下图显示了 $y = \text{f}(x)$ 的图像。该曲线穿过 $x$ 轴的点为 $(-2, 0)$ 和 $(1, 0)$，并且有一个最大值点，其坐标为 $(2, 2)$。直线 $x = -1$ 和 $y = 1$ 是该图像的渐近线。 在可能的情况下，指出所有渐近线的方程、与坐标轴的交点坐标以及驻点坐标，并画出以下函数的图像：

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).