[在此问题中，所给图像的草图未按比例绘制] $y = |\mathrm{f}(x)|$ 和 $y^2 = \mathrm{f}(x)$ 的图像如下所示。 在不同的图上，分别画出下列函数的图像草图， 写出所有渐近线方程以及与坐标轴交点的坐标。

1. Identify the transformation from $y^2 = \mathrm{f}(x)$ to $y^2 = \mathrm{f}(-2x)$.

The equation $y^2 = \mathrm{f}(-2x)$ replaces $x$ with $-2x$. This corresponds to a horizontal scaling by a factor of $\frac{1}{2}$ and a reflection in the $y$-axis.

2. Determine the new asymptote and $x$-intercept.

Applying the transformation to the key features of $y^2 = \mathrm{f}(x)$:

• Vertical asymptote: $x = 3 \implies -2x = 3 \implies x = -\frac{3}{2}$
• $x$-intercept (vertex): $x = 5 \implies -2x = 5 \implies x = -\frac{5}{2}$, giving the coordinate $(-\frac{5}{2}, 0)$

3. Determine the shape and domain of the new graph.

The original graph exists for $3 < x \le 5$ and opens to the left from $(5, 0)$.
For the transformed graph, the domain is $3 < -2x \le 5 \implies -\frac{5}{2} \le x < -\frac{3}{2}$.
The new graph consists of two branches starting at $(-\frac{5}{2}, 0)$ and opening to the right towards the vertical asymptote $x = -\frac{3}{2}$, maintaining symmetry about the $x$-axis.

1. Determine the intervals where $\mathrm{f}(x)$ is positive or negative from the $y^2 = \mathrm{f}(x)$ graph.

The relation $y^2 = \mathrm{f}(x)$ only yields real values of $y$ when $\mathrm{f}(x) \ge 0$.
Given that the graph of $y^2 = \mathrm{f}(x)$ consists of branches starting at $(5,0)$ and opening to the left towards the asymptote $x=3$, this curve only exists for $3 < x \le 5$.
Furthermore, it is explicitly defined "only for $x > 3$", ruling out any regions $x < 3$.
Therefore, we deduce:

• $\mathrm{f}(x) \ge 0$ for $3 < x \le 5$
• $\mathrm{f}(x) \le 0$ for $x < 3$ and $x > 5$

2. Relate $y = \mathrm{f}(x)$ to the provided graph of $y = |\mathrm{f}(x)|$.

Using the signs of $\mathrm{f}(x)$, we construct $y = \mathrm{f}(x)$ from $y = |\mathrm{f}(x)|$:

• For $3 \le x \le 5$: Since $\mathrm{f}(x) \ge 0$, $\mathrm{f}(x) = |\mathrm{f}(x)|$. The sketch of $y = \mathrm{f}(x)$ in this region is identical to the graph of $y = |\mathrm{f}(x)|$.
• For $x < 3$ and $x > 5$: Since $\mathrm{f}(x) \le 0$, $\mathrm{f}(x) = -|\mathrm{f}(x)|$. The sketch is obtained by reflecting the corresponding parts of the graph of $y = |\mathrm{f}(x)|$ in the $x$-axis.

Note that the sharp cusp on $y = |\mathrm{f}(x)|$ at $x = 5$ unfolds into a smooth curve crossing the $x$-axis in $y = \mathrm{f}(x)$, which matches $\mathrm{f}(x)$ changing sign from positive to negative.

3. Determine the exact asymptotes and intercepts.

Applying the reflections in the $x$-axis for $x < 3$ and $x > 5$ to the key features of the $|\mathrm{f}(x)|$ graph:

• The horizontal asymptote $y = 1$ reflects to become $y = -1$.
• The $y$-intercept $(0, 2)$ reflects to become $(0, -2)$.
• The known $x$-intercept $(5, 0)$ remains $(5, 0)$.
• The smooth local minimum touching the $x$-axis at the negative root remains an $x$-intercept, becoming a local maximum.
• The vertical asymptotes $x = 1$ and $x = 3$ are unchanged.