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

(This page is Part 2. Click here for Part 1.)

## 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.

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)

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

$$\cssId{s66}{-\frac{b}{2a} = -\frac{3}{2(5)} = -\frac{3}{10}}$$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:

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!