How do you work out the turning points of a cubic function?

What you have solved for above are the roots of this function (you have located the points where the curve crosses the x-axis).

What you are looking for are the turning points, or where the slop of the curve is equal to zero.
You need to establish the derivative of the equation:
y' = 3x^2 + 10x + 4
Then you need to solve for zeroes using the quadratic equation, yielding x = -2.9, -0.5
Plug them back into your original equation, and you get the points
(-2.9,6.1) and (-0.5,-0.9)

first expand the expression:
(x^2 + x)(x + 4) = x^3 + 4x^2 + x^2 + 4x = x^3 + 5x^2 + 4x

now note that the turning points of the curve have a gradient of zero as they flatten out, so if we differenciate the differenciation (equation to get the gradient by substituting x) must equal zero, the gradient of the required places, which must be 2, if u have noticed on the graph.
therefore:
differentiating the equation f(x)= x^3 + 5x^2 + 4x  will give us:
(dy/dx) = 2x^2 + 10x + 4
when the gradient is zero (the places we want):
2x^2 + 10x + 4 = 0
simplify and solve quadratically:
x^2 + 5x +2 = 0
therefore, 2 points will be given to you, which are the x-points:
 (to 2 significant figures:)x1=-0.438, x2=-4.56, therefore the coordinates are:

( -0.438 , 0 )  or ( -4.56 , 0 )

i hope this helped