A curve $C$ has equation $y = \frac{x^2 - 4x + 4}{x - 1}$.

1. Find the equations of the asymptotes.

By algebraic manipulation or long division:
$y = \frac{x^2 - 4x + 4}{x - 1} = \frac{(x-1)(x-3) + 1}{x - 1} = x - 3 + \frac{1}{x - 1}$
As $x \to 1$, $y \to \pm\infty$, so the vertical asymptote is $x = 1$.
As $x \to \pm\infty$, $y \to x - 3$, so the oblique asymptote is $y = x - 3$.

2. Find the points of intersection with the axes and the turning points.

Set $x = 0$: $y = \frac{4}{-1} = -4$. The $y$-intercept is $(0, -4)$.
Set $y = 0$: $x^2 - 4x + 4 = 0 \implies (x-2)^2 = 0 \implies x = 2$. The $x$-intercept is $(2, 0)$.
Using a graphing calculator (or by setting $\frac{dy}{dx} = 1 - \frac{1}{(x-1)^2} = 0$), the turning points are:

• Local maximum at $(0, -4)$
• Local minimum at $(2, 0)$
  Note that the turning points coincide with the axial intercepts.

3. Sketch the curve, clearly labelling all key features.

Draw the required features on a Cartesian plane:

• Dotted lines for asymptotes $x = 1$ and $y = x - 3$.
• The left branch approaches $y = x - 3$ from below as $x \to -\infty$, turns at the local maximum $(0, -4)$, and approaches $x = 1$ downwards.
• The right branch approaches $x = 1$ upwards from the right, turns at the local minimum $(2, 0)$ where it touches the $x$-axis, and approaches $y = x - 3$ from above as $x \to \infty$.

1. Substitute the expression for the curve $C$ into the given equation to identify the suitable graph.

Let $y = \frac{x^2 - 4x + 4}{x - 1}$. The given equation becomes:
$(x - 2)^2 + \frac{1}{b^2}(y + 1)^2 = 4$
Dividing by 4 yields:
$\frac{(x - 2)^2}{2^2} + \frac{(y + 1)^2}{(2b)^2} = 1$
This represents an ellipse centred at $(2, -1)$ with a horizontal semi-axis of $2$ and a vertical semi-axis of $2b$.

2. Identify the critical points of $C$ from part (a) to determine intersection boundaries.

From part (a), the curve $C$ has a local minimum at $(2, 0)$ and a local maximum at $(0, -4)$.
For the equation to have no real roots, the drawn ellipse must not intersect the curve $C$.
The upper branch of $C$ lies entirely in the region $y \ge 0$, and the lower branch lies entirely in the region $y \le -4$.

3. Formulate an inequality for $b$ to ensure the ellipse does not intersect the upper branch of $C$.

The highest point of the ellipse occurs at its centre's $x$-coordinate, $x = 2$, with a $y$-coordinate of $-1 + 2b$.
To avoid intersection with the upper branch of $C$ (which has its minimum at $y = 0$), the ellipse's highest point must be strictly below it:
$-1 + 2b < 0$
$2b < 1 \implies b < \frac{1}{2}$

4. Verify the condition against the lower branch and state the final range of values.

We must also ensure it does not intersect the lower branch. The lowest point of the ellipse is at $y = -1 - 2b$.
Given $0 < b < \frac{1}{2}$, we have:
$-1 - 2b > -1 - 1 = -2$
Because the lower branch of $C$ is at $y \le -4$, the ellipse (whose lowest point is $> -2$) will never intersect the lower branch. Thus, no real roots exist for this range.
The condition is satisfied for:
$0 < b < \frac{1}{2}$