What is the the vertex and the axis of symmetry of the parabola: y = x^2 − 16x + 63?

We need to determine the vertex and the the axis of
symmetry of the parabola y = x^2 - 16x + 63

For the general
equation of a parabola y = ax^2 + bx + c, the x-coordinate of the vertex is given by
-b/2a.

Substituting the values for the given parabola we
have 16/2 = 8

For x = 8, y = 8^2 - 16*8 + 63 = 64 - 128 +
63 = -1

The vertex of the given parabola is (8,
-1)

The axis of symmetry of a parabola is given by x =
-b/2a

For the given parabola it is x =
8

The axis of symmetry of the parabola is x =
8 and the vertex is (8, -1)