Show that the curve y =f(x) has exactly two turning points, where f(x)= x^3 - 3x^2 - 24x - 28

This question is about turning points of a curve, these occur when the gradient of a line is zero. In its current form, the equation of the curve tells us nothing about the gradient of the curve; it must be differentiated, recalling the power rule for differentiation:dy/dx = (3)x^(3-1) - (2)3x^(2-1) - (1)24x^(1-1) - (0)28dy/dx = 3x^2 - 6x -24Remembering that turning points occur when dy/dx = 0, we must show that there are two solutions to 3x^2 - 6x -24 = 0 to answer the question. We can do this by using the quadratic discriminant, b^2 - 4ac for quadratic expressions of the form ax^2 + bx + c = 0.First, simplify the expression by dividing by 3 (this is allowed as we are not dividing by zero or something that could be zero) to x^2 - 2x - 8 = 0 and compare it to the general form of the quadratic. This tells us that a = 1, b = -2 and c = -8. Now substitute the values into the discriminant:(-2)^2 - 4(1)(-8) = 4 - (-32) = 36 > 0As the value of the discriminant is positive, the quadratic has two distinct real roots and the curve y = f(x) has exactly two turning points.