Question

The curve for which dy/dx = 3x^2+ax+b, where a and b are constants, has stationary points at (1,0) , (-3,32). Find the values of a and b

Answer 1

at each starionary point, dy/dx = 0, so when dy/dx = 3x^2 + ax + b
at (1,0) , 0 = 0 + 0 + b
b = 0
at (-3,32), 0 = 27 - 3a + 0
a = 9

a = 9, b = 0

check:
dy/dx = 3x^2 + 9x
at stationary points, 3x^2 + 9x = 0
3x(x+3) = 0
such points have x = 0 or x = -3, as shown by the given points.