The curve $C$ has the equation $y = \dfrac{ax^{2} - 3x + b}{x - c}$ where $a$, $b$, and $c$ are real constants. It is given that the equations of asymptotes of $C$ are $x = 2$ and $y = x - 1$. Also, $C$ passes through the point $P\left(0, -\frac{3}{2}\right)$.

1. Explain why $a = 1$ using the oblique asymptote.

The rational function is $y = \dfrac{ax^2 - 3x + b}{x - c}$.
Performing polynomial long division, the quotient begins with the term $\dfrac{ax^2}{x} = ax$.
Since the oblique asymptote is given as $y = x - 1$, the linear quotient must be $x - 1$.
Comparing the coefficient of $x$, we deduce that $a = 1$.

2. State the value of $c$ by identifying the vertical asymptote.

The vertical asymptote occurs where the denominator is zero, so $x - c = 0 \implies x = c$.
Given that the vertical asymptote is $x = 2$, we have:
$c = 2$

3. Show that $b = 3$ by substituting the given point $P$.

The curve passes through the point $P\left(0, -\frac{3}{2}\right)$.
Substituting $x = 0$, $y = -\frac{3}{2}$, and $c = 2$ into the curve equation $y = \dfrac{x^2 - 3x + b}{x - c}$:
$-\frac{3}{2} = \dfrac{0 - 0 + b}{0 - 2}$
$-\frac{3}{2} = \dfrac{b}{-2}$
$b = \left(-\frac{3}{2}\right)(-2) = 3 \text{ (shown)}$

1. Determine the axial intercepts of the curve.

From part (i), $a = 1$, $b = 3$, and $c = 2$, so the equation of the curve $C$ is:
$y = \frac{x^2 - 3x + 3}{x - 2}$
To find the $y$-intercept, let $x = 0$:
$y = \frac{3}{-2} = -1.5$
The point $(0, -1.5)$ is the $y$-intercept (also given as point $P$).

To find any $x$-intercepts, let $y = 0$:
$x^2 - 3x + 3 = 0$
Checking the discriminant, $\Delta = (-3)^2 - 4(1)(3) = 9 - 12 = -3$.
Since $\Delta < 0$, there are no real roots, meaning the curve has no $x$-intercepts.

2. Find the coordinates of the stationary points.

Using the asymptotes $y = x - 1$ and $x = 2$, or by algebraic division, $C$ can be rewritten as:
$y = x - 1 + \frac{1}{x - 2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 1 - \frac{1}{(x - 2)^2}$
At stationary points, $\frac{dy}{dx} = 0$:
$1 - \frac{1}{(x - 2)^2} = 0$
$(x - 2)^2 = 1$
$x - 2 = 1 \quad \text{or} \quad x - 2 = -1$
$x = 3 \quad \text{or} \quad x = 1$
Substitute the $x$-values back into the equation of the curve to find the corresponding $y$-coordinates:
When $x = 3$, $y = 3 - 1 + \frac{1}{1} = 3$.
When $x = 1$, $y = 1 - 1 + \frac{1}{-1} = -1$.
Therefore, the stationary points are $(1, -1)$ and $(3, 3)$.

3. Sketch the curve with all indicated key features.

Draw the coordinate axes.

• Draw and label the vertical asymptote $x = 2$ and oblique asymptote $y = x - 1$ with dashed lines.
• Mark the $y$-intercept at $(0, -1.5)$.
• Plot and label the local maximum turning point at $(1, -1)$ and the local minimum turning point at $(3, 3)$.
• Sketch the two branches of the curve such that they pass through the respective points and asymptotically approach the dashed lines.

1. Find the derivative $\frac{dy}{dx}$ of the curve.

From part (i), $a=1$, $b=3$ and $c=2$, so the equation of the curve is:
$y = \frac{x^2 - 3x + 3}{x - 2}$
Using the quotient rule:
$\frac{dy}{dx} = \frac{(x - 2)(2x - 3) - (x^2 - 3x + 3)(1)}{(x - 2)^2}$

2. Calculate the gradient of the tangent at $P\left(0, -\frac{3}{2}\right)$.

Substitute $x = 0$ into $\frac{dy}{dx}$:
$\left.\frac{dy}{dx}\right|_{x=0} = \frac{(-2)(-3) - (3)(1)}{(-2)^2} = \frac{6 - 3}{4} = \frac{3}{4}$

3. Find the gradient of the normal and the exact equation of the normal.

The gradient of the normal $m_n$ is the negative reciprocal of the tangent gradient:
$m_n = -\frac{1}{3/4} = -\frac{4}{3}$
Using the point-gradient form for $P\left(0, -\frac{3}{2}\right)$:
$y - \left(-\frac{3}{2}\right) = -\frac{4}{3}(x - 0)$
$y = -\frac{4}{3}x - \frac{3}{2}$