Completing
the Square:
Deriving the Vertex Formula (page
2 of 2)

Since
you always do exactly the same procedure each time you find the vertex
form, the procedure can be done symbolically (using y = ax^{2} + bx + c instead of putting in numbers), so you end up with a formula that you
can use instead of doing the completing-the-square process each time.
Here's what it looks like:

Factor
out whatever is multiplied on the squared term.

Make
room on the left-hand side, and put a copy of "a"
in front of this space.

Take half of the
coefficient of the x-term
(divide it by two) (and don't forget its sign!); square this,
and add it inside the parentheses on both sides.

Multiply
through the parentheses on the left.

Simplify
on the left (converting to common denominators, if necessary),
and convert to squared form on the right.

Move
the loose numbers back over to the right-hand side.

Convert
to vertex form, as necessary.

Then
the the vertex (h, k) for any
given quadratic y = ax^{2} + bx + cobeys the formula:

ADVERTISEMENT

Practically
speaking, you can just memorize that h = –b/ (2a) and then plug your value for "h"
back in to "y =" to calculate "k".
If you're allowed to use this formula, you can then more quickly find
the vertex, because simply calculating h = –b/ (2a) and then finding k is a lot faster than completing the square. On the other hand, even if
you're not "supposed" to use this formula, you can still use
it to check your work. If you're not sure if you completed the square
correctly on a test, you can always use the formulaic answer to verify
(or correct) whatever you've come up with.

However,
if you've memorized the Quadratic
Formula, you can
probably easily memorize the formula here for k.
The discriminant in the Quadratic Formula is b^{2} – 4ac. Take the
negative of this to get the reverse of the subtraction:

–(b^{2} – 4ac) = 4ac – b^{2}

The
formula for h has
a denominator of 2a;
double this for a denominator of 4a.
Put them together, and this gives you the formula for k.

Stapel, Elizabeth.
"Completing the Square: Deriving the Vertex Formula." Purplemath. Available from http://www.purplemath.com/modules/sqrvertx2.htm.
Accessed