The gradient of a curve is given by 3x\(^2\) - 8x + 2. P(0, 4)

\(\frac{dy}{dx}\) = 3x\(^2\) - 8x + 2

\(\int\)\(\frac{dy}{dx}\) = \(\int\)[3x\(^2\) - 8x + 2]

y = \(\frac{3x^2}{3} - \frac{8x^2}{2} + 2x + c\)

y = x\(^3\) - 4x\(^2\) + 2x + c

x = 0, y = 4

4 =  \(\frac{3 \times 0^2}{3} - \frac{8 \times 0^2}{2} + 2 \times 0 + c\)

c = 4.

Thus, y = x\(^3\) - 4x\(^2\) + 2x + 4