曲线 $C$ 的方程为 $y=f(x)$，其中 $f(x)=\frac{ax^2+bx+c}{x+d},$ 且 $a, b, c$ 和 $d$ 是常数，且 $a \ne 0$。

1. Express $f(x)$ in quotient-remainder form to find the equation of its oblique asymptote.

Using algebraic division, we express $f(x)$ as a quotient and a remainder:
$ax^2+bx+c = ax(x+d) + (b-ad)x + c$
$= ax(x+d) + (b-ad)(x+d) + c - d(b-ad)$

Thus, $f(x)$ can be written as:
$f(x) = ax + (b - ad) + \frac{c - d(b-ad)}{x+d}$

As $x \to \pm\infty$, the fractional part approaches $0$, so the equation of the oblique asymptote is $y = ax + b - ad$.

2. Compare the theoretical asymptote with the given asymptote $y = x + 1$ to find $a$ and show $b = d + 1$.

Given that the asymptote is $y = x + 1$, we can compare the coefficients with $y = ax + b - ad$.

Comparing the $x$-coefficients:
$a = 1$

Comparing the constant terms:
$b - ad = 1$

Substituting $a = 1$:
$b - d(1) = 1 \implies b = d + 1 \text{ (shown)}$

3. Find the first derivative of $f(x)$.

Substitute $a = 1$ and $b - d = 1$ back into the earlier expression for $f(x)$:
The remainder simplifies to $c - d(1) = c - d$.
$f(x) = x + 1 + \frac{c-d}{x+d}$

To find when $f(x)$ is increasing, we evaluate the first derivative $f'(x)$:
$f'(x) = 1 - \frac{c-d}{(x+d)^2}$

4. Use the condition that $f$ is an increasing function to deduce that $c < d$.

Since $f$ is an increasing function for all $x > -d$, we require $f'(x) > 0$ (or $f'(x) \ge 0$) for all $x > -d$:
$1 - \frac{c-d}{(x+d)^2} > 0$
$\frac{c-d}{(x+d)^2} < 1$

As $x \to -d^+$, $\frac{1}{(x+d)^2} \to \infty$.
If $c - d > 0$, the term $\frac{c-d}{(x+d)^2}$ would approach $+\infty$, which eventually violates the inequality $< 1$. Thus, we must have $c - d \le 0$.

Additionally, since $C$ is a curve with an oblique asymptote, $f(x)$ cannot simply be the straight line $y = x + 1$. This means the remainder $c - d \neq 0$.

Therefore, we must have:
$c - d < 0 \implies c < d \text{ (shown)}$

1. Substitute known constants to find the full equation of $C$.

From part (a), $a = 1$ and $b = d + 1$.
Given $d = 2$, we have $b = 3$.
Given $c = 1$, the function is:
$y = \frac{x^2+3x+1}{x+2}$
By polynomial division or algebraic manipulation:
$y = \frac{x(x+2) + (x+2) - 1}{x+2} = x + 1 - \frac{1}{x+2}$

2. Identify the equations of the asymptotes.

From the rewritten form $y = x + 1 - \frac{1}{x+2}$:

• The vertical asymptote occurs when the denominator is zero: $x = -2$.
• As $x \to \pm\infty$, $\frac{1}{x+2} \to 0$, so the oblique asymptote is $y = x + 1$.

3. Determine the coordinates of the axes intercepts.

To find the $y$-intercept, let $x = 0$:
$y = \frac{0+0+1}{0+2} = \frac{1}{2}$
To find the $x$-intercepts, let $y = 0$:
$x^2 + 3x + 1 = 0$
Using the quadratic formula:
$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(1)}}{2} = \frac{-3 \pm \sqrt{5}}{2}$
So the $x$-intercepts are approximately $-0.382$ and $-2.618$.

4. Analyze the shape and position of the branches to draw the final sketch.

Note that $f'(x) = 1 + \frac{1}{(x+2)^2} > 0$ for all $x \neq -2$, so there are no stationary points and the function is strictly increasing on both sides of the vertical asymptote.
As $x \to -2^+$, $-\frac{1}{x+2} \to -\infty$, meaning the curve approaches the asymptote downwards on the right side.
As $x \to \infty$, the curve stays strictly below $y = x+1$.
Combining this with the intercepts, sketch the standard hyperbola branches: one bounded in the top-left region and the other in the bottom-right relative to the intersection of the asymptotes.

1. Identify the equation of the suitable graph to sketch.

Let $y = f(x) = \frac{x^2+3x+1}{x+2}$.
The given equation is:
$y^2 + 4x^2 = 16$

Dividing throughout by $16$ gives:
$\frac{x^2}{4} + \frac{y^2}{16} = 1$
This is the equation of an ellipse centred at the origin $(0,0)$, with $x$-intercepts at $(\pm 2, 0)$ and $y$-intercepts at $(0, \pm 4)$.

2. Analyse the intersections of the ellipse and curve $C$ to find the number of roots.

By sketching the ellipse on the same diagram as curve $C$:

• At $x=0$, curve $C$ is at $y = 0.5$ and the ellipse has points at $y = \pm 4$.

• At $x=2$, curve $C$ is at $y = 2.75$ and the ellipse is at $y = 0$.
  Since $C$ is continuous and strictly increasing from $y=0.5$ to $y=2.75$ for $x \in (0, 2)$, and the upper half of the ellipse decreases from $y=4$ to $y=0$, they must intersect exactly once in the interval $(0, 2)$.

• As $x \to -2^+$, curve $C \to -\infty$. The ellipse intersects the $x$-axis at $(-2, 0)$.

• Between $x = -2$ and $x = 0$, $C$ increases continuously to $y = 0.5$, crossing the lower half of the ellipse (which connects $(-2, 0)$ to $(0, -4)$). This gives exactly $1$ more point of intersection in the interval $(-2, 0)$.

For $x < -2$ and $x > 2$, the ellipse is not defined.
Thus, there are exactly $2$ points of intersection.

3. State the number of real roots.

Since there are $2$ points of intersection, there are exactly $2$ real roots to the equation.