Answer: \(\frac13\lt k\lt 3\).

Answer: \(\frac13\lt k\lt 3\).

At an intersection, the two expressions for \(y\) are equal:

\(9kx+1=kx^2+3x(2k+1)+4\).

Rearrange into a quadratic in \(x\):

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

For the line not to meet the curve, this quadratic must have no real roots, so its discriminant must be negative:

\((3-3k)^2-4(k)(3)\lt 0\).

Simplifying,

\(9(1-k)^2-12k\lt 0\),

\(9k^2-30k+9\lt 0\),

\(3k^2-10k+3\lt 0\).

Factorise:

\((3k-1)(k-3)\lt 0\).

Therefore \(k\) lies between the two critical values:

\(\frac13\lt k\lt 3\).