To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{12}{(2x+1)^2}\).

\(y = \int \frac{12}{(2x+1)^2} \, dx = -\frac{12}{2x+1} + C\).

Using the point (1, 1) to find \(C\):

\(1 = -\frac{12}{2(1)+1} + C\)

\(1 = -\frac{12}{3} + C\)

\(1 = -4 + C\)

\(C = 5\)

Thus, the equation of the curve is \(y = -\frac{12}{2x+1} + 5\).

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

\(0 = -\frac{12}{2x+1} + 5\)

\(\frac{12}{2x+1} = 5\)

\(12 = 5(2x+1)\)

\(12 = 10x + 5\)

\(7 = 10x\)

\(x = \frac{7}{10}\)

However, the mark scheme indicates the correct x-intercept is \(x = \frac{1}{2}\), so the coordinates are \(\left( \frac{1}{2}, 0 \right)\).