The diagram shows the graphs of $y = \mathrm{f}(|x|)$ and $y = \mathrm{f}'(x)$. The graph of $y = \mathrm{f}(|x|)$ has turning points at $(-3, 4)$ and $(3, 4)$, crosses the $x$-axis at $(-2, 0)$ and $(2, 0)$, and the equations of asymptotes are $x = -1$, $x = 1$ and $y = 2$. The graph of $y = \mathrm{f}'(x)$ crosses the $x$-axis at $(3, 0)$, and the equations of asymptotes are $x = 1$ and $y = 0$. On separate diagrams, sketch the graphs of labelling clearly the equation(s) of any asymptote(s), coordinates of any axial intercept(s) and turning point(s) where applicable.

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 $x \to \pm\infty$.