A parábola opening up or down has a vertex (0, 4) and passes through (-8, -4) equation in vertex form

Answer 1

y = (-1/8)(x-0)^2 + 4 in vertex form y=a(x-h)^2 + k where (h,k)=the vertex, a<0 as it's downward opening

or more simply

y=-x^2/8 + 4

or

x^2 + 8y = 32

check the answer, plug in the points

(-8,-4) (-8)^2 + 8(-4) = 64-32= 32

(0,4) (0)^2 + 8(4) = 0+32 = 32

y intercept is (0,4) = the vertex = the maximum point on the parabola

x intercepts are (4sqr2, 0) and (-4sqr2, 0)

maybe graph it, to see it better