The graph of $y = f(x)$ is shown below - AQA - A-Level Maths Pure - Question 6 - 2020 - Paper 3

To sketch the graph of $y = 2f(x) - 4$, we will stretch the original graph vertically by a factor of 2 and then shift it down by 4 units:

• For the point $(2, 6)$, after stretching, it becomes $(2, 12)$, and after shifting down, it becomes $(2, 8)$.
• The point $(0, 2)$ will stretch to $(0, 4)$ and then shift to $(0, 0)$.
• The point $(-1, 0)$ will stretch to $(-1, 0)$ and then shift to $(-1, -4)$.
• The point $(-2, -2)$ will stretch to $(-2, -4)$ and then shift to $(-2, -6)$.

The final graph will include points $(2, 8)$, $(0, 0)$, $(-1, -4)$, and $(-2, -6)$.

To sketch the graph of $y = f'(x)$, we first identify the intervals where the original function $f(x)$ is increasing or decreasing:

• For $x < -2$, $f(x)$ is a horizontal line, hence $f'(x) = 0$.
• From $(-2, 0)$, $f(x)$ is increasing, hence $f'(x) > 0$.
• At $x = 0$, $f(x)$ has a local minimum, so $f'(0) = 0$.
• For the interval $(0, 2)$, $f(x)$ is increasing, hence $f'(x) > 0$ but less than the gradient before.
• After $x = 2$, the graph is horizontal, hence $f'(x) = 0$ again.

Therefore, the graph of $f'(x)$ will show values jumping from 0 to positive values between $x = -2$ to $x = 2$, then falling back to 0.