(i) To find the equation of the curve, integrate the gradient function:

\(\frac{dy}{dx} = \sqrt{1 + 2x} = (1 + 2x)^{1/2}\)

Integrate: \(y = \int (1 + 2x)^{1/2} \, dx\)

Using substitution, let \(u = 1 + 2x\), then \(du = 2 \, dx\), so \(dx = \frac{du}{2}\).

\(y = \int u^{1/2} \cdot \frac{1}{2} \, du = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{3}(1 + 2x)^{3/2} + C\)

\(y = \frac{2}{3}(1 + 2x)^{3/2} + C\)

Use the point \((4, 11)\) to find \(C\):

\(11 = \frac{2}{3}(1 + 2 \cdot 4)^{3/2} + C\)

\(11 = \frac{2}{3}(9)^{3/2} + C\)

\(11 = \frac{2}{3} \cdot 27 + C\)

\(11 = 18 + C\)

\(C = 2\)

Thus, the equation of the curve is \(y = \frac{2}{3}(1 + 2x)^{3/2} + 2\).

(ii) To find the y-intercept, set \(x = 0\):

\(y = \frac{2}{3}(1 + 2 \cdot 0)^{3/2} + 2\)

\(y = \frac{2}{3}(1)^{3/2} + 2\)

\(y = \frac{2}{3} + 2\)

\(y = \frac{2}{3} + \frac{6}{3}\)

\(y = \frac{8}{3}\)

The curve intersects the y-axis at \((0, \frac{8}{3})\).