Answer: (i) \(d=-2\).

(ii) \(a=1\), \(b=-1\), and \(c=4\).

(iii) At all points on \(C\),

\(y\leq3-2\sqrt6\quad\text{or}\quad y\geq3+2\sqrt6.\)

The vertical asymptote occurs where the denominator is zero. Since the vertical asymptote is \(x=2\),

\(x+d=0\quad\text{at}\quad x=2,\)

so \(d=-2\).

The curve cuts the \(y\)-axis at \((0,-2)\). Hence

\(-2=\frac{c}{-2},\)

so \(c=4\).

Now divide by \(x-2\):

\(\frac{ax^2+bx+c}{x-2}=ax+(b+2a)+\frac{c+2b+4a}{x-2}.\)

The oblique asymptote is \(y=x+1\), so

\(a=1,\qquad b+2a=1.\)

Thus \(b=-1\), and with \(c=4\),

\(y=\frac{x^2-x+4}{x-2}.\)

Rearrange the equation as a quadratic in \(x\):

\(y(x-2)=x^2-x+4,\)

so

\(x^2-(y+1)x+2y+4=0.\)

For real points on the curve, this quadratic must have real roots, so the discriminant is non-negative:

\((y+1)^2-4(2y+4)\geq0.\)

This gives

\(y^2-6y-15\geq0,\)

or

\((y-3)^2\geq24.\)

Therefore

\(|y-3|\geq2\sqrt6,\)

so

\(y\leq3-2\sqrt6\quad\text{or}\quad y\geq3+2\sqrt6.\)