下图显示了 $y = \mathrm{f}(x)$ 的图象。该图象的渐近线为 $x = 2$ 和 $y = 2$。图象与 $x$ 轴交于 $(1, 0)$，与 $y$ 轴交于 $(0, 1)$，且有一个转折点 $(4, 1)$。 在不同的图上分别画出下列函数的图象，清晰地标出与两坐标轴的任何交点的坐标、任何转折点，以及任何渐近线的方程（如果可能）。

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 $x \to \infty$.