The graph of $y = \text{f}(x)$ is given above. Sketch, on separate diagrams, the graphs of Your sketches should show clearly the coordinates of any stationary points, axial intercepts and the equations of any asymptotes.

1. Establish the coordinate mapping for the transformation

The transformation from $y = \text{f}(x)$ to $y = \text{f}(2x + 3)$ maps each point $(x, y)$ on the original graph to a new point $(x', y)$.
The relationship between the $x$-coordinates is given by:
$2x' + 3 = x \implies x' = \frac{x - 3}{2}$

2. Determine the new equations of the asymptotes

• Vertical asymptote $x = -1 \implies x' = \frac{-1 - 3}{2} = -2$
• Vertical asymptote $x = 3 \implies x' = \frac{3 - 3}{2} = 0$
• The horizontal asymptote is unaffected by horizontal transformations, so it remains $y = 2$.

3. Calculate the new coordinates of the stationary point and intercepts

• The $x$-intercept at $(0, 0) \implies x' = \frac{0 - 3}{2} = -1.5$. The new intercept is $(-\frac{3}{2}, 0)$.
• The $x$-intercept at $(2, 0) \implies x' = \frac{2 - 3}{2} = -0.5$. The new intercept is $(-\frac{1}{2}, 0)$.
• The local maximum at $(1, \frac{2}{3}) \implies x' = \frac{1 - 3}{2} = -1$. The new stationary point is $(-1, \frac{2}{3})$.
  Note: Since $x = 0$ is now a vertical asymptote, there is no $y$-intercept.

1. Deduce the equations of the asymptotes for the derivative graph.

As $x \to \pm\infty$, the graph of $y = \text{f}(x)$ approaches the horizontal asymptote $y = 2$, meaning the gradient flattens out and approaches $0$. Hence, the horizontal asymptote for $y = \text{f}'(x)$ is $y = 0$.
As $x \to -1$ and $x \to 3$, the graph of $y = \text{f}(x)$ approaches vertical asymptotes, so the curve becomes infinitely steep. Hence, the vertical asymptotes remain $x = -1$ and $x = 3$.

2. Identify the x-intercept of the derivative graph from the stationary point of f(x).

The graph of $y = \text{f}(x)$ has a local maximum (stationary point) at $(1, \frac{2}{3})$. At this maximum turning point, the gradient of the tangent is $0$.
Hence, $y = \text{f}'(x)$ has an $x$-intercept at $(1, 0)$.

3. Determine the sign and behaviour of the derivative graph in each region to complete the sketch.

Analyse the gradient of $y = \text{f}(x)$ across the three intervals:

• For $x < -1$: $\text{f}(x)$ is increasing, so $\text{f}'(x) > 0$. The gradient increases from $0$ (at $-\infty$) to $+\infty$ (near $x = -1$).
• For $-1 < x < 1$: $\text{f}(x)$ is increasing from $-\infty$, so $\text{f}'(x) > 0$. The gradient decreases from steepest at $+\infty$ (near $x = -1$) to $0$ (at $x = 1$).
• For $1 < x < 3$: $\text{f}(x)$ is decreasing towards $-\infty$, so $\text{f}'(x) < 0$. The gradient becomes increasingly negative, dropping from $0$ to $-\infty$ (near $x = 3$).
• For $x > 3$: $\text{f}(x)$ is decreasing from $+\infty$, so $\text{f}'(x) < 0$. The gradient starts very steep at $-\infty$ (near $x = 3$) and becomes less steep, approaching $0$ (at $\infty$).

The concavity of $y = \text{f}(x)$ does not change sign within any of the three branches, meaning there are no stationary points for $y = \text{f}'(x)$. Sketch the three branches satisfying these gradient changes.

1. Find the vertical asymptotes and $x$-intercepts of the reciprocal graph

The $x$-intercepts of $y = \text{f}(x)$ occur at $x = 0$ and $x = 2$. These become vertical asymptotes for $y = \frac{1}{\text{f}(x)}$, yielding the lines $x = 0$ and $x = 2$.

The vertical asymptotes of $y = \text{f}(x)$ are $x = -1$ and $x = 3$. As $x \to -1$ and $x \to 3$, $|\text{f}(x)| \to \infty$, which means $\frac{1}{\text{f}(x)} \to 0$. Thus, $y = \frac{1}{\text{f}(x)}$ has $x$-intercepts at $(-1, 0)$ and $(3, 0)$.

2. Find the new horizontal asymptote and stationary point

The horizontal asymptote of $y = \text{f}(x)$ is $y = 2$. Therefore, the horizontal asymptote of the reciprocal graph is $y = \frac{1}{2}$.

The local maximum point of $y = \text{f}(x)$ is $(1, \frac{2}{3})$. Since taking the reciprocal of positive values inverses the nature of the stationary point, $y = \frac{1}{\text{f}(x)}$ will have a local minimum at $(1, \frac{3}{2})$.

3. Determine the limits and asymptotic behaviour to shape the branches

For $x < -1$ and $x > 3$, $y = \text{f}(x)$ is positive and greater than $2$ (approaching $2$ as $x \to \pm\infty$). Thus, $\frac{1}{\text{f}(x)}$ is positive, less than $\frac{1}{2}$, and approaches the horizontal asymptote $y = \frac{1}{2}$ from below.

Between the vertical asymptotes $x = -1$ and $x = 3$, $\text{f}(x)$ changes sign as it passes through $x = 0$ and $x = 2$. Consequently:

• As $x \to -1^-$, $\text{f}(x) \to \infty \implies y \to 0^+$
• As $x \to -1^+$, $\text{f}(x) \to -\infty \implies y \to 0^-$
  This shows the curve cuts through $(-1, 0)$ smoothly. A similar crossing from negative to positive occurs at $(3,0)$.