<p>Well, the vertex of a parabola can easily be found in general. Letting f(x) = ax^2 + bx + c, f’(x) = 2ax + b, setting f’(x) = 0 gives x = -b/2a. Limits on Math II are pretty basic, but if you had never seen limits before taking calculus, more power to you.</p>
<p>A similar scenario I’ve seen occurred when I had to find the minimum value of x + (1/x) (or something like that, idr) where x > 0. You can do this with calculus without too much effort, but AM-GM gives a slick, one-second solution, giving x = 1 as the value that minimizes x + 1/x.</p>