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

To sketch the graph of $y = 2f(x) - 4$, follow these steps:

1. Vertical stretch: Multiply the y-values of the original graph by 2. For example:

   • $(-1, 0) \rightarrow (-1, 0)$
   • $(0, 2) \rightarrow (0, 4)$
   • $(2, 6) \rightarrow (2, 12)$
   • $(-2, -2) \rightarrow (-2, -4)$

2. Vertical shift down: Subtract 4 from all y-values. Updating those points:

   • $(-1, 0) \rightarrow (-1, -4)$
   • $(0, 4) \rightarrow (0, 0)$
   • $(2, 12) \rightarrow (2, 8)$
   • $(-2, -4) \rightarrow (-2, -8)$

3. Draw the new graph: Connect these points smoothly to show the transformed graph.

To sketch $y = f'(x)$ which represents the derivative of $f(x)$, consider the following steps:

1. Identify intervals of increase and decrease:

   • The function $f(x)$ increases from $(-1, 0)$ and $(0, 2)$, hence $f'(x) > 0$ in these intervals.
   • The function decreases in the interval $(2, + ext{infinity})$, hence $f'(x) < 0$ here.

2. Identify critical points: From the graph, find the points where $f(x)$ has horizontal tangents and changes direction, which indicates where $f'(x) = 0$. For instance, at $(0, 2)$ and where the curve begins to decrease.

3. Sketch the derivative:

   • For $x < -2$, $f'(x) = 0$ leading to a horizontal line.
   • For $-2 < x < 0$, $f'(x)$ is positive, represented by above the x-axis.
   • At $x = 0$, mark a transition point.
   • For $0 < x < 2$, $f'(x)$ remains positive.
   • For $x > 2$, $f'(x)$ becomes negative until the end.