The diagram below shows the graph of $y = \text{f}(x)$. The curve passes through the $x$-axis at $(-2, 0)$ and $(1, 0)$, and has a maximum point with coordinates $(2, 2)$. The lines $x = -1$ and $y = 1$ are asymptotes to the graph. Stating the equations of any asymptotes and the coordinates of any points of intersection with the axes and stationary points, where possible, sketch the graphs of

1. Identify the geometrical transformations required

The equation $y = 3\text{f}(x + a)$ indicates two transformations on the graph of $y = \text{f}(x)$:

1. A translation by $a$ units in the negative $x$-direction (shift left by $a$).
2. A vertical stretch by a scale factor of $3$ parallel to the $y$-axis.

2. Find the equations of the new asymptotes

Original vertical asymptote: $x = -1$
Translated left by $a$:
$x = -1 - a$

Original horizontal asymptote: $y = 1$
Stretched vertically by 3:
$y = 1 \times 3 \implies y = 3$

3. Find the coordinates of the new stationary point

Original maximum point: $(2, 2)$
Translated left by $a$ and stretched vertically by 3:
$(x, y) \to (2 - a, 2 \times 3) \implies (2 - a, 6)$
Since $1 < a < 2$, we have $0 < 2 - a < 1$, so the new maximum point lies in the first quadrant.

4. Determine the new intersections with the axes

Original $x$-intercepts: $(-2, 0)$ and $(1, 0)$
Translated left by $a$ (vertical stretch does not affect $x$-intercepts):
$(-2 - a, 0) \quad \text{and} \quad (1 - a, 0)$
Since $1 < a < 2$, we have $-1 < 1 - a < 0$, meaning both $x$-intercepts lie on the negative $x$-axis.

The new $y$-intercept occurs at $x = 0$, giving $y = 3\text{f}(a)$. From the original graph, for $1 < x < 2$, $\text{f}(x) > 0$. Hence $\text{f}(a) > 0$, and the curve intersects the positive $y$-axis (the exact coordinate cannot be determined).

5. Sketch the final transformed graph

Sketch the two branches based on the transformed features:

• Left branch: Bounded by $x = -1 - a$ and $y = 3$, passing through $(-2-a, 0)$.
• Right branch: Starts near the bottom of $x = -1 - a$, crosses the $x$-axis at $(1-a, 0)$, crosses the positive $y$-axis, peaks at the maximum $(2-a, 6)$, and approaches $y = 3$ from above as $x \to \infty$.

1. Determine the equations of the asymptotes for $y = \text{f}'(x)$.

Since the graph of $y = \text{f}(x)$ has a vertical asymptote at $x = -1$, the gradient is undefined and its magnitude approaches infinity. Thus, $x = -1$ is a vertical asymptote for $y = \text{f}'(x)$.
As $x \to \pm \infty$, the curve approaches the horizontal asymptote $y = 1$, meaning the curve flattens out and the gradient approaches $0$. Thus, $y = 0$ (the $x$-axis) is a horizontal asymptote for $y = \text{f}'(x)$.

2. Determine the coordinates of intersection with the axes.

At the maximum point $(2, 2)$ on $y = \text{f}(x)$, the gradient is $0$. Therefore, $\text{f}'(2) = 0$, giving an $x$-intercept at $(2, 0)$.
The exact $y$-intercept cannot be calculated from the given information, but it must be positive since $\text{f}(x)$ is strictly increasing at $x = 0$.

3. Analyse the sign of the gradient in each interval to deduce the shape of the derivative graph.

For $x < -1$, $\text{f}(x)$ is strictly decreasing, so $\text{f}'(x) < 0$. The curve comes from $y = 0$ as $x \to -\infty$ and plunges to $-\infty$ as $x \to -1^-$.
For $-1 < x < 2$, $\text{f}(x)$ is strictly increasing, so $\text{f}'(x) > 0$. The curve falls from $\infty$ as $x \to -1^+$, and decreases to cross the $x$-axis at $(2, 0)$.
For $x > 2$, $\text{f}(x)$ is strictly decreasing, so $\text{f}'(x) < 0$. The curve dips below the $x$-axis. Since it must approach the horizontal asymptote $y = 0$ as $x \to \infty$, it must reach a local minimum point before turning upwards. (The exact coordinates of this minimum cannot be determined).