The curve C has equation y = 2x^3 - 5x^2 - 4x + 2 - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 2

To find the turning points, we set ( \frac{dy}{dx} = 0 ):

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

Factoring gives:

$2(3x + 1)(x - 2) = 0,$ which results in ( x = 2 ) and ( x = -\frac{1}{3} ).

Now, we substitute these x-values back into the original equation to find the corresponding y-coordinates:

For ( x = 2 ):

$y = 2(2)^3 - 5(2)^2 - 4(2) + 2 = 2(8) - 5(4) - 8 + 2 = 16 - 20 - 8 + 2 = -10.$

Thus, one turning point is ( (2, -10) ).

For ( x = -\frac{1}{3} ):

$y = 2(-\frac{1}{3})^3 - 5(-\frac{1}{3})^2 - 4(-\frac{1}{3}) + 2 = 2( -\frac{1}{27}) - 5( \frac{1}{9}) + \frac{4}{3} + 2.$

Calculate this value to find the second turning point.

To find ( \frac{d^2y}{dx^2} ), we differentiate ( \frac{dy}{dx} ):

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

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

1. For ( x = 2 ):
   ( \frac{d^2y}{dx^2} = 12(2) - 10 = 24 - 10 = 14 ) (positive, indicating a local minimum).

2. For ( x = -\frac{1}{3} ):
   ( \frac{d^2y}{dx^2} = 12(-\frac{1}{3}) - 10 = -4 - 10 = -14 ) (negative, indicating a local maximum).

In summary, the turning point at ( (2, -10) ) is a minimum and the turning point at ( (-\frac{1}{3}, y) ) is a maximum.