The diagram below shows the graph of y = 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 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).