下图显示了 y = f(x) 的图像。曲线在 (-2, 0) 和 (1, 0) 处穿过 x 轴，并且具有坐标为 (2, 2) 的极大值点。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，画出以下函数的图像：

1. Identify the geometrical transformations represented by the equation.

The equation is $y = 3\mathrm{f}(x + a)$. Compared to the original graph of $y = \mathrm{f}(x)$, this represents:

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

2. Determine the new equations of the asymptotes.

Applying the transformations to the original asymptotes:

• The vertical asymptote $x = -1$ is translated $a$ units left, becoming $x = -1 - a$.
• The horizontal asymptote $y = 1$ is unaffected by the horizontal translation but is stretched vertically by a factor of 3, becoming $y = 3(1) = 3$.

3. Find the coordinates of the transformed $x$-intercepts and stationary point, and verify their graphical positions.

Applying the transformations to the known points:

• For the $x$-intercepts $(-2, 0)$ and $(1, 0)$, the $y$-coordinates remain $0$. The $x$-coordinates are translated to give new intercepts at $(-2 - a, 0)$ and $(1 - a, 0)$.
• For the maximum point $(2, 2)$, the $x$-coordinate is shifted by $a$ and the $y$-coordinate is multiplied by $3$. The new maximum is at $(2 - a, 6)$.

Given the condition $1 < a < 2$:

• $1 - a$ is between $-1$ and $0$, so the intercept $(1 - a, 0)$ must be drawn on the negative $x$-axis.
• $2 - a$ is between $0$ and $1$, so the maximum point $(2 - a, 6)$ must be drawn in the first quadrant (to the right of the $y$-axis).
• Since the maximum is to the right of the $y$-axis and the $x$-intercept is to the left, the curve's $y$-intercept will be positive.

1. Determine the coordinates of the intersection with the $x$-axis.

The graph of $y = f(x)$ has a maximum point at $(2, 2)$, which means the gradient at $x = 2$ is zero.
Therefore, $f'(2) = 0$, giving an $x$-intercept at $(2, 0)$ for the graph of $y = f'(x)$.

2. Determine the equations of the asymptotes.

As $x \to \pm\infty$, the graph of $f(x)$ approaches the horizontal asymptote $y = 1$, meaning the gradient approaches $0$. Thus, $y = 0$ (the $x$-axis) is a horizontal asymptote for $f'(x)$.
As $x \to -1$, $f(x)$ approaches the vertical asymptote $x = -1$. Because the curve falls to $-\infty$ on both sides of $x = -1$, the gradient $f'(x) \to -\infty$ as $x \to -1^-$ and $f'(x) \to +\infty$ as $x \to -1^+$. Thus, $x = -1$ is a vertical asymptote for $f'(x)$.

3. Determine the signs of $f'(x)$ and the presence of stationary points to deduce the final shape.

By observing the increasing/decreasing nature of $f(x)$:

• For $x < -1$, $f(x)$ is decreasing $\implies f'(x) < 0$ (curve below $x$-axis).
• For $-1 < x < 2$, $f(x)$ is increasing $\implies f'(x) > 0$ (curve above $x$-axis).
• For $x > 2$, $f(x)$ is decreasing $\implies f'(x) < 0$ (curve below $x$-axis).

Since $f'(2) = 0$ and $f'(x) \to 0$ as $x \to \infty$, the continuous graph of $f'(x)$ must reach a local minimum at some point where $x > 2$ before approaching the $x$-axis from below. (Exact coordinates of this stationary point and the $y$-intercept cannot be deduced).