y = 2x^3 + 15x^2 + 24x + 10 Find the stationary points on this curve and determine their nature

First differentiate the function to obtain
dy/dx = 6x^2 + 30x + 24
Then set this equal to 0 and solve to find the x values of the stationary points
6x^2 + 30x + 24 = 0
6(x^2 + 5x + 4) = 0
6(x +1)(x+4) = 0
therefore x = -1, -4
then to find the y values sub back into the original function to get
y = (2 x -1^3) + (15 x -1^2) + (24 x -1) + 10 = -1
y = (2 x -4^3) + (15 x -4^2) + (24 x -4) + 10 = 26
so the stationary points are (-1, -1) and (-4, 26)
to determine heir nature find the second differential by differentiating dy/dx = 6x^2 + 30x +24
which is 12x + 30
sub in the x values of the stationary points to obtain
(12 x -1) + 30 = 18 this is positive so (-1, -1) is a minimum
(12 x -4) + 30 = -18 this is negative so (-4, 26) is a maximum