The diagram below shows the graph of $y = \mathrm{f}(x)$. The curve cuts the axes at $A(0, 1.5)$ and $B(-3, 0)$. The asymptotes of the curve are $x = -2$ and $y = 1$. Sketch, on separate diagrams, the graphs of indicating clearly the asymptotes, axial intercepts and the points corresponding to $A$ and $B$ where possible.

1. Identify the geometric transformations that map $y = \mathrm{f}(x)$ to $y = -\mathrm{f}(1+x)$.

The transformation from $\mathrm{f}(x)$ to $-\mathrm{f}(x+1)$ consists of two steps:

1. A horizontal translation of $1$ unit to the left (mapping $\mathrm{f}(x)$ to $\mathrm{f}(x+1)$).
2. A reflection about the $x$-axis (mapping $\mathrm{f}(x+1)$ to $-\mathrm{f}(x+1)$).

2. Find the new equations of the asymptotes.

Applying the transformations to the original asymptotes:

• Vertical asymptote $x = -2$: translating left by $1$ gives $x = -3$. The reflection in the $x$-axis does not affect vertical lines.
  New vertical asymptote: $x = -3$.
• Horizontal asymptote $y = 1$: translation left by $1$ does not affect horizontal lines. Reflection in the $x$-axis maps $y = 1$ to $y = -1$.
  New horizontal asymptote: $y = -1$.

3. Determine the new coordinates of the specific points $A$ and $B$.

Applying the transformations to the given points:

• For point $A(0, 1.5)$:
  Translate left by $1$: $(0 - 1, 1.5) = (-1, 1.5)$
  Reflect in $x$-axis: $(-1, -1.5)$
  So, $A$ maps to $A'(-1, -1.5)$.

• For point $B(-3, 0)$:
  Translate left by $1$: $(-3 - 1, 0) = (-4, 0)$
  Reflect in $x$-axis: $(-4, 0)$ since it lies on the $x$-axis.
  So, $B$ maps to the new $x$-intercept $B'(-4, 0)$.

1. Determine the new asymptotes for the reciprocal graph.

For $y = \frac{1}{\mathrm{f}(x)}$:

• As $x \to \pm\infty$, $\mathrm{f}(x) \to 1$. Therefore, $y \to \frac{1}{1} = 1$. The horizontal asymptote remains $y = 1$.
• The $x$-intercept of $y = \mathrm{f}(x)$ is at $x = -3$ (point $B$). As $\mathrm{f}(x) \to 0$, $\frac{1}{\mathrm{f}(x)} \to \pm\infty$. Thus, $x = -3$ is the new vertical asymptote.

2. Determine the new axial intercepts and map points $A$ and $B$.

• 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 creates a new $x$-intercept at $(-2, 0)$.
• The $y$-intercept is at $A(0, 1.5)$. Thus, the new $y$-intercept is at $\left(0, \frac{1}{1.5}\right) = \left(0, \frac{2}{3}\right)$. This is the point corresponding to $A$.
• Point $B(-3, 0)$ corresponds to the newly formed vertical asymptote $x = -3$.

3. Determine the shape of the curve to complete the sketch.

• For $x < -3$: $\mathrm{f}(x)$ is positive and $0 < \mathrm{f}(x) < 1$, so $\frac{1}{\mathrm{f}(x)} > 1$. The graph approaches $y=1$ as $x \to -\infty$ and goes to $+\infty$ as $x \to -3^-$.
• For $x > -3$: As $x \to -3^+$, $\mathrm{f}(x) \to 0^-$, so $\frac{1}{\mathrm{f}(x)} \to -\infty$. The curve rises smoothly, crosses the $x$-axis at $(-2, 0)$, crosses the $y$-axis at $(0, \frac{2}{3})$, and approaches $y=1$ from below as $x \to \infty$.

1. Determine the asymptotes of $y = \mathrm{f}'(x)$ based on the behaviour of $y = \mathrm{f}(x)$.

Observe the limits of the gradient of $\mathrm{f}(x)$:

• As $x \to \pm \infty$, the graph of $y = \mathrm{f}(x)$ flattens out and approaches the horizontal asymptote $y = 1$. This means the gradient $\mathrm{f}'(x) \to 0$. Therefore, $y = 0$ (the $x$-axis) is a horizontal asymptote for $y = \mathrm{f}'(x)$.
• As $x \to -2$, the tangent to the curve becomes increasingly steep towards the vertical, meaning $\mathrm{f}'(x) \to -\infty$. Therefore, $x = -2$ is a vertical asymptote for $y = \mathrm{f}'(x)$.

2. Determine the shape, sign, and key points of the derivative graph.

• The original curve $y = \mathrm{f}(x)$ is strictly decreasing on both intervals $(-\infty, -2)$ and $(-2, \infty)$. Consequently, $\mathrm{f}'(x) < 0$ for all valid $x$, meaning the entire graph of $y = \mathrm{f}'(x)$ must lie below the $x$-axis.
• Since $\mathrm{f}'(x) \neq 0$, there are no $x$-intercepts.
• The point corresponding to $A(0, 1.5)$ is at $x = 0$, giving a negative $y$-intercept on the derivative graph.
• The point corresponding to $B(-3, 0)$ is at $x = -3$, which will sit at some negative $y$-value on the left branch.
• Since the exact functional equation is not given, we simply mark the $x$-coordinates $x = 0$ (as the $y$-intercept) and $x = -3$ on the graph to correspond to $A$ and $B$.