图中给出了 $y = \mathrm{f}(|x|)$ 和 $y = \mathrm{f}'(x)$ 的图像。 $y = \mathrm{f}(|x|)$ 的图像在 $(-3, 4)$ 和 $(3, 4)$ 处有转折点，交 $x$ 轴于 $(-2, 0)$ 和 $(2, 0)$，其渐近线方程为 $x = -1$，$x = 1$ 和 $y = 2$。$y = \mathrm{f}'(x)$ 的图像交 $x$ 轴于 $(3, 0)$，其渐近线方程为 $x = 1$ 和 $y = 0$。 在不同的图中，分别草拟以下函数的图像，清晰标明任何渐近线方程、任何坐标轴交点以及转折点（若有）的坐标。

1. Determine the shape of the graph for $x \ge 0$ using the graph of $y = \mathrm{f}(|x|)$.

Since $|x| = x$ for $x \ge 0$, the graph of $y = \mathrm{f}(x)$ for $x \ge 0$ is exactly the right half of the given graph of $y = \mathrm{f}(|x|)$.
From the given $y = \mathrm{f}(|x|)$ graph, we extract the following features for $x \ge 0$:

• The curve passes through the origin $(0,0)$.
• There is a vertical asymptote at $x = 1$, with the curve falling to $-\infty$ as $x \to 1^-$.
• For $x > 1$, the curve rises from $-\infty$, crossing the $x$-axis at $(2,0)$.
• There is a local maximum turning point at $(3,4)$.
• The curve approaches the horizontal asymptote $y = 2$ from above as $x \to \infty$.

2. Determine the shape of the graph for $x < 0$ using the graph of $y = \mathrm{f}'(x)$.

For $x < 0$, we refer to the given derivative graph $y = \mathrm{f}'(x)$.
The graph of $y = \mathrm{f}'(x)$ is entirely below the $x$-axis for $x < 1$, which means $\mathrm{f}'(x) < 0$. Thus, $y = \mathrm{f}(x)$ must be strictly decreasing for all $x < 1$.
Furthermore, $\mathrm{f}'(x) \to 0$ as $x \to -\infty$, indicating that $y = \mathrm{f}(x)$ approaches a horizontal asymptote. Matching the horizontal asymptote $y = 2$ on the right side, $\mathrm{f}(x) \to 2$ as $x \to -\infty$.
Therefore, for $x < 0$, the curve of $y = \mathrm{f}(x)$ is a smooth, monotonically decreasing curve that approaches $y = 2$ from below as $x \to -\infty$ and passes through $(0,0)$. The asymptote $x = -1$ seen in the $\mathrm{f}(|x|)$ graph is omitted as it was purely a reflection of $x = 1$ due to the absolute value function.

1. Determine the vertical asymptote(s) and axial intercept(s) of the reciprocal graph.

• Vertical Asymptotes: These occur where the original function $\mathrm{f}'(x) = 0$. From the given graph, $\mathrm{f}'(x)$ crosses the $x$-axis at $(3, 0)$. Thus, $x = 3$ is a vertical asymptote.
• $x$-intercepts: These occur where $\mathrm{f}'(x) \to \pm\infty$. The given graph has a vertical asymptote at $x = 1$. Thus, $\frac{1}{\mathrm{f}'(x)} = 0$ at $x = 1$, giving an $x$-intercept at $(1, 0)$.
• $y$-intercept: Occurs at $x=0$. Since $\mathrm{f}'(0)$ is a finite negative value, the $y$-intercept is negative (exact value unknown).

2. Determine the behavior and shape of the graph for $x < 3$.

• For $x < 1$: $\mathrm{f}'(x) < 0$. As $x \to -\infty$, $\mathrm{f}'(x) \to 0^-$, meaning $y \to -\infty$. The curve rises from the bottom left, crosses the negative $y$-axis, and smoothly passes through $(1, 0)$.
• For $1 < x < 3$: $\mathrm{f}'(x) > 0$. As $x \to 3^-$, $\mathrm{f}'(x) \to 0^+$, meaning $y \to \infty$. The curve continues upward from $(1, 0)$ to the vertical asymptote $x = 3$.

3. Determine the behavior, shape, and turning points for $x > 3$.

• For $x > 3$: $\mathrm{f}'(x) < 0$. As $x \to 3^+$, $\mathrm{f}'(x) \to 0^-$, so $y \to -\infty$. As $x \to \infty$, $\mathrm{f}'(x) \to 0^-$, meaning $y \to -\infty$.
• The graph of $\mathrm{f}'(x)$ has a local minimum in the fourth quadrant. Correspondingly, its reciprocal $y = \frac{1}{\mathrm{f}'(x)}$ will have a local maximum. Since the values are strictly negative in this region, the turning point (local maximum) lies below the $x$-axis. Coordinates are marked as a turning point but remain unlabeled as exact values are not provided.

1. Identify the geometrical transformations connecting $y = \mathrm{f}(|x|)$ to $y = -\mathrm{f}(|x-1|)$.

Let $g(x) = \mathrm{f}(|x|)$. The target function is $y = -\mathrm{f}(|x-1|) = -g(x-1)$.
This represents two transformations applied to the given graph of $y = \mathrm{f}(|x|)$:

1. A translation of $1$ unit in the positive $x$-direction ($x \to x - 1$).
2. A reflection in the $x$-axis ($y \to -y$).

2. Apply transformations to the asymptotes and intercepts.

Asymptotes:

• The original vertical asymptotes are at $x = -1$ and $x = 1$.
  Translating them by $+1$ gives $x = 0$ and $x = 2$. Reflection in the $x$-axis does not affect vertical lines.
• The original horizontal asymptote is at $y = 2$.
  Translation does not affect it. Reflection in the $x$-axis moves it to $y = -2$.

Axial Intercepts:

• The original $x$-intercepts are at $(-2, 0)$, $(0, 0)$, and $(2, 0)$.
  Translating by $+1$ gives $x$-intercepts at $(-1, 0)$, $(1, 0)$, and $(3, 0)$. Reflection in the $x$-axis leaves points on the $x$-axis unchanged.
• There is no $y$-intercept because $x = 0$ is a vertical asymptote.

3. Apply transformations to the turning points and describe the final shape.

Turning Points:

• The original local maximums are at $(-3, 4)$, $(0, 0)$, and $(3, 4)$.
• Applying the translation ($+1$ to $x$-coordinates) gives $(-2, 4)$, $(1, 0)$, and $(4, 4)$.
• Applying the reflection (negating $y$-coordinates) turns them into local minimums at $(-2, -4)$, $(1, 0)$, and $(4, -4)$.

The final graph consists of three branches separated by the vertical asymptotes, opening upwards towards $+\infty$ near the asymptotes, and approaching $y = -2$ from below as $x \to \pm\infty$.