A curve C has equation $y = f(x)$, where $f(x) = \frac{2(x-1)^2}{x-3}$.

1. Find the axial intercepts.

Substitute $x = 0$:
$y = \frac{2(-1)^2}{-3} = -\frac{2}{3}$
The $y$-intercept is $(0, -\frac{2}{3})$.

Substitute $y = 0$:
$2(x-1)^2 = 0 \implies x = 1$
The $x$-intercept is $(1, 0)$.

2. Find the equations of the vertical and oblique asymptotes.

Expand the numerator:
$f(x) = \frac{2x^2 - 4x + 2}{x-3}$
By polynomial long division or algebraic manipulation:
$f(x) = \frac{2x(x-3) + 2(x-3) + 8}{x-3} = 2x + 2 + \frac{8}{x-3}$
The vertical asymptote is $x = 3$.
As $x \to \pm\infty$, $\frac{8}{x-3} \to 0$, so the oblique asymptote is $y = 2x + 2$.

3. Find the coordinates of the stationary points.

Differentiating $f(x)$:
$f'(x) = \frac{d}{dx}\left(2x + 2 + \frac{8}{x-3}\right) = 2 - \frac{8}{(x-3)^2}$
Setting $f'(x) = 0$ for stationary points:
$2 - \frac{8}{(x-3)^2} = 0$
$(x-3)^2 = 4$
$x - 3 = 2 \quad \text{or} \quad x - 3 = -2$
$x = 5 \quad \text{or} \quad x = 1$
Substitute into $f(x)$:
When $x = 5$, $y = 2(5) + 2 + \frac{8}{2} = 16$.
When $x = 1$, $y = 2(1) + 2 + \frac{8}{-2} = 0$.
The stationary points are $(1, 0)$ and $(5, 16)$.

4. Sketch the curve C.

Draw a set of coordinate axes. Mark the $y$-intercept at $(0, -\frac{2}{3})$ and $x$-intercept at $(1, 0)$.
Draw and label the vertical asymptote $x = 3$ and oblique asymptote $y = 2x + 2$ with dashed lines.
Plot the local maximum turning point at $(1, 0)$ and the local minimum turning point at $(5, 16)$.
Sketch the two smooth branches of the hyperbola, ensuring the curve correctly approaches the asymptotes.

1. Sketch the graph of $y = f(|x|)$.

The graph of $y = f(|x|)$ is an even function. It is obtained by retaining the part of the curve $C$ where $x \ge 0$, and reflecting it in the $y$-axis.

Key features to show on the sketch:

• Axial intercept: $(0, -\frac{2}{3})$
• Stationary points: Local maxima at $(1, 0)$ and $(-1, 0)$; Local minima at $(5, 16)$ and $(-5, 16)$
• Vertical asymptotes: $x = 3$ and $x = -3$
• Oblique asymptotes: $y = 2x + 2$ (for $x > 0$) and $y = -2x + 2$ (for $x < 0$)

2. Identify the suitable curve representing the given equation.

The given equation is:
$x^2 + \left[ \frac{2(|x|-1)^2}{|x|-3} - k \right]^2 = a^2$

Substitute $y = f(|x|) = \frac{2(|x|-1)^2}{|x|-3}$ into the equation:
$x^2 + (y - k)^2 = a^2$

This is the equation of a circle with centre $(0, k)$ and radius $a$. Thus, the suitable curve to add to the diagram is this circle.

3. Use symmetry to deduce the condition for an odd number of roots.

The roots of the equation correspond to the points of intersection between the curve $y = f(|x|)$ and the circle $x^2 + (y - k)^2 = a^2$.

Both the graph of $y = f(|x|)$ and the circle are symmetrical about the $y$-axis ($x = 0$). Any intersection point $(x, y)$ with $x \neq 0$ will correspond to a symmetric intersection point $(-x, y)$, meaning these intersections always occur in pairs (an even number of roots).

For the total number of intersections (roots) to be odd, there must be exactly an odd number of intersections on the axis of symmetry, $x = 0$. Since the curve only intersects the $y$-axis at one point, the circle must pass through this single $y$-intercept.

4. Solve for $a$ in terms of $k$.

From the graph, the only point on $y = f(|x|)$ where $x = 0$ is $(0, -\frac{2}{3})$.

For the circle to intersect the curve at this point, we substitute $x = 0$ and $y = -\frac{2}{3}$ into the circle's equation:
$0^2 + \left(-\frac{2}{3} - k\right)^2 = a^2$
$\implies k \pm a = -\frac{2}{3}$

We are given that $a > 0$ and $k > 0$, which implies $k + a > 0$.
Because $-\frac{2}{3}$ is negative, $k + a$ cannot be $-\frac{2}{3}$. Therefore, we must take the subtraction case:
$k - a = -\frac{2}{3}$
$a = k + \frac{2}{3}$