Find vertex and the equation for axis of symmetry of the par

Find vertex and the equation for axis of symmetry of the parabola -2x^2+16x-24. Have to show work.

Solution

y=-2x^2+16x-24

y=-2(x^2 -8x +12)

y=-2(x^2 -8x +12+4 -4)

y=-2(x^2 -8x +16) -2(-4)

y=-2(x^2 -8x +16) +8

y=-2(x-4)² +8

vertex of parabola y =a(x-h)² +k is (h,k) and axis of symmetry is x -h =0

vertex of parabola y=-2(x-4)² +8 is (4,8) and axis of symmetry is x - 4 =0