A curve C has equation $y = e^x + x^4 + 8x + 5$ (a) Show that the x coordinate of any turning point of C satisfies the equation $x^2 = 2 - e^{-x}$ (b) On the axes given on page 5, sketch, on a single diagram, the curves with equations (i) $y = x^3$ (ii) $y = 2 - e^{-x}$ On your diagram give the coordinates of the points where each curve crosses the y-axis and state the equation of any asymptotes - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 6

For sketching the curves:

1. For $y = x^3$, it will pass through the origin and will have increasing values as x goes from negative to positive.
2. For $y = 2 - e^{-x}$, observe that it crosses the y-axis at $(0, 2 - e^0) = (0, 1)$ and has a horizontal asymptote as $x \to \infty$ where it approaches $y=2$.

Make sure both curves are well-labeled on the diagram, and mark their intersection clearly if applicable.

From the sketch, we note that the curve for $y = x^2$ intersects the curve $y = 2 - e^{-x}$ at only one point, indicating that there is a unique solution to the equation $x^2 = 2 - e^{-x}$. The distinct shapes and positions of these curves in relation to each other confirm the single crossing point.

Using the iteration formula:
$x_{n+1} = (-2 - e^{-x_n})^{\frac{1}{3}}$

1. Start with $x_0 = -1$:

$x_1 = (-2 - e^{1})^{\frac{1}{3}} \approx -1.26376$

2. Now using $x_1$ to find $x_2$:

$x_2 = (-2 - e^{-1.26376})^{\frac{1}{3}} \approx -1.26126$

Thus, the values are approximately $x_1 \approx -1.26376$ and $x_2 \approx -1.26126$.

The turning point of the curve C is found at the coordinate derived from $x_2$:

At $x \approx -1.26$, substituting into the original equation gives:

$y = e^{-1.26} + (-1.26)^4 + 8(-1.26) + 5$

Calculating gives a corresponding y value.

The turning point coordinates are approximately $(-1.26, y)$, with $y$ calculated to two decimal places.