[Kevin Sevcik]

Well.... the standard to vertex thing is straight factoring. I'll demonstrate so you can see where those formulae come from:

We're trying to do this:
ax^2 + bx + c --> a(x-h)^2 + k

From here I'm just working with the left side:
(1) a(x^2 + b/a*x + c/a)

Now, we're looking for a perfect square. (x-h)^2 = x^2 - 2hx + h^2
so you match up the similar termsand get:
-2hx = b/a*x
h=-b/2a

plug this into (1) to get the squared term:

a(x^2 - 2hx + h^2 - h^2 + c/a)

note: the -2hx goes in cause we made it equivalent to b/a*x. the h^2 and -h^2 cancel out so you can add them in. you've got a perfect square in there now, so you factor it.
a((x - h)^2 - h^2 + c/a)
a(x - h)^2 + a(-h^2 + c/a)
a(x - h)^2 - ah^2 + c

now you just have that weird term out at the end. that's where k goes, so you set them equal to each other.
k = -a(h)^2 + c
k = -a(-b/2a)^2 + c
k = -b^2/4a + c

so there's your formula for h and k and how you get them.
h=-b/2a
k = -b^2/4a + c

This makes perfect sense to me, which undoubtedly means it's completely incomprehensible. so let me know what specifically isn't clear.