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

(b) For all points on \(C\), \(y\le \frac{25}{12}\).

(c) The stationary point is \(\left(7,\frac{25}{12}\right)\).

(d) The curve meets the coordinate axes at \(\left(-\frac12,0\right)\), \((2,0)\) and \((0,-2)\). It meets the horizontal asymptote \(y=2\) at \((4,2)\), and has vertical asymptote \(x=1\).

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

Asymptotes

Since \(x^{2}-2x+1=(x-1)^2\), the denominator is zero when \(x=1\), so there is a vertical asymptote \(x=1\).

The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of the leading coefficients. Hence \(y=2\).

Showing that \(y\le \frac{25}{12}\)

Rearrange to form a quadratic in \(x\):

\(y(x^2-2x+1)=2x^2-3x-2\)

so

\(yx^2-2yx+y=2x^2-3x-2\),

hence

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

For points on the curve, this quadratic must have real roots in \(x\), so its discriminant is non-negative:

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

Expanding:

\(4y^2-12y+9-4(y^2-4)\ge 0\)

\(4y^2-12y+9-4y^2+16\ge 0\)

\(25-12y\ge 0\).

Therefore \(12y\le 25\), so \(y\le \frac{25}{12}\).

Stationary points

Differentiate using the quotient rule:

\(y' = \dfrac{(x^2-2x+1)(4x-3)-(2x^2-3x-2)(2x-2)}{(x^2-2x+1)^2}\).

For a stationary point, \(y'=0\), so

\((x^2-2x+1)(4x-3)-(2x^2-3x-2)(2x-2)=0\).

Simplifying gives

\(x^2-8x+7=0\)

so

\((x-7)(x-1)=0\).

Thus \(x=7\) or \(x=1\), but \(x=1\) is not on the curve since it is a vertical asymptote. So \(x=7\).

Now substitute into the equation of the curve:

\(y=\dfrac{2(7)^2-3(7)-2}{(7)^2-2(7)+1}=\dfrac{98-21-2}{49-14+1}=\dfrac{75}{36}=\dfrac{25}{12}\).

So the stationary point is \(\left(7,\frac{25}{12}\right)\).

Points needed for the sketch

To find the intersections with the coordinate axes:

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

\(2x^2-3x-2=0\)

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

so the intercepts are \(\left(-\frac12,0\right)\) and \((2,0)\).

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

\(y=\dfrac{-2}{1}=-2\),

so the intercept is \((0,-2)\).

To find where the curve meets the horizontal asymptote \(y=2\), solve

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

Then

\(2x^2-3x-2=2x^2-4x+2\),

so

\(x=4\).

Hence the curve meets the asymptote \(y=2\) at \((4,2)\).

The sketch therefore has a left-hand branch for \(x\lt 1\), passing through \(\left(-\frac12,0\right)\) and \((0,-2)\), approaching \(y=2\) as \(x\to -\infty\), and falling to \(-\infty\) as \(x\to 1^-\).

For \(x\gt 1\), the right-hand branch comes down from \(+\infty\) as \(x\to 1^+\), passes through \((2,0)\) and \((4,2)\), reaches the stationary point \(\left(7,\frac{25}{12}\right)\), and then approaches \(y=2\) from above as \(x\to \infty\).