1. **State the problem:** We have a cubic curve given by the equation $$y = x^3 + ax + b$$ and it has a stationary point at $$(1, 13)$$. We need to find the values of $a$ and $b$. 2. **Recall the formula for stationary points:** Stationary points occur where the derivative of the function is zero. The derivative of $$y = x^3 + ax + b$$ is: $$y' = 3x^2 + a$$ 3. **Use the stationary point condition:** At the stationary point $x=1$, the derivative must be zero: $$0 = 3(1)^2 + a = 3 + a$$ 4. **Solve for $a$:** $$a = -3$$ 5. **Use the point on the curve:** The stationary point is also on the curve, so when $x=1$, $y=13$: $$13 = (1)^3 + a(1) + b = 1 + a + b$$ 6. **Substitute $a = -3$ into the equation:** $$13 = 1 - 3 + b$$ 7. **Simplify and solve for $b$:** $$13 = -2 + b$$ $$b = 13 + 2 = 15$$ **Final answer:** $$a = -3, \quad b = 15$$