已知曲线 $C$ 的方程为 $y = \dfrac{-4x^2 - 6x - 11}{2x + 2}$，其中 $x \in \mathbb{R}$ 且 $x \ne -1$。

1. Rearrange the given equation to form a quadratic equation in $x$.

$y = \frac{-4x^2 - 6x - 11}{2x + 2}$
$y(2x + 2) = -4x^2 - 6x - 11$
$2xy + 2y = -4x^2 - 6x - 11$
$4x^2 + (2y + 6)x + (2y + 11) = 0$

2. Apply the discriminant condition for real roots to establish an inequality in $y$.

For $x$ to have real values, the discriminant must be non-negative ($b^2 - 4ac \ge 0$):
$(2y + 6)^2 - 4(4)(2y + 11) \ge 0$

3. Expand and simplify the discriminant inequality.

$4y^2 + 24y + 36 - 16(2y + 11) \ge 0$
$4y^2 + 24y + 36 - 32y - 176 \ge 0$
$4y^2 - 8y - 140 \ge 0$
$y^2 - 2y - 35 \ge 0$

4. Factorise and solve for the valid range of $y$, then state the set of values $y$ cannot take.

$(y - 7)(y + 5) \ge 0$
Hence, the possible values for $y$ are $y \le -5$ or $y \ge 7$.
The set of values that $y$ cannot take is the complement of this range:
$-5 < y < 7$

1. Determine the equations of the asymptotes.

By algebraic division, rewrite the rational function:
$y = \frac{-4x^2 - 6x - 11}{2x + 2} = \frac{(-2x - 1)(2x + 2) - 9}{2x + 2} = -2x - 1 - \frac{9}{2x + 2}$
The vertical asymptote occurs when the denominator is zero:
$2x + 2 = 0 \implies x = -1$
As $x \to \pm\infty$, the fractional term $\frac{9}{2x+2} \to 0$, giving the oblique asymptote:
$y = -2x - 1$

2. Find the coordinates of the stationary points.

Differentiate $y$ with respect to $x$ to find the stationary points:
$\frac{dy}{dx} = \frac{d}{dx}\left(-2x - 1 - 9(2x+2)^{-1}\right) = -2 + \frac{18}{(2x+2)^2}$
Set $\frac{dy}{dx} = 0$:
$\frac{18}{(2x+2)^2} = 2 \implies (2x+2)^2 = 9$
$2x+2 = 3 \quad \text{or} \quad 2x+2 = -3$
$x = 0.5 \quad \text{or} \quad x = -2.5$
Substitute the $x$-values back into the equation of $C$:
When $x = 0.5$, $y = -2(0.5) - 1 - \frac{9}{3} = -5$.
When $x = -2.5$, $y = -2(-2.5) - 1 - \frac{9}{-3} = 7$.
The stationary points are $(0.5, -5)$ and $(-2.5, 7)$.

(Note: Students can also deduce the $y$-values $-5$ and $7$ from the boundary values established in part (i), then solve for the corresponding $x$-values.)

3. Find the point(s) where the curve crosses the axes.

Substitute $x = 0$ to find the $y$-intercept:
$y = \frac{-11}{2} = -5.5$
The curve crosses the $y$-axis at $(0, -5.5)$.

Substitute $y = 0$ to find any $x$-intercepts:
$-4x^2 - 6x - 11 = 0$
Check the discriminant $\Delta$:
$\Delta = (-6)^2 - 4(-4)(-11) = 36 - 176 = -140$
Since $\Delta < 0$, there are no real solutions, hence the curve does not cross the $x$-axis.

4. Sketch the curve $C$.

Draw the coordinate axes and plot the key features:

• Draw and label the dashed lines for the asymptotes $x = -1$ and $y = -2x - 1$.
• Plot the local maximum point at $(0.5, -5)$ and the local minimum point at $(-2.5, 7)$.
• Plot the $y$-intercept at $(0, -5.5)$.
• Sketch the two continuous branches of the rational function approaching the asymptotes smoothly.

1. Apply a substitution to simplify the given equation.

Let $X = x + 1$. Then $2x + 2 = 2(x + 1) = 2X$.
By completing the square for the expression $-4x^2 - 8x - 13$:
$-4(x^2 + 2x) - 13 = -4(x^2 + 2x + 1) - 9 = -4X^2 - 9$
Substitute these into the given equation:
$X^2(2X)^2 - a^2(-4X^2 - 9)^2 = a^2(2X)^2$

2. Expand and simplify the equation into a polynomial in $X$.

Expand the terms:
$4X^4 - a^2(16X^4 + 72X^2 + 81) = 4a^2X^2$
$4X^4 - 16a^2X^4 - 72a^2X^2 - 81a^2 - 4a^2X^2 = 0$
Group the terms by powers of $X$:
$4(1 - 4a^2)X^4 - 76a^2X^2 - 81a^2 = 0$

3. Analyze the roots of the quadratic in $X^2$ to find the condition for no real roots.

Let $Z = X^2$. Since $x \ne -1$, we require $Z > 0$ for $x$ to be real.
The equation is a quadratic in $Z$:
$4(1 - 4a^2)Z^2 - 76a^2Z - 81a^2 = 0$
For the original equation to have no root, this quadratic must have no positive real root for $Z$. Notice that since $a > 0$, the coefficients $-76a^2$ and $-81a^2$ are both strictly negative.

• If $1 - 4a^2 > 0$, the signs of the coefficients are $(+, -, -)$, meaning the product of roots is negative, so there is exactly one positive real root.
• If $1 - 4a^2 \le 0$, all coefficients are $\le 0$. For any $Z > 0$, the polynomial's value is strictly negative and can never equal $0$, so there are no positive real roots.
  Thus, for no root, we must have:
  $1 - 4a^2 \le 0 \implies a^2 \ge \frac{1}{4}$
  Since $a$ is a positive constant, $a \ge \frac{1}{2}$.

1. Express the given rational function in quotient-remainder form.

$y = \frac{-4x^2 - 6x - 11}{2x + 2}$
By algebraic manipulation or polynomial long division:
$y = \frac{-2x(2x + 2) - 2x - 11}{2x + 2}$
$y = -2x + \frac{-(2x + 2) - 9}{2x + 2}$
$y = -2x - 1 - \frac{9}{2x + 2}$

2. Relate the simplified expression to the original function.

Let $f(x) = 2x + \frac{9}{2x}$. We want to express the simplified $y$ in terms of $f(x)$.
Rewrite $y$ by completing the form of $f(x+1)$:
$y = -\left(2x + 1 + \frac{9}{2(x+1)}\right)$
$y = -\left(2x + 2 + \frac{9}{2(x+1)} - 1\right)$
$y = -\left(2(x+1) + \frac{9}{2(x+1)}\right) + 1$
$y = -f(x+1) + 1$

3. State the sequence of transformations.

The equation $y = -f(x+1) + 1$ can be mapped from $y = f(x)$ by:

1. A reflection in the $x$-axis, transforming $y = f(x)$ into $y = -f(x)$.
2. A translation by the vector $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$, transforming $y = -f(x)$ into $y - 1 = -f(x - (-1)) \implies y = -f(x+1) + 1$.