Given: f(x) = ax 3 + bx 2 + cx + d with the following properties: f(-2) = f(1) = 0, f'(1) = f'(-1) = 0, f(-1) = 4, f(0) = 2, f'(x) < 0 for −1 < x < 1. Draw the graph of f. Indicate clearly the co-ordinates of the turning points and the intercepts.

The graph of $f(x)$ is a cubic function that:

• Passes through the x-intercepts $(-2, 0)$ and $(1, 0)$.
• Passes through the y-intercept $(0, 2)$.
• Has a local maximum at $(-1, 4)$.
• Has a local minimum at $(1, 0)$.
• Increases for $x < -1$, decreases for $-1 < x < 1$, and increases for $x > 1$.

(Since I cannot draw a graph, I've provided a detailed description of its key features and shape.)
The turning points are $\boxed{(-1, 4) \text{ and } (1, 0)}$.
The intercepts are $\boxed{(-2, 0), (1, 0) \text{ (x-intercepts) and } (0, 2) \text{ (y-intercept)}}$.

The function $g(x) = f(x-3)$ represents a horizontal shift of $f(x)$ by 3 units to the right.
The x-intercepts of $f(x)$ are where $f(x)=0$, which are $x=-2$ and $x=1$.
For $g(x)$, the x-intercepts occur when $g(x)=0$, meaning $f(x-3)=0$.
So, we set the argument of $f$ equal to its x-intercepts:

The x-intercepts of $g$ are $\boxed{x=1 \text{ and } x=4}$.

The equation $x^3 - 3x + 2 = p$ is equivalent to $f(x) = p$.
This means we are looking for the values of $p$ (a horizontal line $y=p$) that intersect the graph of $f(x)$ at exactly one point.
From part 8.1, we know the turning points of $f(x)$ are:

• Local maximum: $(-1, 4)$
• Local minimum: $(1, 0)$ For the line $y=p$ to intersect the cubic graph at exactly one point, $p$ must be either above the local maximum or below the local minimum. Therefore, $p > 4$ or $p < 0$. The values of $p$ for which the equation will have exactly one root are $\boxed{p < 0 \text{ or } p > 4}$.