1. **State the problem:** We have the curve given by the equation $$y = \frac{9}{2}x^2 - 3x + 1$$ and we want to find the coordinates of the point where the gradient (slope) is 33. 2. **Find the gradient function:** The gradient of the curve at any point is the derivative of $$y$$ with respect to $$x$$. $$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{9}{2}x^2 - 3x + 1 \right)$$ Using the power rule and constant rule: $$\frac{dy}{dx} = \frac{9}{2} \times 2x - 3 = 9x - 3$$ 3. **Set the gradient equal to 33 and solve for $$x$$:** $$9x - 3 = 33$$ Add 3 to both sides: $$9x - \cancel{3} + 3 = 33 + 3$$ $$9x = 36$$ Divide both sides by 9: $$\frac{\cancel{9}x}{\cancel{9}} = \frac{36}{9}$$ $$x = 4$$ 4. **Find the corresponding $$y$$ coordinate:** Substitute $$x=4$$ into the original equation: $$y = \frac{9}{2} (4)^2 - 3(4) + 1$$ Calculate step-by-step: $$y = \frac{9}{2} \times 16 - 12 + 1 = 72 - 12 + 1 = 61$$ 5. **Final answer:** The point on the curve where the gradient is 33 is $$\boxed{(4, 61)}$$. Since the $$y$$ value is an integer, no decimal rounding is needed.