Since -(x - 16)<sup>2</sup> = -x<sup>2</sup> + 32x - 256, not 16 - x<sup>2</sup>, your last function cannot be equivalent, and thus cannot be the completed square.
The following is the usual process for completing the square to find the vertex, as illustrated using a different function:
. . . . .Given, a function:
. . . . .y = 3x<sup>2</sup> + 6x + 5
. . . . .Separate the constant term:
. . . . .y = (3x<sup>2</sup> + 6x) + 5
. . . . .Factor out whatever is multiplied on the square:
. . . . .y = 3(x<sup>2</sup> + 2x) + 5
. . . . .Find half of the linear coefficient:
. . . . .(1/2)(2) = 1
. . . . .Square, and add inside and subtract outside,
. . . . .remembering to account for the factoring:
. . . . .y = 3(x<sup>2</sup> + 2x + 1) + 5 - 3(1)
. . . . .Simplify:
. . . . .y = 3(x<sup>2</sup> + 2x + 1) + 2
. . . . .Convert to squared form:
. . . . .y = 3(x + 1)<sup>2</sup> + 2
. . . . .y = 3(x - (-1))<sup>2</sup> + 2
. . . . .Read off the vertex:
. . . . .(h, k) = (-1, 2)
Follow this process with your exercise. If you get stuck, please reply showing all of your work and reasoning. Thank you.
Eliz.