Answer: The asymptotes are \(x=-1\) and \(y=x-3\).

For the case where the \(x\)-axis is tangent to \(C\), \(\lambda=1\).

When \(\lambda=-4\), the curve meets the \(x\)-axis at \((1-\sqrt{5},0)\) and \((1+\sqrt{5},0)\).

Write the curve in a form that separates the linear part from the remainder:

\(y=\frac{x^{2}-2x+\lambda}{x+1}.\)

Perform division of \(x^{2}-2x+\lambda\) by \(x+1\):

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

So

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

This immediately shows that as \(x\to\pm\infty\), the fraction tends to \(0\), so one asymptote is

\(y=x-3.\)

Also, the expression is undefined when \(x=-1\), and as \(x\to-1\) the term \(\frac{\lambda+3}{x+1}\) becomes unbounded, so the other asymptote is

\(x=-1.\)

Both asymptotes are independent of \(\lambda\).

For the \(x\)-axis to be tangent to \(C\), the equation \(y=0\) must have a repeated root. Set

\(\frac{x^{2}-2x+\lambda}{x+1}=0 \quad \Rightarrow \quad x^{2}-2x+\lambda=0.\)

For tangency, this quadratic must have discriminant zero:

\((-2)^2-4\lambda=0\)

\(4-4\lambda=0\)

\(\lambda=1.\)

Then

\(x^{2}-2x+1=(x-1)^2,\)

so the curve touches the \(x\)-axis at \((1,0)\). The sketch should show the two asymptotes \(x=-1\) and \(y=x-3\), with the curve touching the \(x\)-axis at \((1,0)\).

Now take \(\lambda=-4\). The \(x\)-intercepts satisfy

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

Solving gives

\(x=\frac{2\pm\sqrt{4+16}}{2}=1\pm\sqrt{5}.\)

Hence the points of intersection with the \(x\)-axis are

\((1-\sqrt{5},0)\quad \text{and} \quad (1+\sqrt{5},0).\)

For the sketch, draw the asymptotes \(x=-1\) and \(y=x-3\), then draw the two branches passing through these intercepts with the usual hyperbola-like behaviour near the vertical asymptote.