The function $f$ is defined as $f : x \mapsto 6x - x^{2}$. For the rest of this question, the domain of f is assumed to be $[3,\infty)$.

1. Find the maximum point and axis of symmetry of the quadratic function by completing the square.

$f(x) = 6x - x^2 = -(x^2 - 6x) = -[(x - 3)^2 - 9] = 9 - (x - 3)^2$
The turning point is $(3, 9)$, which is a maximum point.

2. Sketch the graph of $y = f(x)$ for the restricted domain $[3, \infty)$.

When $f(x) = 0$, $x(6-x) = 0 \implies x=0$ or $x=6$.
Since the domain is $[3, \infty)$, the graph starts at the vertex $(3, 9)$ and passes through the $x$-axis at $(6, 0)$.
(Student sketches a decreasing half-parabola starting at a solid dot at $(3, 9)$, passing through $(6, 0)$, and continuing downwards)

3. State the mathematical condition for the inverse function to exist.

For the inverse function $f^{-1}$ to exist, $f(x)$ must be a one-to-one function. Graphically, this means it must pass the horizontal line test.

4. Explain why the least value of $a$ is 3.

From the graph, the line of symmetry is $x = 3$. If the domain was $[a, \infty)$ where $a < 3$, the graph would contain the turning point $(3, 9)$ and points on both sides of the line of symmetry.
A horizontal line (e.g., $y = 8$) would intersect the graph more than once, meaning the function would fail the horizontal line test and no longer be one-to-one.
Hence, to keep the function strictly decreasing and one-to-one, the domain must start at or after the turning point, making the least value of $a$ exactly 3.

1. Set $y = f(x)$ and complete the square to isolate the terms involving $x$.

Let $y = 6x - x^2$.
To complete the square:
$y = -(x^2 - 6x)$
$y = -((x - 3)^2 - 9)$
$y = 9 - (x - 3)^2$

2. Rearrange the equation to make $x$ the subject.

$(x - 3)^2 = 9 - y$
$x - 3 = \pm \sqrt{9 - y}$

3. Determine the correct sign for the square root based on the domain of $f$.

Since the given domain of $f$ is $[3, \infty)$, we have $x \ge 3$, which means $x - 3 \ge 0$.
Therefore, we must reject the negative root:
$x - 3 = \sqrt{9 - y}$
$x = 3 + \sqrt{9 - y}$

4. Write the inverse function in the requested similar mapping form.

Replacing $y$ with $x$ and expressing the result in mapping notation:
$f^{-1} : x \mapsto 3 + \sqrt{9 - x}$

(Note: The domain of $f^{-1}$ corresponds to the range of $f$. Since $f(x) = 9 - (x - 3)^2$ for $x \ge 3$, the range of $f$ is $(-\infty, 9]$. The full formal answer can include this domain as $x \le 9$.)

1. State the relationship between the graphs of f and its inverse

The graph of $y=f^{-1}(x)$ is the reflection of the graph of $y=f(x)$ in the line $y=x$. So sketch $y=x$ (a dashed straight line through the origin) as the axis of reflection.

2. Reflect the key points of f to plot f inverse

From part (i)/(ii): $f$ has endpoint (maximum) at $(3,9)$ with range $(-\infty,9]$. Reflecting in $y=x$ swaps coordinates, so $y=f^{-1}(x)=3+\sqrt{9-x}$ has its endpoint at $(9,3)$, with domain $(-\infty,9]$ and range $[3,\infty)$. Draw $f^{-1}$ as the mirror image of $f$: a curve starting at $(9,3)$ and rising to the left.

3. Show the intersection on the line y=x

The two curves meet on the line $y=x$, where $f(x)=x$: $6x-x^2=x\Rightarrow x^2-5x=0\Rightarrow x=0$ or $x=5$. Since $x\ge 3$, the curves meet at $(5,5)$. Mark this point clearly to emphasise the reflective symmetry.