The curve $C$ has equation $y = \frac{x^2 + ax + b}{x + 1}$, where $a$ and $b$ are constants. It is given that $C$ has a minimum point at $(2, 0)$.

1. Form an equation in $a$ and $b$ using the given point on the curve.

Since the curve passes through the minimum point $(2, 0)$, we substitute $x = 2$ and $y = 0$ into $y = \frac{x^2 + ax + b}{x + 1}$:
$0 = \frac{2^2 + 2a + b}{2 + 1}$
$0 = \frac{4 + 2a + b}{3}$
$4 + 2a + b = 0$
$2a + b = -4 \quad \text{--- (1)}$

2. Find $\frac{dy}{dx}$ using the quotient rule.

$\frac{dy}{dx} = \frac{(x + 1)\frac{d}{dx}(x^2 + ax + b) - (x^2 + ax + b)\frac{d}{dx}(x + 1)}{(x + 1)^2}$
$\frac{dy}{dx} = \frac{(x + 1)(2x + a) - (x^2 + ax + b)(1)}{(x + 1)^2}$
$\frac{dy}{dx} = \frac{2x^2 + ax + 2x + a - x^2 - ax - b}{(x + 1)^2}$
$\frac{dy}{dx} = \frac{x^2 + 2x + a - b}{(x + 1)^2}$

3. Use the condition for a stationary point to form a second equation.

At the minimum point, the derivative is zero, so $\frac{dy}{dx} = 0$. Substituting $x = 2$:
$\frac{2^2 + 2(2) + a - b}{(2 + 1)^2} = 0$
$\frac{8 + a - b}{9} = 0$
$8 + a - b = 0$
$a - b = -8 \quad \text{--- (2)}$

4. Solve the simultaneous equations to find $a$ and $b$.

Adding equation (1) and equation (2):
$(2a + b) + (a - b) = -4 + (-8)$
$3a = -12$
$a = -4$

Substitute $a = -4$ into equation (1):
$2(-4) + b = -4$
$-8 + b = -4$
$b = 4$

Thus, $a = -4$ and $b = 4$ (shown).

1. Substitute the given values of $a$ and $b$ to find the equation of $C$ and its axial intercepts.

Substitute $a = -4$ and $b = 4$ from part (i):
$y = \frac{x^2 - 4x + 4}{x + 1} = \frac{(x-2)^2}{x+1}$
For the $y$-intercept, let $x = 0$:
$y = \frac{4}{1} = 4$
For the $x$-intercept, let $y = 0$:
$(x-2)^2 = 0 \implies x = 2$
The axial intercepts are $(0, 4)$ and $(2, 0)$.

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

By algebraic long division (or algebraic manipulation):
$y = \frac{x(x+1) - 5x + 4}{x+1} = \frac{x(x+1) - 5(x+1) + 9}{x+1} = x - 5 + \frac{9}{x + 1}$
The vertical asymptote occurs where the denominator is zero:
$x + 1 = 0 \implies x = -1$
As $x \to \pm \infty$, $\frac{9}{x+1} \to 0$. Therefore, the oblique asymptote is:
$y = x - 5$

3. Determine the coordinates of the turning points analytically or via a graphing calculator.

Differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = 1 - \frac{9}{(x + 1)^2}$
At the turning points, $\frac{dy}{dx} = 0$:
$1 - \frac{9}{(x + 1)^2} = 0 \implies (x + 1)^2 = 9$
$x + 1 = \pm 3 \implies x = 2 \quad \text{or} \quad x = -4$
When $x = 2$, $y = 0$ (this confirms the given minimum point).
When $x = -4$, $y = -4 - 5 + \frac{9}{-4 + 1} = -9 - 3 = -12$.
The turning points are the minimum at $(2, 0)$ and the maximum at $(-4, -12)$.

4. Sketch the graph of $C$ incorporating all identified features.

Draw the axes and the asymptotes $x = -1$ and $y = x - 5$.
Sketch the two branches of the rational curve:

• The upper right branch approaches $x = -1$ as $x \to -1^+$ and $y = x - 5$ as $x \to \infty$, passing through $(0,4)$ and turning at the local minimum $(2,0)$.
• The lower left branch approaches $x = -1$ as $x \to -1^-$ and $y = x - 5$ as $x \to -\infty$, turning at the local maximum $(-4,-12)$.

1. Rearrange the given cubic equation to express it in terms of the function for curve $C$.

Substitute $a = -4$ and $b = 4$ into the given equation:
$x^3 - 4x^2 + 5x + 1 = 0$
Rearranging to isolate terms that match the numerator of $C$:
$x^3 - 4x^2 + 4x + x + 1 = 0$
$x(x^2 - 4x + 4) = -(x + 1)$
Dividing both sides by $x(x + 1)$ (verifying that $x = 0$ and $x = -1$ are not roots):
$\frac{x^2 - 4x + 4}{x + 1} = -\frac{1}{x}$
The appropriate graph to add is $y = -\frac{1}{x}$.

2. Deduce the number of real roots by analyzing the intersections of the two graphs.

Imagine or sketch $y = -\frac{1}{x}$ on the same axes as $C$:

• For $x > 0$, the graph of $C$ is strictly non-negative ($y \ge 0$) while $y = -\frac{1}{x} < 0$, meaning no intersection.
• For $x < -1$, the graph of $C$ is entirely below the $x$-axis (its local maximum is at $y = -12$) while $y = -\frac{1}{x} > 0$, meaning no intersection.
• For $-1 < x < 0$, as $x \to -1^+$, $C(x) \to \infty$ and $y = -\frac{1}{x} \to 1$. As $x \to 0^-$, $C(x) \to 4$ and $y = -\frac{1}{x} \to \infty$.

Since $C(x)$ is continuous and monotonic decreasing on this interval while $y = -\frac{1}{x}$ is continuous and monotonic increasing, the two curves cross exactly once in $(-1, 0)$.
Therefore, there is $1$ point of intersection, meaning the original equation has $1$ real root.