[In this question, sketches of the given graphs are not drawn to scale] The graphs of $y = |\mathrm{f}(x)|$ and $y^2 = \mathrm{f}(x)$ are given below. On separate diagrams, draw sketches of the graphs of stating the equations of any asymptotes and the coordinates of any points of intersection with the axes.

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.