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)