When trying to find stationary points, the first thing you should think about is differentiating. At a stationary point, the gradient of a curve or function is equal to zero. Therefore if we differentiate the equation of the curve and set it equal to zero, we can solve for x to find where the stationary points are. You can then substitute the values of x back into the original equation to find the values of y, respectively. Differentiating using the "bring the power down, and subtract one from the power" method gives us f'(x) = 3x2+12x-36. We then set this equal to zero, and can divide by 3 on both sides since it is a common factor, leaving x2+4x-12 = 0We can factorise this quadratic to obtain (x+6)(x-2)=0, giving two solutions at x = -6 and x = 2.Then sub in these values to f(x), which gives 216 and -40, respectively. The stationary points are therefore (-6, 216) and (2, -40).