}

Now the part in the parentheses is a "perfect square" and the expression can be written as:

(3)
\begin{equation} (x+2)^2-2 \end{equation}

The equation is now in standard/vertex form with the vertex at (-2,-2). This visual method is nice and works for some examples. For less "nice" problems we need something a bit more formal and abstract.

(8)
\begin{align} a \left ( x+ \frac{b}{2a} \right )^ 2 +c - \frac{b^2}{4a} \end{align}

The square has been completed! Now to translate that to vertex form we find that

(9)
\begin{align} h = -\frac{b}{2a} \end{align}
(10)
\begin{align} k = c - \left ( \frac{b^2}{4a} \right ) \end{align}

Notice that the value of h is the same as for the axis of symmetry… Math its just one big circle of logic, it comes back on itself.