﻿ Quadratic Functions and the Completing the Square Technique (Part 2)

# Quadratic Functions and the Completing the Square Technique (Part 2)

## Caution! Completing the Square Changes the Expression!

Be careful—when you add $\,{(\frac{b}{2})}^2\,$ to complete the square, you are changing the expression that you started with!

In other words, the expression you started with (which wasn't a perfect square) and the expression you end up with (which is a perfect square) are different.

Sometimes you just want to rename an expression in a form that involves a perfect square; you don't want to change the original expression. In this situation, when you add $\,{(\frac{b}{2})}^2\,,$ you also have to subtract it at the same time. This is illustrated in the next example.

## Example(Renaming as an Expression Involving a Perfect Square Trinomial)

Question: Rename $\ x^2+6x\$ as an expression involving a perfect square trinomial.

Solution: Note that $\,(\frac{6}{2})^2 = 3^2 = 9\,.$

 $x^2+6x\,$ starting expression $\ \ =x^2+6x+9-9$ add zero in an appropriate form $\ \ =(x^2+6x+9)-9$ regroup $\ \ =(x+3)^2-9$ rename the perfect square

The technique of completing the square only works when the coefficient of the $\,x^2\,$ term is $\,1\,.$ Any other coefficient must be factored out before completing the square, as shown next.

## Going From Standard Form to Vertex Form

In what follows, the completing the square technique is applied to the standard form, $\,ax^2+bx+c\,,$ to change it to vertex form, $\,a{(x-h)}^2+k\,$:

 $ax^2+bx+c\,$           (original expression) $\quad = (ax^2+bx)+c$           (group first two terms) $\quad =a(x^2+\frac{b}{a}x)+c$           (factor $\,a\ne 0\,$ out of the first two terms) $\quad =a\color{red}{\bigl(}x^2+\frac{b}{a}x \color{blue}{+(\frac{b}{2a})^2 - (\frac{b}{2a})^2} \color{red}{\bigr)}+c$           (add zero in an appropriate form           inside the parentheses;           note that $\frac{b}{a}\div 2 =\frac{b}{a}\cdot \frac{1}{2} =\frac{b}{2a}$) $\quad =a\left(x^2+\frac{b}{a}x+{(\frac{b}{2a})}^2 \right) - a{(\frac{b}{2a})}^2 + c$           (distributive law) $\quad =a{(x+\frac{b}{2a})}^2 + \text{stuff}$           (rename as a perfect square)

Notice that the $x$-value of the vertex is $\,-\frac{b}{2a}\,.$ This is worth recording and memorizing.

Don't bother memorizing the yucky formula for the $\,y$-value of the vertex. Once you have the $x$-value, it's easy to compute the corresponding $y$-value.

VERTEX OF A QUADRATIC FUNCTION IN STANDARD FORM

Let $\,a\ne 0\,.$

The vertex of the parabola $\ y=ax^2+bx+c\$ has $x$-value equal to $\displaystyle\,-\frac{b}{2a}\,.$

In this context, the expression $\,-\frac{b}{2a}\,$ is often called the vertex formula.

To conclude, we look at two different methods for going from standard form to vertex form:

## Example(Going from Standard Form to Vertex Form)

Question: Write $\,y=5x^2+3x-1\,$ in vertex form and give the coordinates of the vertex.
First Solution: Use the technique of completing the square to put the function in vertex form: \begin{align} \cssId{s57}{y}\ &\cssId{s58}{= 5x^2+3x-1}\cr\cr &\cssId{s59}{= 5(x^2 +\frac{3}{5}x) - 1}\cr\cr &\cssId{s60}{= 5\left(x^2 +\frac{3}{5}x+(\frac{3}{10})^2 - (\frac{3}{10})^2\right)-1}\cr\cr &\cssId{s61}{= 5\bigl(x +\frac{3}{10}\bigr)^2-5\cdot \frac{9}{100}-1}\cr\cr &\cssId{s62}{= 5\bigl(x+\frac{3}{10}\bigr)^2 - \frac{29}{20}} \end{align} Thus, the vertex is $\,\bigl(-\frac{3}{10},-\frac{29}{20}\bigr)\,.$

Second Solution: Use the vertex formula to find the $x$-value of the vertex:

$$\cssId{s66}{-\frac{b}{2a} = -\frac{3}{2(5)} = -\frac{3}{10}}$$
Find the corresponding $y$-value:
\begin{align} y &= \,5(-\frac{3}{10})^2+3(-\frac{3}{10})-1\cr\cr &= -\frac{29}{20} \end{align}

Substitute the values of $\,a\,,$ $\,h\,$ and $\,k\,$ into $\ y = a(x-h)^2 + k\,.$

With $\,h = -\frac{3}{10}\,,$ $\,k = -\frac{29}{20}\,,$ and $\,a = 5\,,$ vertex form is:

$$y=5\bigl(x +\frac{3}{10}\bigr)^2 -\frac{29}{20}$$

Finally, you could zip up to WolframAlpha and type in:

vertex of y = 5x^2 + 3x - 1

Using the coordinates of the vertex, it's then easy to write the vertex form yourself!