1. **State the problem:** We have a curve $C$ defined parametrically by $x = t^2 + t$ and $y = 4t^2 - t^3$ for $t \geq 0$. We need to find the gradient of $C$ at the point where it intersects the positive $x$-axis. 2. **Find where the curve intersects the positive $x$-axis:** On the $x$-axis, $y=0$. So, set $y=4t^2 - t^3 = 0$. 3. **Solve for $t$:** $$t^2(4 - t) = 0$$ This gives $t=0$ or $t=4$. Since $t \geq 0$, both are valid, but $t=0$ corresponds to the origin, and $t=4$ corresponds to the positive $x$-axis intersection. 4. **Find the gradient formula for parametric curves:** The gradient $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. 5. **Calculate derivatives:** $$\frac{dx}{dt} = 2t + 1$$ $$\frac{dy}{dt} = 8t - 3t^2$$ 6. **Evaluate derivatives at $t=4$:** $$\frac{dx}{dt}\bigg|_{t=4} = 2(4) + 1 = 8 + 1 = 9$$ $$\frac{dy}{dt}\bigg|_{t=4} = 8(4) - 3(4)^2 = 32 - 48 = -16$$ 7. **Calculate the gradient at $t=4$:** $$\frac{dy}{dx} = \frac{-16}{9}$$ **Final answer:** The gradient of the curve $C$ at the point where it intersects the positive $x$-axis is $\boxed{-\frac{16}{9}}$.