The curve with the equation: y=x^2 - 32sqrt(x) + 20 has a stationary point P. Find the coordinates of P.

A stationary point implies that the gradient at this point will be equal to 0, it is a turning point in the graph. So we need to find the value of X at which dy/dx = 0. [where dy/dx is the gradient.]So we first differentiate the equation Y with respect to X giving us: dy/dx = 2x + 16x^(-0.5) [using simple differentiation, i.e. bring the power down, and take 1 away from the power]We would now equate this to 0, as this will tell us when the gradient is equal to 0.so 2x + 16x^(-0.5) = 0Now solve for x: equation simplifies to x^(1.5) = 8, which tells us that X=4We now have X, we need to work out the Y value at this point. So we substitue X=4 into the equation of the curvey=4^2 - 32sqrt(4) + 20, which tells us that Y= -28. So point P(4,-28)