Question 8: a) Sketch the graph of the function y = x(x + 2)(2 − x). b) Calculate the point of inflexion for the function y = x³ − 2x² + x + 1.

Set $y=0$:

This gives $x=0$, $x+2=0 \Rightarrow x=-2$, and $2-x=0 \Rightarrow x=2$.
The x-intercepts are at $(-2,0)$, $(0,0)$, and $(2,0)$.

Set $x=0$:

The y-intercept is at $(0,0)$.

Expand the function to find the leading term:

The leading term is $-x^3$. Since the degree is odd (3) and the leading coefficient is negative (-1):

• As $x \to \infty$, $y \to -\infty$ (the graph goes down on the right side).
• As $x \to -\infty$, $y \to \infty$ (the graph goes up on the left side).

1. Plot the x-intercepts at $(-2,0)$, $(0,0)$, and $(2,0)$.
2. Starting from the top-left (as $x \to -\infty$, $y \to \infty$), draw the curve downwards.
3. Pass through the x-intercept $(-2,0)$.
4. The curve will then turn and rise, passing through the y-intercept and x-intercept $(0,0)$.
5. It will continue to rise to a local maximum, then turn downwards.
6. Pass through the x-intercept $(2,0)$.
7. Continue downwards to the bottom-right (as $x \to \infty$, $y \to -\infty$). The graph will be a cubic curve that starts high on the left, dips below the x-axis, crosses the origin, rises above the x-axis, and then dips below the x-axis again as it moves to the right.

To combine these terms, find a common denominator, which is 27:

The point of inflexion is $\boxed{\left(\frac{2}{3}, \frac{29}{27}\right)}$.