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 7 - 2010 - Paper 5

To find where the curve crosses the x-axis, set $y = 0$.
Starting with the equation: $0 = (2x^2 - 5x + 2)e^{-x}$
Since $e^{-x} > 0$, we can set: $2x^2 - 5x + 2 = 0$
Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)}$
This simplifies to: $x = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}$
Calculating gives:

1. $x = 2$
2. $x = \frac{1}{2}$
   Thus, the curve crosses the x-axis at $x = 2$ and at $x = \frac{1}{2}$.

To differentiate $y = (2x^2 - 5x + 2)e^{-x}$, we use the product rule:

If $u = 2x^2 - 5x + 2$ and $v = e^{-x}$, then: $\frac{dy}{dx} = u'v + uv'$
Calculating $u'$ and $v'$ gives: $u' = 4x - 5, \quad v' = -e^{-x}$
Thus: $\frac{dy}{dx} = (4x - 5)e^{-x} + (2x^2 - 5x + 2)(-e^{-x})$
This simplifies to: $\frac{dy}{dx} = e^{-x}[(4x - 5) - (2x^2 - 5x + 2)]$
Further simplification results in: $\frac{dy}{dx} = e^{-x}[-2x^2 + 9x - 7].$

Turning points occur where $\frac{dy}{dx} = 0$.
Setting the numerator to zero gives: $-2x^2 + 9x - 7 = 0.$
Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{9^2 - 4(-2)(-7)}}{2(-2)}$
Calculating: $x = \frac{9 \pm \sqrt{81 - 56}}{-4} = \frac{9 \pm \sqrt{25}}{-4} = \frac{9 \pm 5}{-4}$
Thus:

1. $x = \frac{4}{-4} = -1$
2. $x = \frac{14}{-4} = -\frac{7}{2}$
   Now, substitute these x-values into the original equation to find the y-coordinates.
   For $x = -1$: $y = (2(-1)^2 - 5(-1) + 2)e^{1} = (2 + 5 + 2)e^{1} = 9e$ For $x = -\frac{7}{2}$: Calculate similarly to find $y$. Thus, turning points are at $(-1, 9e)$ and at the coordinates found for $x = -\frac{7}{2}$.