The diagram below shows the graph of $y = \mathrm{f}(x)$. The asymptotes of the graph are $x = 2$ and $y = 2$. The graph crosses the $x$-axis at $(1, 0)$ and the $y$-axis at $(0, 1)$, and has a turning point at $(4, 1)$. On separate diagrams, sketch the following graphs, indicating clearly the coordinates of any points of intersections with both axes and any turning point(s), and the equations of any asymptotes where possible.

1. Determine the effect of the $|x|$ transformation on the key features of the graph.

The transformation $x \mapsto |x|$ means we keep the part of the graph $y = \mathrm{f}(x)$ where $x \ge 0$, and reflect it in the $y$-axis to generate the part for $x < 0$.

• The portion for $x < 0$ of the original graph is discarded.
• $y$-intercept: Remains at $(0, 1)$.
• Turning point: $(4, 1)$ is retained, and its reflection $(-4, 1)$ is added.
• Vertical asymptote: $x = 2$ is retained, and its reflection $x = -2$ is added.
• Horizontal asymptote: $y = 2$ is retained (now approached as both $x \to \infty$ and $x \to -\infty$).

2. Apply the vertical translation to find the final key features for the sketch.

The $+ 1$ translates the graph of $y = \mathrm{f}(|x|)$ vertically upwards by $1$ unit.

• $y$-intercept: $(0, 1) \to (0, 2)$. Because the graph for $x>0$ is decreasing from $x=0$, reflecting it creates a sharp peak (local maximum) at $(0,2)$.
• Minimum turning points: $(4, 1) \to (4, 2)$ and $(-4, 1) \to (-4, 2)$.
• Vertical asymptotes: Remain at $x = 2$ and $x = -2$.
• Horizontal asymptote: $y = 2 \to y = 3$.
• $x$-intercepts: The original graph passed through $(1,0)$, which translates to $(1,1)$. The new graph crosses the $x$-axis closer to the vertical asymptotes (between $1$ and $2$, and between $-1$ and $-2$), but their exact coordinates cannot be algebraically determined so they do not need to be labelled.

1. Determine the equations of the new asymptotes.

• The $x$-intercept of $y = \mathrm{f}(x)$ is at $(1, 0)$. As $x \to 1$, $\mathrm{f}(x) \to 0$, so $\frac{1}{\mathrm{f}(x)} \to \pm\infty$. This gives a new vertical asymptote at $x = 1$.
• The horizontal asymptote of $y = \mathrm{f}(x)$ is $y = 2$. As $x \to \pm\infty$, $\mathrm{f}(x) \to 2$, so $\frac{1}{\mathrm{f}(x)} \to \frac{1}{2}$. This gives a new horizontal asymptote at $y = 0.5$.

2. Determine the new intercepts and turning point.

• The vertical asymptote of $y = \mathrm{f}(x)$ is $x = 2$. As $x \to 2$, $\mathrm{f}(x) \to \pm\infty$, so $\frac{1}{\mathrm{f}(x)} \to 0$. This gives an $x$-intercept at $(2, 0)$.
• The $y$-intercept of $y = \mathrm{f}(x)$ is $(0, 1)$, so $\frac{1}{\mathrm{f}(0)} = \frac{1}{1} = 1$. The $y$-intercept remains at $(0, 1)$.
• The local minimum of $y = \mathrm{f}(x)$ is at $(4, 1)$. Taking the reciprocal of the $y$-coordinate gives a local maximum at $\left(4, \frac{1}{1}\right) = (4, 1)$.

3. Sketch the graph using the deduced features.

• For $x < 1$: The curve approaches $y = 0.5$ from above as $x \to -\infty$, increases through $(0, 1)$, and goes to $+\infty$ as $x \to 1^-$.
• For $x > 1$: The curve comes from $-\infty$ as $x \to 1^+$, crosses the $x$-axis at $(2, 0)$, reaches a local maximum at $(4, 1)$, and decreases asymptotically to $y = 0.5$ from above as $x \to \infty$.