Completing
the Square: Finding the Vertex (page
1 of 2)

The vertex form of a quadratic
is given by y = a(x – h)^{2} + k,
where (h, k) is the vertex.
The "a"
in the vertex form is the same "a"
as
in y = ax^{2} + bx + c (that is, both a's
have exactly the same value). The sign on "a"
tells you whether the quadratic opens up or opens down. Think of it this
way: A positive
"a"
draws a smiley, and a negative "a"
draws a frowny. (Yes, it's a silly picture to have in your head, but it
makes is very easy to remember how the leading coefficient works.)

In the vertex form of the
quadratic, the fact that (h, k) is the vertex
makes sense if you think about it for a minute, and it's because the quantity
"x – h" is
squared, so its value is always zero or greater; being squared, it can
never be negative.

Suppose that "a"
is positive, so a(x – h)^{2} is zero or positive and, whatever x-value
you choose, you're always taking k and adding a(x – h)^{2} to it. That is, the smallest value y can be is just k;
otherwise y will equal kplus something positive. When does y equal only k?
When x – h, the
squared part, is zero; in other words, when x
= h. So the lowest
value that y can have, y
= k, will only
happen if x
= h. And the lowest
point on a positive quadratic is of course the vertex.

If, on the other hand,
you suppose that "a"
is negative, the exact same reasoning holds, except that you're always
taking k and subtracting the squared part from it, so the highest value y can achieve is y
= k at x
= h. And the highest
point on a negative quadratic is of course the vertex.

They don't usually give
you the quadratic in vertex form; instead, they usually give you the quadratic
in the regular y = ax^{2} + bx + c format. How do you convert from the regular format to the vertex format?
By using the technique of completing the square. Here's an example:

Factor
out whatever is multiplied on the squared term.

Remember that "factor"
does not mean to "make disappear" or "divide off
onto the other side"; "factor" means "divide
out front".

Create
space on the left-hand side, and, if "a"
is anything other than 1,
put a copy of "a"
in front of this space. You'll need this space and the copy of
"a"
to keep your equation "balanced".

Take half of the
coefficient of the x-term
from inside the right-hand side parentheses (that is, divide it
by two, not forgetting its sign). Square the result, and add it
to both sides inside the parentheses.

Multiply
out the "a times the squared coefficient" part on the left-hand side,
and convert the right-hand side to squared form. (This is where
you use that sign you kept track of earlier, putting that sign
in the middle of the squared expression.)

Simplify
some more, as necessary.

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

To
be thorough, reformat into vertex form, and read off the vaues
of "h"
and "k".

In this case, since a = 3, and three is
positive, then this is a right-side-up parabola, and the vertex, (h, k) = ( ^{–1}/_{3} , ^{–4}/_{3} ),
is the lowest point on the graph.

Why did I reformat things
in that last line? Because the formula for the vertex form is y = a(x – h)^{2} + k.
In other words, I needed to make clear the value that was subtracted to
get "x + ^{1}/_{3}"?
To get a "plus" in the simplified form, I had to have subtracted
a negative:

x –
( ^{–1}/_{3} ) = x + ^{1}/_{3}

The negatives cancelled
to give me the "plus". And what was added to get " ^{–
4 }/ _{3}"?
I had to have added a negative number, ^{–
4 }/ _{3},
because +
( ^{– 4 }/ _{3 }) = ^{– 4 }/ _{3}.

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Warning: It's easy to confuse
yourself at that final stage, by trying to read off the vertex as " (h, k) = (whatever
number is inside the squared part, whatever the other number is) ",
without noticing the fact that the h-part
is subtracted and the k-part
is added. If you take care to ensure that you have your quadratic
completely converted to vertex form by being careful of the signs, then
you'll be able to avoid one of the most commonly-made mistakes for these
problems. Make sure you practice this until you can consistently interpret
your results correctly.

By the way, did you notice
that the vertex coordinates weren't whole numbers? Instructors are starting
to figure out that students are guessing the vertex from the pretty pictures
in their graphing calculators, and they know that students often have
the idea that all answers are always either whole numbers or "neat"
fractions. For instance, if the calculator screen estimates a vertex as
being at (0.48,
0.98), many students
will assume that the answer must "really" be (0.5,
1), instead of, say, ( ^{12}/_{25}, ^{49}/_{50} ).

So, in order to check that
students really do know how to find the vertex (and not just guess a decimal
approximation from a picture), teachers are giving more complicated exercises.
If you have been told that you should know this technique for finding
the vertex, rest assured that your teacher has ways of checking whether
you have really learned this. Don't plan on using calculator cheats.