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

For the function $y = 2f(x) - 4$, we first apply a vertical stretch by a factor of 2, followed by a downward translation of 4 units.

1. Vertical Stretch: Multiply the y-coordinates of each point from the original graph by 2.

   • $(2, 6)$ becomes $(2, 12)$
   • $(0, 2)$ becomes $(0, 4)$
   • $(-1, 0)$ stays at $(-1, 0)$
   • $(-2, -2)$ becomes $(-2, -4)$

2. Downward Translation: Subtract 4 from each y-coordinate:

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

Plot these new points, connecting them according to the transformations applied.

To sketch the graph of the derivative $y = f'(x)$, we need to analyze the slopes between key points of the original function $f(x)$.

1. Between points $(-2, -2)$ and $(-1, 0)$, the slope is positive, therefore $f'(x) > 0$.
2. The slope is zero at $(0, 2)$ where the function has a horizontal tangent.
3. Between $(0, 2)$ and $(2, 6)$, the slope is also positive as the function is increasing.
4. After $(2, 6)$, the function appears to flatten out, indicating that the derivative approaches zero.

The graph consists of:

• A positive line segment between $x = -2$ and $x = -1$
• A horizontal line at $y = 0$ between $x = -1$ and $x = 0$
• Another positive line segment between $x = 0$ and $x = 2$
• A horizontal line at $y = 0$ again after $x = 2$.

Mark these sections clearly, indicating where $f'(x)$ is positive and where it changes from positive to zero.