Let y = x^3 + ax^2 + bx. If (x,y) = (2,64) is a point on the curve and the slope of the tangent at x=-1 is 3, what is the value of a+b?

Answer 1

dy/dx = 3x^2 + 2ax + b
when x = -1, dy/dx = 3

3 - 2a + b = 3
b = 2a

also the point (2,64) lies on the curve, so
64 = 8 + 4a + 2b
4a + 2b = 56
2a + b = 28
but b = 2a
2a + 2a = 28
a = 7
b = 14

thus a+b = 21