For $f(x) = ax^2 + bx +c$, why is it written $a(x-h)^2 + k$

Question

I'm going to have to teach how to graph quadratic equations. Since we've already done a lot of work with the Quadratic Formula, the students are more or less familiar with the standard notation of a parabola, $ax^2 + bx +c$

The alternate form for the parabola, which is used to quickly identify the vertex point, is $a(x-h)^2 +h$. It's been quite some time since I learned this personally, so I can't remember the full reason and I can't seem to find it on the internet anywhere, but why is it $-h$ instead of $+h$?

Here were my algebra steps - I am assuming there was an algebra step that I either forgot to do or a sign switch that was done for s&g to make the reading easier on the final form.

$ax^2 + bx +c$

=$a(x^2 + \frac{b}{a}x +\frac{c}{a})$

=$a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} + \frac{c}{a})$

=$a[(x+\frac{b}{2a})^2 + \frac{4ac-b^2}{4a^2}]$

=$a(x+\frac{b}{2a})^2 + \frac{4ac-b^2}{4a}$

Why do they do this last bit?

$a(x - (-\frac{b}{2a}))^2 + \frac{4ac-b^2}{4a}$

In order to have a form similar to the first term of the quadratic formula? The last bit is just a substitution of $h = -\frac{b}{2a}, k = \frac{4ac-b^2}{4a}$ , I'm just wondering why they chose $a(x-h)^2 +k$ instead of $a(x+h)^2 +k$ (assuming that an appropriate alternation to the value of h was made)

To restate: why did they do the last bit of substitution to make the vertex-form $(x-h)^2$ instead of $(a+h)^2$ ?

Answer 1

This is done precisely so that we can make the statement that

(For $a \neq 0$,) the vertex of the quadratic $$y = a(x - h)^2 + k$$ is simply $(h, k)$.

If we instead use the form $$y = a (x + h')^2 + k,$$ the vertex would be at $(-h', k)$, which of contains a minus sign.