The curve $C$ has equation $y = \frac{2x^2 - 10x + 17}{x - 3}, \quad x \in \mathbb{R}, \quad x \ne 3$.

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

$y = \frac{2x^2 - 10x + 17}{x - 3}$

$y(x - 3) = 2x^2 - 10x + 17$

$xy - 3y = 2x^2 - 10x + 17$

$2x^2 - (10 + y)x + (17 + 3y) = 0$

2. Apply the discriminant condition for $x$ to have real roots and simplify into a quadratic inequality in $y$.

Since the curve exists for real values of $x$, the quadratic equation must have real roots. Thus, the discriminant $\Delta \ge 0$.

$[-(10 + y)]^2 - 4(2)(17 + 3y) \ge 0$

$100 + 20y + y^2 - 8(17 + 3y) \ge 0$

$100 + 20y + y^2 - 136 - 24y \ge 0$

$y^2 - 4y - 36 \ge 0$

3. Find the critical values by solving the corresponding quadratic equation.

Solving $y^2 - 4y - 36 = 0$ using the quadratic formula:

$y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-36)}}{2(1)}$

$y = \frac{4 \pm \sqrt{16 + 144}}{2}$

$y = \frac{4 \pm \sqrt{160}}{2}$

$y = \frac{4 \pm 4\sqrt{10}}{2}$

$y = 2 \pm 2\sqrt{10}$

4. Determine the solution intervals for the quadratic inequality.

The quadratic curve for $y^2 - 4y - 36 \ge 0$ is a U-shaped parabola. The inequality is satisfied for values outside the roots.

$y \le 2 - 2\sqrt{10} \quad \text{or} \quad y \ge 2 + 2\sqrt{10}$

1. Rewrite the function to find its asymptotes.

By algebraic manipulation or algebraic long division:
$y = \frac{2x^2 - 10x + 17}{x - 3} = \frac{2x(x - 3) - 4x + 17}{x - 3}$
$y = \frac{2x(x - 3) - 4(x - 3) + 5}{x - 3} = 2x - 4 + \frac{5}{x - 3}$
As $x \to \pm\infty$, $y \to 2x - 4$. As $x \to 3$, $y \to \pm\infty$.
Therefore, the vertical asymptote is $x = 3$ and the oblique asymptote is $y = 2x - 4$.

2. Find the exact coordinates of the turning points.

Differentiating with respect to $x$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 2 - \frac{5}{(x - 3)^2}$
At the stationary points, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$\frac{5}{(x - 3)^2} = 2 \implies (x - 3)^2 = \frac{5}{2}$
$x - 3 = \pm\sqrt{\frac{5}{2}} = \pm\frac{\sqrt{10}}{2} \implies x = 3 \pm \frac{\sqrt{10}}{2}$
From part (i), or by direct substitution, the corresponding $y$-values are $y = 2 \pm 2\sqrt{10}$.
The minimum point is $\left(3 + \frac{\sqrt{10}}{2}, 2 + 2\sqrt{10}\right)$ and the maximum point is $\left(3 - \frac{\sqrt{10}}{2}, 2 - 2\sqrt{10}\right)$.

3. Find the axial intercepts.

When $x = 0$:
$y = \frac{17}{-3} = -\frac{17}{3}$
The $y$-intercept is $\left(0, -\frac{17}{3}\right)$.
To find $x$-intercepts, set $2x^2 - 10x + 17 = 0$. The discriminant is $(-10)^2 - 4(2)(17) = -36 < 0$, so there are no $x$-intercepts.

4. Sketch the curve with all required features accurately positioned and labelled.

Draw axes and the asymptotes ($x=3$ and $y=2x-4$). Sketch the left branch passing through $\left(0, -\frac{17}{3}\right)$, turning at the maximum point $\left(3 - \frac{\sqrt{10}}{2}, 2 - 2\sqrt{10}\right)$ and approaching $x=3$ downwards. Sketch the right branch coming from $x=3$ downwards, turning at the minimum point $\left(3 + \frac{\sqrt{10}}{2}, 2 + 2\sqrt{10}\right)$, and going upwards along $y=2x-4$.