The curve $C$ has equation $y = \frac{3x^2 + ax + 3}{x + 1}$, where $a$ is a constant, $a \ne 6$.

1. Form the equation to find the $x$-intercepts of the curve.

At the $x$-axis, $y = 0$.
$\frac{3x^2 + ax + 3}{x + 1} = 0$
$3x^2 + ax + 3 = 0$

2. Apply the condition for the equation to have no real roots.

For the curve not to cut the $x$-axis, the quadratic equation must have no real roots. Therefore, its discriminant must be strictly less than zero.
$b^2 - 4ac < 0$
$a^2 - 4(3)(3) < 0$
$a^2 - 36 < 0$

3. Solve the quadratic inequality for $a$.

$(a - 6)(a + 6) < 0$
$-6 < a < 6$

1. Perform polynomial division to express the equation in a form suitable for finding asymptotes.

$y = \frac{3x^2 - 6x + 3}{x + 1}$

Using algebraic long division:
$3x^2 - 6x + 3 = 3x(x + 1) - 9x + 3 = 3x(x + 1) - 9(x + 1) + 12$
$\therefore y = 3x - 9 + \frac{12}{x + 1}$

2. State the equations of the asymptotes.

From $y = 3x - 9 + \frac{12}{x + 1}$:
As $x \to -1$, $y \to \pm\infty$, so the vertical asymptote is $x = -1$.
As $x \to \pm\infty$, $\frac{12}{x + 1} \to 0$, so the oblique asymptote is $y = 3x - 9$.

3. Find the coordinates of the intersections with the axes.

$y$-intercept: Let $x = 0$,
$y = \frac{3(0)^2 - 6(0) + 3}{0 + 1} = 3 \implies (0, 3)$

$x$-intercept: Let $y = 0$,
$\frac{3x^2 - 6x + 3}{x + 1} = 0 \implies 3(x^2 - 2x + 1) = 0 \implies 3(x - 1)^2 = 0$
$x = 1 \implies (1, 0)$

Intersections with the axes are $(0, 3)$ and $(1, 0)$.

4. Differentiate the function and equate to zero to find the turning points.

Using $y = 3x - 9 + 12(x + 1)^{-1}$:
$\frac{dy}{dx} = 3 - 12(x + 1)^{-2} = 3 - \frac{12}{(x + 1)^2}$

For turning points, $\frac{dy}{dx} = 0$:
$3 - \frac{12}{(x + 1)^2} = 0 \implies (x + 1)^2 = 4$
$x + 1 = 2 \quad \text{or} \quad x + 1 = -2$
$x = 1 \quad \text{or} \quad x = -3$

When $x = 1$, $y = 3(1) - 9 + \frac{12}{2} = 0$.
When $x = -3$, $y = 3(-3) - 9 + \frac{12}{-2} = -9 - 9 - 6 = -24$.

The turning points are $(1, 0)$ and $(-3, -24)$.

5. Sketch the curve $C$ incorporating all identified features.

The sketch must show:

• A dotted vertical line at $x = -1$.
• A dotted diagonal line for $y = 3x - 9$.
• The right branch located above the oblique asymptote and to the right of $x=-1$, with a minimum turning point touching the $x$-axis at $(1,0)$ and crossing the $y$-axis at $(0,3)$.
• The left branch located below the oblique asymptote and to the left of $x=-1$, with a maximum turning point at $(-3,-24)$.

1. Identify the equation of the new graph to sketch.

Substitute $y = \frac{3x^2 - 6x + 3}{x + 1}$ into the given equation:
$k^2(x-1)^2 + (y+12)^2 = k^2$
Since $k$ is a positive constant, dividing through by $k^2$ gives:
$(x-1)^2 + \frac{(y+12)^2}{k^2} = 1$
This represents an ellipse centered at $(1, -12)$ with a horizontal semi-axis of $1$ and a vertical semi-axis of $k$.

2. Determine the condition for the ellipse to intersect the curve $C$ at exactly two distinct points.

The ellipse only exists for horizontal values bounded by $(x-1)^2 \le 1$, meaning $0 \le x \le 2$. Thus, it will never intersect the left branch of $C$ (where $x < -1$).
From part (b), for $x > -1$, curve $C$ has a minimum turning point at $(1, 0)$.
At the boundaries $x = 0$ and $x = 2$, the ellipse points are strictly at $y = -12$, which are below curve $C$ (where $y = 3$ and $y = 1$ respectively).
The highest point of the ellipse occurs at $x = 1$, yielding $y = -12 + k$.
For the equation to have exactly 2 real and distinct roots, the upper half of the ellipse must intersect curve $C$ at exactly 2 points in the interval $(0, 2)$.
Since $C$ is concave up and the upper half of the ellipse is concave down, they will intersect exactly twice if and only if the peak of the ellipse is strictly above the minimum of $C$:
$-12 + k > 0$
$k > 12$