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

(ii) The turning points are \((-1,5)\) and \((3,-3)\).

(iii) The curve intersects the \(y\)-axis at \((0,6)\), and it has no intersection with the \(x\)-axis.

(iv) Sketch a hyperbola-like curve with vertical asymptote \(x=1\) and oblique asymptote \(y=2-x\), passing through \((0,6)\), with turning points at \((-1,5)\) and \((3,-3)\).

Write the function in a form that is easier to analyse by division:

\(y=\dfrac{x^{2}-3x+6}{1-x}=\dfrac{x^{2}-3x+6}{-(x-1)}\).

Now divide \(x^{2}-3x+6\) by \(x-1\):

\(x^{2}-3x+6=(x-1)(x-2)+4\).

So

\(y=-\left(x-2+\dfrac{4}{x-1}\right)=2-x-\dfrac{4}{x-1}\).

(i) From this form, the vertical asymptote is where the denominator is zero, so \(x=1\). As \(x\to\pm\infty\), the term \(-\dfrac{4}{x-1}\to 0\), so the oblique asymptote is \(y=2-x\).

(ii) Differentiate:

\(y'=-1-4\dfrac{d}{dx}(x-1)^{-1}=-1+4(x-1)^{-2}=-1+\dfrac{4}{(x-1)^2}.\)

Set \(y'=0\):

\(-1+\dfrac{4}{(x-1)^2}=0\Rightarrow \dfrac{4}{(x-1)^2}=1\Rightarrow (x-1)^2=4\).

Hence \(x-1=\pm 2\), so \(x=-1\) or \(x=3\).

Find the corresponding \(y\)-values:

For \(x=-1\), \(y=2-(-1)-\dfrac{4}{-2}=3+2=5\).

For \(x=3\), \(y=2-3-\dfrac{4}{2}=-1-2=-3\).

So the turning points are \((-1,5)\) and \((3,-3)\).

(iii) The \(y\)-intercept is found by setting \(x=0\):

\(y=\dfrac{0-0+6}{1}=6\), so the curve crosses the \(y\)-axis at \((0,6)\).

For the \(x\)-intercepts, set \(y=0\):

\(x^{2}-3x+6=0\).

The discriminant is

\(\Delta=(-3)^2-4\cdot 1\cdot 6=9-24=-15\).

Since \(\Delta\lt 0\), there are no real roots, so there is no intersection with the \(x\)-axis.

(iv) The sketch should show two branches separated by the vertical asymptote \(x=1\), both approaching the slant asymptote \(y=2-x\). The left branch passes through \((0,6)\) and has a turning point at \((-1,5)\). The right branch has a turning point at \((3,-3)\). Since there are no \(x\)-intercepts, neither branch crosses the \(x\)-axis.