Answer: (i) The asymptotes are \(x=-1\), \(x=2\) and \(y=0\).

(ii) There is no point on \(C\) for which \(\frac{1}{3}\lt y\lt 3\).

(iii) The turning points are \((1,3)\) and \(\left(5,\frac{1}{3}\right)\).

(iv) A sketch should show the vertical asymptotes \(x=-1\) and \(x=2\), the horizontal asymptote \(y=0\), and the intercepts \((0,\frac{9}{2})\) and \((3,0)\), with branches passing through the turning points above.

(i) Since the degree of the numerator is less than the degree of the denominator, the curve has horizontal asymptote \(y=0\).

Vertical asymptotes occur where the denominator is zero:

\((x-2)(x+1)=0,\)

so \(x=2\) and \(x=-1\). Therefore the asymptotes are \(x=-1\), \(x=2\) and \(y=0\).

(ii) Start from

\(y=\frac{3x-9}{(x-2)(x+1)}.\)

Multiplying through by the denominator gives

\(y(x-2)(x+1)=3x-9.\)

Since \((x-2)(x+1)=x^2-x-2\), this becomes

\(yx^2-(y+3)x+9-2y=0.\)

For a fixed value of \(y\), this is a quadratic in \(x\). There is no point on \(C\) when this quadratic has no real roots, so its discriminant is negative:

\((y+3)^2-4y(9-2y)\lt 0.\)

Simplifying,

\(y^2+6y+9-36y+8y^2\lt0,\)

so

\(9y^2-30y+9\lt0.\)

Divide by \(3\):

\(3y^2-10y+3\lt0.\)

Factorising gives

\((3y-1)(y-3)\lt0.\)

Hence

\(\frac13\lt y\lt3.\)

So there is no point on \(C\) for which \(\frac13\lt y\lt3\).

(iii) Differentiate using the quotient rule:

\(\frac{dy}{dx}=\frac{3(x^2-x-2)-(3x-9)(2x-1)}{(x^2-x-2)^2}.\)

At a turning point, the numerator is zero:

\(3(x^2-x-2)-(3x-9)(2x-1)=0.\)

Expanding carefully,

\(3x^2-3x-6-(6x^2-21x+9)=0,\)

so

\(-3x^2+18x-15=0.\)

Divide by \(-3\):

\(x^2-6x+5=0=(x-1)(x-5).\)

Thus \(x=1\) or \(x=5\). Substituting in the curve gives

\(y(1)=\frac{3(1)-9}{(1-2)(1+1)}=3,\qquad y(5)=\frac{15-9}{(5-2)(5+1)}=\frac13.\)

Therefore the turning points are \((1,3)\) and \(\left(5,\frac13\right)\).

(iv) For the sketch, show the asymptotes \(x=-1\), \(x=2\), and \(y=0\). The intercepts are \((0,\frac92)\) and \((3,0)\). The right-hand branch passes through \(\left(5,\frac13\right)\), and the middle branch passes through \((1,3)\); the remaining branch lies to the left of \(x=-1\) and tends to the horizontal asymptote.