dh/dx = 0 at critical points
dh/dx = 3 x^2 + 2 b x = 0
x (3x+2b) = 0
so
at (2,4), 3x+2b = 0 = 6 + 2b
so b = -3
so
h = x^3 -3 x^2 + d
at x = 2, h = -4
-4 = 8 - 12 + d
d = 0
so
h = x^3 -3 x^2
The function h(x) = x^3 + bx^2 + d has a critical point at (2, -4). Determine the constants b and d and find the equation of h(x).
IM SO CONFUSED :(
1 answer