The graph of $y = |\mathrm{f}(x)|$ has a turning point at $(0,2)$ and passes through the points $(-4,0)$ and $(4,0)$. The graph of $y = \mathrm{f}'(x)$ has a point of inflexion at the origin. On separate diagrams, sketch the graphs of stating the equations of any asymptotes, the coordinates of any stationary points and points of intersection with the axes.

1. Identify the geometrical transformations mapping the original graph to the new graph.

To sketch $y = \mathrm{f}'(-2x)$ from $y = \mathrm{f}'(x)$, we apply two transformations:

1. A horizontal scaling by a scale factor of $\frac{1}{2}$ parallel to the $x$-axis.
2. A reflection in the $y$-axis.

2. Determine the transformed asymptotes and axis intersections.

Under the transformation $x \mapsto -\frac{x}{2}$:

• The vertical asymptotes $x = -2$ and $x = 2$ map to $x = 1$ and $x = -1$.
• The horizontal asymptote $y = 0$ (the $x$-axis) remains unchanged at $y = 0$.
• The origin $(0,0)$, which is the only point of intersection and a point of inflexion, remains at $(0,0)$.

3. Determine the shape and position of the new curve branches.

Due to the reflection in the $y$-axis and scaling:

• The branch initially in $x < -2$ ($y > 0$) moves to $x > 1$ ($y > 0$).
• The branch initially in $x > 2$ ($y < 0$) moves to $x < -1$ ($y < 0$).
• The central branch ($-2 < x < 2$) passing through $(0,0)$ with a negative gradient reflects horizontally to become an increasing curve on $-1 < x < 1$, passing through $(0,0)$ with a positive gradient.

1. Determine the coordinates and nature of the stationary point for the middle branch.

From the graph of $y = \mathrm{f}'(x)$, we have $\mathrm{f}'(x) > 0$ for $-2 < x < 0$ and $\mathrm{f}'(x) < 0$ for $0 < x < 2$, with $\mathrm{f}'(0) = 0$.
This indicates that $\mathrm{f}(x)$ is increasing then decreasing, giving a local maximum at $x = 0$.
Since the graph of $y = |\mathrm{f}(x)|$ has a minimum at $(0, 2)$ and goes to $+\infty$ at $x = \pm 2$, $\mathrm{f}(x)$ must be negative in this interval to reflect into a minimum on the modulus graph.
Thus, the turning point is a local maximum at $(0, -2)$, and the curve approaches $-\infty$ as $x \to -2^+$ and $x \to 2^-$.

2. Determine the properties of the outer branches.

For $x < -2$, $\mathrm{f}'(x) > 0$, so $\mathrm{f}(x)$ is strictly increasing. Since it passes through the $x$-axis at $(-4, 0)$ and goes towards $+\infty$, it must be negative for $x < -4$. As $y = |\mathrm{f}(x)|$ has a horizontal asymptote $y = 1$, $\mathrm{f}(x)$ must approach a negative value, giving the horizontal asymptote $y = -1$ as $x \to -\infty$.
Similarly, for $x > 2$, $\mathrm{f}'(x) < 0$, so $\mathrm{f}(x)$ is strictly decreasing. It passes through $(4, 0)$, coming from $+\infty$, so it is negative for $x > 4$. Thus, it also approaches the horizontal asymptote $y = -1$ as $x \to \infty$.

3. Summarise the features for the final sketch.

The sketch of $y = \mathrm{f}(x)$ requires:

• Vertical asymptotes at $x = -2$ and $x = 2$.
• A horizontal asymptote at $y = -1$.
• $x$-intercepts at $(-4, 0)$ and $(4, 0)$.
• A $y$-intercept and local maximum at $(0, -2)$.