The diagram shows the curve $y = f(x)$, with turning points at $A(3, 2.5)$ and $B(0, 1)$. The lines $y = 1$ and $x = -2$ are asymptotes of the curve.

1. Identify the geometrical transformations applied to the original graph.

The equation $y = f(x-2) + 2$ represents a translation of the graph of $y = f(x)$ by $2$ units in the positive $x$-direction (to the right) and $2$ units in the positive $y$-direction (upwards).

2. Find the coordinates of the new turning points.

Point $A(3, 2.5) \rightarrow A'(3+2, 2.5+2) = (5, 4.5)$
Point $B(0, 1) \rightarrow B'(0+2, 1+2) = (2, 3)$

3. Find the equations of the new asymptotes.

Vertical asymptote: $x = -2 \rightarrow x = -2 + 2 \implies x = 0$ (the $y$-axis)
Horizontal asymptote: $y = 1 \rightarrow y = 1 + 2 \implies y = 3$

1. Determine the asymptotes of y = f'(x)

The vertical asymptote of $f$ is unchanged, so $x=-2$ is still a vertical asymptote of $f'(x)$; near it $f$ increases steeply to $+\infty$, so $f'(x)\to-\infty$ as $x\to-2^+$.

Since $y=1$ is a horizontal asymptote of $f$, the curve flattens as $x\to+\infty$, so $f'(x)\to0$. Hence the horizontal asymptote of $f'(x)$ is $y=0$ (the $x$-axis).

2. Convert turning points of f into x-intercepts of f'

At a turning point the gradient is zero, so $f'(x)=0$ at the $x$-coordinates of $A$ and $B$.

• $B(0,1)\Rightarrow$ $x$-intercept $(0,0)$.
• $A(3,2.5)\Rightarrow$ $x$-intercept $(3,0)$.

3. Use sign of the gradient on each interval to sketch f'

Read off the gradient of $f$ on each interval (B is a minimum, A is a maximum):

• $-2<x<0$: $f$ decreasing $\Rightarrow f'(x)<0$ (below axis, from $-\infty$ up to $0$ at $x=0$).
• $0<x<3$: $f$ increasing $\Rightarrow f'(x)>0$ (above axis, a positive hump).
• $x>3$: $f$ decreasing towards $y=1$ $\Rightarrow f'(x)<0$, approaching $0^-$.

Sketch a curve rising from $-\infty$ near $x=-2$, crossing the axis at $(0,0)$, peaking, crossing back at $(3,0)$, then staying just below the $x$-axis and tending to $y=0$.

1. Set $y = 0$ to find the value of the parameter $t$ at the $x$-axis intercept.

$y = t^3 - 3t = 0$
$t(t^2 - 3) = 0$
Since the domain restricts $t > 0$, we have $t^2 = 3 \implies t = \sqrt{3}$.

2. Find the original $x$-coordinate, apply the reflection in the $y$-axis, and state the final exact coordinates.

Original $x$-coordinate at $t = \sqrt{3}$:
$x = (\sqrt{3})^2 - 2(\sqrt{3}) = 3 - 2\sqrt{3}$
When a curve is reflected in the $y$-axis, its $y$-coordinates remain unchanged and its $x$-coordinates are negated.
Reflected $x$-coordinate:
$-x = -(3 - 2\sqrt{3}) = 2\sqrt{3} - 3$
The exact coordinates of the point are $(2\sqrt{3} - 3, 0)$.