Figure 1 shows a sketch of the curve C with the equation $y = (2x^2 - 5x + 2)e^{-x}$ - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 5

To determine where C crosses the x-axis, we need to solve for when $y = 0$: $0 = (2x^2 - 5x + 2)e^{-x}.$
Since $e^{-x}$ is never zero, we can focus on the quadratic: $2x^2 - 5x + 2 = 0.$
Using the quadratic formula, the roots are: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}.$
Thus, we have:

1. $x = \frac{8}{4} = 2$
2. $x = \frac{2}{4} = 0.5$.

So, C crosses the x-axis at $x = 2$ and the other point is $x = 0.5$.

To find the derivative $\frac{dy}{dx}$, we will use the product rule, since $y$ is a product of two functions: $y = (2x^2 - 5x + 2)e^{-x}.$
By the product rule, we have: $\frac{dy}{dx} = u'v + uv',$
where $u = (2x^2 - 5x + 2)$ and $v = e^{-x}$.
Calculating each part: $u' = 4x - 5$
and
$v' = -e^{-x}.$
Therefore: $\frac{dy}{dx} = (4x - 5)e^{-x} + (2x^2 - 5x + 2)(-e^{-x})$
which simplifies to: $\frac{dy}{dx} = e^{-x}[(4x - 5) - (2x^2 - 5x + 2)].$
Further simplifying gives: $\frac{dy}{dx} = e^{-x}[-2x^2 + 9x - 7].$

To find the turning points, we set $\frac{dy}{dx} = 0$: $e^{-x}[-2x^2 + 9x - 7] = 0.$
Since $e^{-x}$ is never zero, we solve the quadratic: $-2x^2 + 9x - 7 = 0.$
Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{9^2 - 4(-2)(-7)}}{2(-2)} = \frac{9 \pm \sqrt{81 - 56}}{-4} = \frac{9 \pm \sqrt{25}}{-4} = \frac{9 \pm 5}{-4}.$
Thus, we have:

1. $x = \frac{14}{-4} = -3.5$
2. $x = \frac{4}{-4} = -1.$

Finally, substituting these values back into the original equation to find the corresponding $y$ coordinates:

1. At $x = -3.5$,
   $y = (2(-3.5)^2 - 5(-3.5) + 2)e^{3.5} = (24.5 + 17.5 + 2)e^{3.5} = 44e^{3.5}.$
2. At $x = -1$,
   $y = (2(-1)^2 - 5(-1) + 2)e^{1} = (2 + 5 + 2)e^{1} = 9e^{1}.$

Therefore, the exact coordinates of the turning points are
$(-3.5, 44e^{3.5})$ and $(-1, 9e^{1}).$