Answer: (i) The asymptotes are

\(x=\frac{1}{k}\) and \(y=\frac{x}{k}\).

(ii) The stationary points are

\(\left(\frac{2}{k},\,\frac{4}{k^2}\right)\) and \((0,0)\).

(iii) The sketch is a rational curve with vertical asymptote \(x=\frac{1}{k}\), oblique asymptote \(y=\frac{x}{k}\), passing through \((0,0)\) and \(\left(\frac{2}{k},\frac{4}{k^2}\right)\), with the branch for \(x<\frac{1}{k}\) lying below the slant asymptote for large negative \(x\) and tending to \(-\infty\) as \(x\to\left(\frac{1}{k}\right)^-\), while the branch for \(x>\frac{1}{k}\) tends to \(+\infty\) as \(x\to\left(\frac{1}{k}\right)^+\) and then approaches the line \(y=\frac{x}{k}\) as \(x\to\infty\).

(i) Write

\(y=\dfrac{x^2}{kx-1}\).

A vertical asymptote occurs when the denominator is zero:

\(kx-1=0 \implies x=\dfrac{1}{k}.\)

For the oblique asymptote, divide \(x^2\) by \(kx-1\):

\(\dfrac{x^2}{kx-1}=\dfrac{1}{k}x+\dfrac{1}{k^2}+\dfrac{1/k^2}{kx-1}\).

Since the final fraction tends to 0 as \(|x|\to\infty\), the slant asymptote is

\(y=\dfrac{x}{k}+\dfrac{1}{k^2}.\)

However, it is cleaner to check by long division carefully:

\(x^2=(kx-1)\left(\dfrac{x}{k}+\dfrac{1}{k^2}\right)+\dfrac{1}{k^2}\),

so

\(y=\dfrac{x}{k}+\dfrac{1}{k^2}+\dfrac{1/k^2}{kx-1}.\)

Hence the oblique asymptote is \(y=\dfrac{x}{k}+\dfrac{1}{k^2}\).

(ii) Differentiate using the quotient rule:

\(y' = \dfrac{(2x)(kx-1)-x^2(k)}{(kx-1)^2}.\)

Simplify the numerator:

\(2x(kx-1)-kx^2 = 2kx^2-2x-kx^2 = kx^2-2x = x(kx-2).\)

So

\(y' = \dfrac{x(kx-2)}{(kx-1)^2}.\)

Stationary points occur when \(y'=0\), so

\(x(kx-2)=0\), giving \(x=0\) or \(x=\dfrac{2}{k}.\)

Now find the corresponding \(y\)-values:

For \(x=0\), \(y=0\).

For \(x=\dfrac{2}{k}\),

\(y=\dfrac{(2/k)^2}{k(2/k)-1}=\dfrac{4/k^2}{2-1}=\dfrac{4}{k^2}.\)

So the stationary points are \((0,0)\) and \(\left(\dfrac{2}{k},\dfrac{4}{k^2}\right)\).

(iii) To sketch the curve, note the key features:

• Vertical asymptote: \(x=\dfrac{1}{k}\).
• Oblique asymptote: \(y=\dfrac{x}{k}+\dfrac{1}{k^2}\).
• Intercept and stationary point: the curve passes through \((0,0)\), which is a stationary point.
• Second stationary point: \(\left(\dfrac{2}{k},\dfrac{4}{k^2}\right)\).

For \(x<\dfrac{1}{k}\), the denominator is negative, so \(y\le 0\) except at \(x=0\) where it is 0. As \(x\to \left(\dfrac{1}{k}\right)^-\), \(y\to -\infty\).

For \(x>\dfrac{1}{k}\), the denominator is positive and the curve goes to \(+\infty\) as \(x\to \left(\dfrac{1}{k}\right)^+\), then turns at \(\left(\dfrac{2}{k},\dfrac{4}{k^2}\right)\), and tends to the oblique asymptote as \(x\to\infty\).

So the sketch consists of two branches separated by the vertical asymptote, with the left branch crossing the origin and the right branch having a turning point at \(\left(\dfrac{2}{k},\dfrac{4}{k^2}\right)\).