In this question, $a$, $b$, $c$ and $d$ are non-zero constants. The curve $C$ has equation $y = \frac{ax^2 + bx - 7}{x - c}$ with oblique asymptote $y = dx + 2$ and vertical asymptote intersecting at $(4, 6)$.

1. Find the value of $c$ using the vertical asymptote.

The asymptotes intersect at $(4, 6)$, which means the vertical asymptote is $x = 4$.
From the denominator of $y = \frac{ax^2 + bx - 7}{x - c}$, the vertical asymptote occurs when $x - c = 0 \implies x = c$.
Therefore, $c = 4$.

2. Find the value of $d$ using the oblique asymptote.

The oblique asymptote $y = dx + 2$ passes through the intersection point $(4, 6)$.
Substitute $x = 4$ and $y = 6$ into the equation of the oblique asymptote:
$6 = d(4) + 2$
$4d = 4 \implies d = 1$

3. Form an equation using the quotient and remainder to compare coefficients and find $a$ and $b$.

Since the oblique asymptote is $y = x + 2$, the function can be written in the form $y = \text{Quotient} + \frac{\text{Remainder}}{\text{Divisor}}$:
$y = x + 2 + \frac{R}{x - 4}$
where $R$ is a constant.
Expressing this as a single fraction:
$y = \frac{(x + 2)(x - 4) + R}{x - 4} = \frac{x^2 - 2x - 8 + R}{x - 4}$
Comparing the numerator with the original function's numerator $ax^2 + bx - 7$:
$ax^2 + bx - 7 = x^2 - 2x + (R - 8)$
By comparing the coefficients of $x^2$ and $x$:
$a = 1$
$b = -2$
(Note: Comparing the constants gives $-7 = R - 8 \implies R = 1$)

1. Determine the equation of the curve and its asymptotes based on the information given for part (i).

The vertical asymptote passes through $(4, 6)$, so the vertical asymptote is $x = 4$, which means $c = 4$.
The oblique asymptote $y = dx + 2$ also passes through $(4, 6)$:
$6 = 4d + 2 \implies 4d = 4 \implies d = 1$.
Thus, the oblique asymptote is $y = x + 2$.

Using long division form, the curve is $y = x + 2 + \frac{R}{x - 4}$.
Combining into a single fraction:
$y = \frac{(x+2)(x-4)+R}{x-4} = \frac{x^2 - 2x - 8 + R}{x - 4}$
Comparing this with $y = \frac{ax^2+bx-7}{x-4}$, we get $a = 1$, $b = -2$, and $-8+R = -7 \implies R=1$.
The equation of the curve is $y = \frac{x^2-2x-7}{x-4}$.

2. Find the coordinates of the axial intercepts.

For the $y$-intercept, set $x = 0$:
$y = \frac{-7}{-4} = \frac{7}{4}$

For the $x$-intercepts, set $y = 0$:
$x^2 - 2x - 7 = 0$
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-7)}}{2} = \frac{2 \pm \sqrt{32}}{2} = 1 \pm 2\sqrt{2}$

The intercepts are $(0, \frac{7}{4})$, $(1 - 2\sqrt{2}, 0)$ and $(1 + 2\sqrt{2}, 0)$.

3. Find the coordinates of the stationary points.

Using $y = x + 2 + (x-4)^{-1}$, differentiate with respect to $x$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - (x-4)^{-2}$

Set $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ for stationary points:
$1 - \frac{1}{(x-4)^2} = 0 \implies (x-4)^2 = 1 \implies x - 4 = \pm 1$
$x = 3 \text{ or } x = 5$

When $x = 3$, $y = 3 + 2 + \frac{1}{-1} = 4$.
When $x = 5$, $y = 5 + 2 + \frac{1}{1} = 8$.

The curve has a local maximum at $(3, 4)$ and a local minimum at $(5, 8)$.

4. Sketch the curve, clearly showing all asymptotes, intercepts, and turning points.

Draw the asymptotes $x = 4$ and $y = x + 2$ with dotted lines.
Sketch the two branches of the hyperbola:

• The left branch is constrained below the oblique asymptote and to the left of the vertical asymptote. It passes through $(0, \frac{7}{4})$, $(1-2\sqrt{2}, 0)$, and $(1+2\sqrt{2}, 0)$, with a local maximum turning point at $(3, 4)$.
• The right branch lies above the oblique asymptote and to the right of the vertical asymptote, with a local minimum turning point at $(5, 8)$.

1. Interpret the given equation in relation to the curve $C$.

Using the values $a=1$, $b=-2$, and $c=4$ established in part (i), the equation becomes:
$\frac{x^2 - 2x - 7}{x - 4} = k$
This represents the points of intersection between the curve $C$ with equation $y = \frac{x^2 - 2x - 7}{x - 4}$ and the horizontal line $y = k$.

2. Find the $y$-coordinates of the turning points of $C$.

The turning points are needed to identify the gap in the range of the function. Using the proper fraction form $y = x + 2 + \frac{1}{x - 4}$, we differentiate to find stationary points:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{1}{(x-4)^2} = 0$
$(x-4)^2 = 1 \implies x = 3 \text{ or } x = 5$
Substituting back to find the corresponding $y$-values:
When $x = 3$, $y = 3 + 2 + \frac{1}{-1} = 4$ (local maximum).
When $x = 5$, $y = 5 + 2 + \frac{1}{1} = 8$ (local minimum).

3. Deduce the range of $k$ for which there are no intersections.

For the equation to have no real roots, the line $y = k$ must not intersect the curve $C$. Based on the sketch from part (ii), the graph of $C$ spans $(-\infty, 4] \cup [8, \infty)$.
Thus, the line $y = k$ will not intersect $C$ strictly between the local maximum and local minimum:
$4 < k < 8$