A curve C has equation $y = e^x + x^4 + 8x + 5$ - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 6

The curve $y = x^3$ is a cubic function that passes through the origin (0, 0) and has a point of inflection at this point. It will appear in Quadrants 1 and 4, increasing steeply in Quadrant 1 and decreasing in Quadrant 4. It does not have any asymptotes.

The curve $y = 2 - e^{-x}$ is an exponential decay curve that approaches the asymptote $y = 2$ as $x$ approaches infinity. It intersects the $y$-axis at the point (0, 1) since: $y(0) = 2 - e^0 = 1.$ The curve approaches the $y$-axis from above and has no crossings with the negative $y$ values.

The graph of $y = x^2$ is a parabola opening upwards, while the graph of $y = 2 - e^{-x}$ is a curve that approaches the horizontal line $y = 2$. As drawn, these two graphs intersect at only one point, indicating that the equation $x^2 = 2 - e^{-x}$ has only one solution, corresponding to a single crossing point.

Using the iteration formula:

$x_{n+1} = (-2 - e^{-x_n})^{1/3}, \, x_0 = -1,$ we can calculate:

1. For $n=0$: $x_1 = (-2 - e^{1})^{1/3} \approx -1.26376$
2. For $n=1$: $x_2 = (-2 - e^{-(-1.26376)})^{1/3} \approx -1.26126$ Thus, rounding to 5 decimal places, $x_1 ext{ is } -1.26376$ and $x_2 ext{ is } -1.26126.$

The $x$ coordinate of the turning point is approximately $-1.26$. To find the $y$ coordinate, substitute this back into the original equation of the curve:

$y = e^{-1.26} + (-1.26)^4 + 8(-1.26) + 5.$ Calculating this will give the $y$ coordinate, which can then be rounded to two decimal places along with the $x$ coordinate.