find th quadratic fuction that has a vertix of (8.-3) and whose graph goest throught the point (-6, -395)

V(8 , -3) , P(-6 , -395).

Y = a(x - h)^2 + K.
-395 = a(-6 - 8)^2 + (-3),
-395 = 196a - 3,
196a = -395 + 3,
196a = -392,
a = -392 / 196 = -2.

Eq: y = -2(x - 8)^2 - 3(Vertex form).

This parabola opens downward, because a < 0.