To find the equation of the curve, we need to integrate the given derivative \(\frac{dy}{dx} = 2 - 8(3x + 4)^{-\frac{1}{2}}\).

Integrate term by term:

\(\int 2 \, dx = 2x + C_1\)

For the second term, let \(u = 3x + 4\), then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

\(\int -8(3x + 4)^{-\frac{1}{2}} \, dx = -8 \int u^{-\frac{1}{2}} \cdot \frac{du}{3}\)

\(= -\frac{8}{3} \int u^{-\frac{1}{2}} \, du\)

\(= -\frac{8}{3} \cdot 2u^{\frac{1}{2}} + C_2\)

\(= -\frac{16}{3} \sqrt{3x+4} + C_2\)

Combine the results:

\(y = 2x - \frac{16}{3} \sqrt{3x+4} + C\)

Given that the curve intersects the y-axis where \(y = \frac{4}{3}\), substitute \(x = 0\) and \(y = \frac{4}{3}\) to find \(C\):

\(\frac{4}{3} = 2(0) - \frac{16}{3} \sqrt{4} + C\)

\(\frac{4}{3} = -\frac{16}{3} \cdot 2 + C\)

\(\frac{4}{3} = -\frac{32}{3} + C\)

\(C = \frac{4}{3} + \frac{32}{3} = 12\)

Thus, the equation of the curve is:

\(y = 2x + \frac{8}{3}\sqrt{3x+4} + 12\)