Derivation of the formula for the vertex of a parabola 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 $x=-\frac{b}{2a}$.