Find both stationary points for y= 4x^(3)-3x^(2)-60x+24. Also find the nature of those points.

y= 4x³- 3x²- 60x+ 24dy/dx = 12x²- 6x- 60 = 2x²- x- 10 = (2x-5)(x+2)
So stationary points are at x= 5/2 and x= -2
d²x/dy² = 24x - 6
24(5/2) - 6 = positive so this is a minimum point
24(-2) - 6 = negative so this is a maximum point