College Algebra
posted by Jean on .
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.