Derivation of the formula for the vertex of a parabola

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

By the vertex I assume you mean the minimum/maximum point of the parabola. Indeed, this result can be discovered easily through a bit of calculus, but there is also a simple purely algebraic way, which I will present here.

Let's consider a generic quadratic expression:

$y = ax^2 + bx + c$

We now complete the square on this formula.

$y = a[x^2 + bx/a + c/a]$

$y = a[(x + b/2a)^2 - (b/2a)^2 + c/a]$

The expression $- (b/2a)^2 + c/a$ is a constant (it does not depend on x), so we can replace it with k for the matter of discussion.

$y = a[(x + b/2a)^2 + k]$

Now, depending on whether a is positive or negative, the parabola given by y will either have a maximum or minimum. Since a and k are fixed, this must occur when $(x + b/2a)^2$ is zero (we know it cannot be less than zero, and it can extend to infinity).

Hence, we know that for $(x + b/2a)^2$ to be zero, $x = -b/2a$. This in turn implies that the function y is at a minimum or a maximum when this is true. Q.E.D.