The critical points of the functions are
(a) y=3 x² +12 x-35
(b) f(x) = -2x³ + 15x² - 36x + 27
How to find the critical points
For the function y = 3x² + 12x - 35, to find critical points, compute the derivative y and set it to zero. Then solve for x
y' = 6x + 12
0 = 6x + 12
-6x = 12
x = -2
For f(x) = -2x³ + 15x² - 36x + 27,
y' = -6x² + 30x - 36
0 = -6x² + 30x - 36
solving the equation gives graphically gives
x = 2 and x = 3