Section A: Pure Mathematics [60 marks] The curve $C$ has equation $y = \frac{x^2 - 2x + 4}{x - b}$, $x \ne b$.

1. Rearrange the given equation into a quadratic equation in terms of $x$.

$y = \frac{x^2 - 2x + 4}{x - b}$
$y(x - b) = x^2 - 2x + 4$
$xy - by = x^2 - 2x + 4$
$x^2 - (y+2)x + (4+by) = 0$

2. Apply the discriminant condition for real roots of $x$.

For $x$ to have real solutions, the discriminant of the quadratic equation must be non-negative ($\Delta \ge 0$):
$(-(y+2))^2 - 4(1)(4+by) \ge 0$
$y^2 + 4y + 4 - 16 - 4by \ge 0$
$y^2 + (4 - 4b)y - 12 \ge 0$

3. Form a quadratic inequality using the given excluded range of $y$ and compare coefficients to find $b$.

Since $y$ cannot take values between $-2$ and $6$, the possible values of $y$ satisfy $y \le -2$ or $y \ge 6$.
This corresponds to the root boundaries $y=-2$ and $y=6$, giving the inequality:
$(y - 6)(y + 2) \ge 0$
$y^2 - 4y - 12 \ge 0$

Comparing the $y$ coefficient of this inequality with the one derived from the discriminant:
$4 - 4b = -4$
$4b = 8$
$b = 2$ (shown)

1. Express both rational functions in their quotient-remainder forms to easily compare them.

Using $b = 2$ from part (i), the equation of $C$ is $y = \frac{x^2 - 2x + 4}{x - 2}$.
By performing algebraic division, we express $C$ as:
$y = \frac{x(x - 2) + 4}{x - 2} = x + \frac{4}{x - 2}$

Similarly, we rewrite the target graph equation $y = \frac{x^2 + 5x + 4}{x}$:
$y = \frac{x^2}{x} + \frac{5x}{x} + \frac{4}{x} = x + 5 + \frac{4}{x}$

2. Relate the two forms using function transformations.

Let $\mathrm{f}(x) = x + \frac{4}{x - 2}$.
To change the fraction's denominator from $x - 2$ to $x$, we replace $x$ with $x + 2$:
$\mathrm{f}(x + 2) = (x + 2) + \frac{4}{(x + 2) - 2} = x + 2 + \frac{4}{x}$

To obtain the target graph $x + 5 + \frac{4}{x}$, we must add $3$ to $\mathrm{f}(x + 2)$:
$\mathrm{f}(x + 2) + 3 = x + 2 + \frac{4}{x} + 3 = x + 5 + \frac{4}{x}$
Hence, the target graph is represented by $y = \mathrm{f}(x + 2) + 3$.

3. State the sequence of transformations explicitly.

The transformation from $y = \mathrm{f}(x)$ to $y = \mathrm{f}(x + 2) + 3$ corresponds to:

1. A translation of $2$ units in the negative $x$-direction (from $x \to x + 2$).
2. A translation of $3$ units in the positive $y$-direction (from adding $3$).

(Note: As these translations are in perpendicular directions, they can be applied in any order, and can also be written as a single translation vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$, though the question explicitly asks for a sequence of 2 transformations).