The curve C has equation y = 2x³ - 5x² - 4x + 2 - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 2

Turning points occur where ( \frac{dy}{dx} = 0 ).

Setting the derivative to zero gives us:

$6x^2 - 10x - 4 = 0.$

This can be factored:
$3(x - 2)(x + \frac{2}{3}) = 0.$

Therefore, the x-coordinates of the turning points are ( x = 2 ) and ( x = -\frac{2}{3} ).

To find the corresponding y-coordinates, substitute back into the original equation:

1. For ( x = 2 ):
   [ y = 2(2)^3 - 5(2)^2 - 4(2) + 2 = -10. ] Thus, one turning point is (2, -10).

2. For ( x = -\frac{2}{3} ):
   [ y = 2(-\frac{2}{3})^3 - 5(-\frac{2}{3})^2 - 4(-\frac{2}{3}) + 2 = \frac{26}{27}. ] Thus, another turning point is ( \left(-\frac{2}{3}, \frac{26}{27}\right) ).

Now, we find the second derivative to analyze the concavity:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(6x^2 - 10x - 4) = 12x - 10.$

To determine the nature of the turning points, evaluate ( \frac{d^2y}{dx^2} ) at the critical points found earlier.

1. For ( x = 2 ):
   [ \frac{d^2y}{dx^2} = 12(2) - 10 = 14 > 0, ]
   which indicates a local minimum at (2, -10).

2. For ( x = -\frac{2}{3} ):
   [ \frac{d^2y}{dx^2} = 12(-\frac{2}{3}) - 10 = -18 < 0, ]
   which indicates a local maximum at ( \left(-\frac{2}{3}, \frac{26}{27}\right) ).