The curve C has equation y = 2x^3 - 5x^2 - 4x + 2 - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 2

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

$6x^2 - 10x - 4 = 0.$
This can be simplified to:
$3(x-2)(x+\frac{2}{3}) = 0.$
Thus, ( x = 2 ) or ( x = -\frac{2}{3} ).

Next, we find the corresponding ( y )-coordinates by substituting these values back into the original equation:

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{2}{3} ):
Substituting $y = 2(-\frac{2}{3})^3 - 5(-\frac{2}{3})^2 - 4(-\frac{2}{3}) + 2 = -\frac{16}{27} - \frac{20}{9} + \frac{8}{3} + 2.$
After finding a common denominator and simplifying, the coordinate can be calculated.

To find the second derivative, we differentiate the first derivative:

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

To determine the nature of the turning points:

We evaluate ( \frac{d^2y}{dx^2} ) at the turning point ( x = 2 ):
$\frac{d^2y}{dx^2} = 12(2) - 10 = 24 - 10 = 14 > 0.$
Since this is positive, the turning point ((2, -10)) is a local minimum.

Now check at ( x = -\frac{2}{3} ):
$\frac{d^2y}{dx^2} = 12(-\frac{2}{3}) - 10 = -8 - 10 = -18 < 0.$
This indicates that the turning point at ((-\frac{2}{3}, y)) is a local maximum.