I'm taking a course on Basic Conic Sections, and one of the ones we are discussing is of a parabola of the form

$$y = a x^2 + b x + c$$

My teacher gave me the formula:

$$x = -\frac{b}{2a}$$

as the $x$ coordinate of the vertex.

I asked her why, and she told me not to tell her how to do her job.

My smart friend mumbled something about it involving calculus, but I've always found him a rather odd fellow and I doubt I'd be able to understand a solution involving calculus, because I have no background in it.  If you use something you know from calculus, explain it to someone who has no background in it.  Because I sure don't.

Is there a purely algebraic or geometrical yet elegant derivation for the $x$ coordinate of a parabola of the above form?

Already so many answers, but I haven't seen my favorite one posted, so here's another.

The vertex occurs on the vertical line of symmetry, which is not affected by shifting up or down.  So subtract $c$ to obtain the parabola $y=ax^2+bx$ having the same axis of symmetry.  Factoring $y=x(ax+b)$, we see that the $x$-intercepts of this parabola occur at $x=0$ and $x=-\frac{b}{a}$, and hence the axis of symmetry lies halfway between, at $x=-\frac{b}{2a}$.
