A curve $C$ has equation $y = \frac{2x^2 - 5ax + 2a^2 + 1}{x - a}$, $x \ne a$.

1. Find the equation of the vertical asymptote

The function is undefined when the denominator is zero.
$x - a = 0 \implies x = a$
The vertical asymptote is $x = a$.

2. Perform algebraic manipulation or polynomial division to find the oblique asymptote

Rewrite the numerator to express $y$ in the form $y = mx + c + \frac{k}{x - a}$:

As $x \to \pm\infty$, the fraction $\frac{1 - a^2}{x - a} \to 0$.
Therefore, the oblique asymptote is $y = 2x - 3a$.

1. Differentiate the curve equation using the simplified form obtained from finding the asymptotes.

From part (i), by performing algebraic division, the curve can be written as:
$y = 2x - 3a + \frac{1 - a^2}{x - a}$

Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 - \frac{1 - a^2}{(x - a)^2}$

2. Equate the derivative to zero and rearrange to isolate $(x - a)^2$.

For stationary points, we set $\frac{dy}{dx} = 0$:
$2 - \frac{1 - a^2}{(x - a)^2} = 0$
$\frac{1 - a^2}{(x - a)^2} = 2$
$(x - a)^2 = \frac{1 - a^2}{2}$

3. Apply the condition for the equation to have no real solutions and solve the inequality for $a$.

For the curve to have no stationary points, the equation must yield no valid real solutions for $x$.
Since $(x - a)^2 \ge 0$ for all real $x$, there are no real solutions if the right-hand side is strictly negative:
$\frac{1 - a^2}{2} < 0$
$1 - a^2 < 0$
$a^2 > 1$

Therefore, the range of values is:
$a < -1 \text{ or } a > 1$

1. Find the equations of the vertical and oblique asymptotes by rewriting the equation of the curve.

By algebraic division:
$\begin{aligned} y &= \frac{2x^2 - 5ax + 2a^2 + 1}{x - a} \\ &= \frac{2x(x - a) - 3ax + 2a^2 + 1}{x - a} \\ &= \frac{2x(x - a) - 3a(x - a) - 3a^2 + 2a^2 + 1}{x - a} \\ &= 2x - 3a + \frac{1 - a^2}{x - a} \end{aligned}$
The equations of the asymptotes are $x = a$ and $y = 2x - 3a$.

2. Determine the $y$-intercept of the curve.

Substitute $x = 0$ into the equation of the curve:
$y = \frac{2(0)^2 - 5a(0) + 2a^2 + 1}{0 - a} = -\frac{2a^2 + 1}{a}$
The $y$-intercept is $\left(0, -\frac{2a^2 + 1}{a}\right)$.

3. Deduce the shape of the branches and outline the sketch.

Since $a > 1$, we have $1 - a^2 < 0$.

• For $x < a$, $x - a < 0$, which implies $\frac{1 - a^2}{x - a} > 0$. The curve lies above the oblique asymptote $y = 2x - 3a$ and approaches $+\infty$ as $x \to a^-$.
• For $x > a$, $x - a > 0$, which implies $\frac{1 - a^2}{x - a} < 0$. The curve lies below the oblique asymptote and approaches $-\infty$ as $x \to a^+$.
  The curve passes through the $y$-intercept $\left(0, -\frac{2a^2 + 1}{a}\right)$, which is on the left branch.