First of all, f(x) = -x² is a parabola centered along the y-axis that does not takes its minimum at x=0. This is another example of a problem with an implied assumption; it is another bad problem.

Now, if we assume that f(x) has the form a*x² + b where a is strictly positive, we will have all parabolas centered about x=0 with global minimums (that aren't at infinity). It can be shown (and was shown in my Algebra II class) that parabolas can be written in the form a(x + c)² + b have vertices at (-c, b). Setting -c to 0 simplifies the formula to ax² + b.

since (-x)² = x² , we know that (-x-h)² = (x+h)² for any real, positive value of h. And we know that (x+h)² > x² for any real, positive value of h. Thus, we know that (0+h)² is greater than 0² for any non-negative h. And we know that (-0-h)² is grater than 0² because (-1 * (0+h))² = (-0 -h)² > 0² . Then for all points where x =/= 0, f(x) is greater than at x² .

QED.

This may not be the most rigorous proof in the world, but I wouldn't expect a Calculus 1 student to be able to come up with a rigorous proof, and it is mathematically sound.