Derive the Vertex Form

Given a polynomial of the form $ax^{2}+bx+c=y$, we can use some intuition to complete the square. Notice there is a new wrinkle, the $x^{2}$ term has a variable, $a$, in front of it.

1. Factor out $a$ from the $x$ containing terms of the equation. $$a\left(x^{2}+\frac{b}{a}x\right)+c=y$$
2. Now “complete the square” for the term inside the parenthesis.
   1. Halve the coefficient of the x term,
   2. and thinking about the squares from the prior section, we can write: $$a\left[\left(x+\frac{b}{2a}x\right)^{2}-\left(\frac{b}{2a}\right)^{2}\right]+c=y$$
3. Get the negative part of the constant term out of the parenthesis. $$a\left(x+\frac{b}{2a}\right)^{2}-a\left(\frac{b}{2a}\right)^{2}+c=y$$
4. And, if you think it helps, tidy up the constant term. $$a\left(x+\frac{b}{2a}\right)^{2}+c-\frac{ab^{2}}{4a^{2}}=y$$ $$y=a\left(x+\frac{b}{2a}\right)^{2}+\frac{4ac}{4a}-\frac{b^{2}}{4a}$$ $$y=a\left(x+\frac{b}{2a}\right)^{2}-\frac{b^{2}-4ac}{4a}$$
5. The vertex form is $y=a\left(x-h\right)^{2}+k$. So if we rearrange to exactly that form: $$y=a\left(x-\left(-\frac{b}{2a}\right)\right)^{2}+\frac{-\left(b^{2}-4ac\right)}{4a}.$$ Then we see that $h=-\frac{b}{2a}$ and $k=-\frac{b^{2}-4ac}{4a}$.