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

The curve C crosses the x-axis where $y = 0$. Setting the equation:

$$(2x^2 - 5x + 2)e^{-x} = 0 \\$$

Since $e^{-x} eq 0$, we solve:

$$2x^2 - 5x + 2 = 0 \\$$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -5$, and $c = 2$:

$$x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} \\$$ $$x = \frac{5 \pm \sqrt{25 - 16}}{4} \\$$ $$x = \frac{5 \pm 3}{4} \\$$

This gives:

1. $x = 2$
2. $x = \frac{1}{2}$

Thus, C crosses the x-axis at $x = 2$ and the other point is at $x = \frac{1}{2}$.

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

$$\frac{dy}{dx} = u'v + uv' \\$$

where $u = (2x^2 - 5x + 2)$ and $v = e^{-x}$.

Calculating derivatives:

• $u' = 4x - 5$
• $v' = -e^{-x}$

Thus:

$$\frac{dy}{dx} = (4x - 5)e^{-x} + (2x^2 - 5x + 2)(-e^{-x}) \\$$ $$= e^{-x} [ (4x - 5) - (2x^2 - 5x + 2) ] \\$$

Combining terms gives:

$$= e^{-x} [ -2x^2 + 9x - 7 ] \\$$

Turning points occur when $\frac{dy}{dx} = 0$:

$$$$

Applying the quadratic formula again:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\$$

where $a = -2$, $b = 9$, and $c = -7$:

$$x = \frac{-9 \pm \sqrt{(9)^2 - 4(-2)(-7)}}{2(-2)} \\$$ $$x = \frac{-9 \pm \sqrt{81 - 56}}{-4} \\$$ $$x = \frac{-9 \pm \sqrt{25}}{-4} \\$$ $$x = \frac{-9 \pm 5}{-4} \\$$

This leads to:

1. $x = 2$
2. $x = 1$

To find the coordinates of the turning points, substitute these x-values back into the original equation:

1. For $x = 2$:
   $y = (2(2)^2 - 5(2) + 2)e^{-2} = (8 - 10 + 2)e^{-2} = 0$
2. For $x = 1$:
   $y = (2(1)^2 - 5(1) + 2)e^{-1} = (2 - 5 + 2)e^{-1} = -1e^{-1}$

Thus, the turning points are (2, 0) and (1, -1/e).