1. **Stating the problem:** We have a cubic function $$y = ax^3 + bx^2 + cx + d$$ that passes through points $(3,0)$ and $(0,6)$, and has a critical point at $(2,2)$. 2. **Using given points to form equations:** - At $(3,0)$: $$a(3)^3 + b(3)^2 + c(3) + d = 0$$ simplifies to $$27a + 9b + 3c + d = 0$$ - At $(0,6)$: $$a(0)^3 + b(0)^2 + c(0) + d = 6$$ simplifies to $$d = 6$$ 3. **Critical point condition:** The derivative is $$y' = 3ax^2 + 2bx + c$$. At critical point $x=2$, $$y'(2) = 0$$, so $$3a(2)^2 + 2b(2) + c = 0$$ which simplifies to $$12a + 4b + c = 0$$. 4. **Point $(2,2)$ lies on the curve:** Substitute into original function: $$a(2)^3 + b(2)^2 + c(2) + d = 2$$ which is $$8a + 4b + 2c + d = 2$$. 5. **Substitute known $d=6$ into equations:** From step 2: $$27a + 9b + 3c + 6 = 0$$ so $$27a + 9b + 3c = -6$$ From step 4: $$8a + 4b + 2c + 6 = 2$$ so $$8a + 4b + 2c = -4$$ 6. **System of three equations with unknowns $a,b,c$:** $$27a + 9b + 3c = -6$$ $$12a + 4b + c = 0$$ $$8a + 4b + 2c = -4$$ 7. **Solve the system:** From second equation: $$c = -12a - 4b$$. Substitute into first: $$27a + 9b + 3(-12a - 4b) = -6$$ $$27a + 9b - 36a - 12b = -6$$ $$-9a - 3b = -6 ightarrow 3a + b = 2$$ Substitute $c = -12a - 4b$ into third: $$8a + 4b + 2(-12a - 4b) = -4$$ $$8a + 4b - 24a - 8b = -4$$ $$-16a - 4b = -4 ightarrow 4a + b = 1$$ 8. **Solve for $a$ and $b$ from:** $$3a + b = 2$$ $$4a + b = 1$$ Subtracting the second from the first: $$(3a + b) - (4a + b) = 2 - 1$$ $$-a = 1 ightarrow a = -1$$ Substitute $a=-1$ into $3a + b = 2$: $$3(-1) + b = 2 ightarrow -3 + b = 2 ightarrow b = 5$$ 9. **Find $c$:** $$c = -12a - 4b = -12(-1) - 4(5) = 12 - 20 = -8$$ 10. **Recall $d=6$ from step 2.** 11. **Determine the type of critical point:** Second derivative $$y'' = 6ax + 2b$$ At $x=2$: $$y''(2) = 6(-1)(2) + 2(5) = -12 + 10 = -2 < 0$$ Since $y''(2)<0$, $(2,2)$ is a local maximum. **Final answer:** $$a = -1, b = 5, c = -8, d = 6$$ and the critical point $(2,2)$ is a local maximum. This corresponds to option C.