Symmetry of the Parabola

The following result is interesting as it can be proved using the quadratic formula.  It involves "Yager's Theorem of Symmetry in Graphs" and the general form of a parabola (ax² + bx + c = 0).

Consider the result for f(x) = ax² + bx + c

                                 If f(x + A) = f(-x + A)

        a(x + A)² + b(x + A) + c = a(-x + A)² + b(-x + A) + c

a(x² + 2Ax + A²) + bx + Ab = a(x² -2Ax + A²) -bx + Ab

                               4Aax + 2bx = 0

/2x):                              2Aa + b = 0

                                                A = -b/2a

Therefore the parabola has an axes of symmetry at x = -b/2a