Answer:

For

\(y=\frac{2x^{2}+kx}{x+1},\)

differentiate using the quotient rule:

\(\frac{dy}{dx}=\frac{(4x+k)(x+1)-(2x^{2}+kx)}{(x+1)^{2}}=\frac{2x^{2}+4x+k}{(x+1)^{2}}.\)

Stationary points require

\(2x^{2}+4x+k=0.\)

For no stationary points, this quadratic must have no real roots:

\(4^{2}-4(2)(k)\lt 0.\)

Thus

\(16-8k\lt 0\quad\Rightarrow\quad k\gt 2.\)

When \(k=4\),

\(y=\frac{2x^{2}+4x}{x+1}=2x+2-\frac{2}{x+1}.\)

So the vertical asymptote is \(x=-1\), and the oblique asymptote is \(y=2x+2\).

The intercepts come from

\(2x^{2}+4x=2x(x+2)=0,\)

so the curve crosses the \(x\)-axis at \((-2,0)\) and \((0,0)\). The curve has one branch below the oblique asymptote for \(x\gt -1\), and one branch above it for \(x\lt -1\), with the vertical asymptote at \(x=-1\).