1. **State the problem:** We have two polynomial functions: $$y_1 = (x + a)^2 + b$$ $$y_2 = cx^3 + d$$ Given that $y_2$ crosses the x-axis at $x = -1$ and the y-axis at $y = 2$, and that $y_1$ and $y_2$ are equal and tangent at $x = 1$, we need to find the values of $a$, $b$, $c$, and $d$. 2. **Use the given conditions:** - Since $y_2$ crosses the x-axis at $x = -1$, then $y_2(-1) = 0$: $$c(-1)^3 + d = -c + d = 0 \implies d = c$$ - Since $y_2$ crosses the y-axis at $y = 2$, then $y_2(0) = d = 2$: $$d = 2$$ From above, $d = c$ and $d = 2$, so: $$c = 2, \quad d = 2$$ 3. **Equality and tangency at $x = 1$:** - The functions are equal at $x=1$: $$y_1(1) = y_2(1)$$ - They are tangent at $x=1$, so their derivatives are equal at $x=1$: $$y_1'(1) = y_2'(1)$$ 4. **Calculate $y_1(1)$ and $y_2(1)$:** $$y_1(1) = (1 + a)^2 + b$$ $$y_2(1) = c(1)^3 + d = c + d = 2 + 2 = 4$$ Set equal: $$(1 + a)^2 + b = 4$$ 5. **Calculate derivatives:** $$y_1'(x) = 2(x + a)$$ $$y_2'(x) = 3cx^2$$ At $x=1$: $$y_1'(1) = 2(1 + a)$$ $$y_2'(1) = 3c(1)^2 = 3c = 3 \times 2 = 6$$ Set equal: $$2(1 + a) = 6 \implies 1 + a = 3 \implies a = 2$$ 6. **Substitute $a=2$ into the equality equation:** $$(1 + 2)^2 + b = 4 \implies 3^2 + b = 4 \implies 9 + b = 4 \implies b = 4 - 9 = -5$$ 7. **Check $y_2$ crosses x-axis at $x=-1$:** $$y_2(-1) = c(-1)^3 + d = -c + d = -2 + 2 = 0$$ This matches the condition. **Final values:** $$a = 2, b = -5, c = 2, d = 2$$ **Answer corresponds to option D.