1. **State the problem:** Find the coefficients $a$, $b$, $c$, and $d$ of the cubic curve $$y = ax^3 + bx^2 + cx + d$$ such that it touches the line $$y = x + 1$$ at the point $(0,1)$ and touches the line $$y = -2x + 10$$ at the point $(3,4)$. 2. **Understand the conditions for tangency:** - The curve passes through the given points: $$y(0) = 1$$ and $$y(3) = 4$$. - The curve's value equals the line's value at these points. - The curve's derivative equals the line's slope at these points (because they are tangent). 3. **Write down the conditions explicitly:** - At $x=0$, curve touches $y = x + 1$: - Curve value: $$a(0)^3 + b(0)^2 + c(0) + d = d = 1$$ - Line value: $$0 + 1 = 1$$ (matches) - Derivative of curve: $$y' = 3ax^2 + 2bx + c$$ - Derivative at $x=0$: $$y'(0) = c$$ - Slope of line: 1 - So, $$c = 1$$ - At $x=3$, curve touches $y = -2x + 10$: - Curve value: $$a(3)^3 + b(3)^2 + c(3) + d = 27a + 9b + 3c + d = 4$$ - Line value: $$-2(3) + 10 = 4$$ (matches) - Derivative at $x=3$: $$y'(3) = 3a(3)^2 + 2b(3) + c = 27a + 6b + c$$ - Slope of line: -2 - So, $$27a + 6b + c = -2$$ 4. **Substitute known values $c=1$ and $d=1$ into the equations:** - From point $(3,4)$ value: $$27a + 9b + 3(1) + 1 = 4$$ $$27a + 9b + 3 + 1 = 4$$ $$27a + 9b + 4 = 4$$ $$27a + 9b = 0$$ - From derivative at $x=3$: $$27a + 6b + 1 = -2$$ $$27a + 6b = -3$$ 5. **Solve the system:** - From $$27a + 9b = 0$$, divide by 9: $$3a + b = 0 \\ b = -3a$$ - Substitute into $$27a + 6b = -3$$: $$27a + 6(-3a) = -3$$ $$27a - 18a = -3$$ $$9a = -3$$ $$a = -\frac{1}{3}$$ - Then, $$b = -3(-\frac{1}{3}) = 1$$ 6. **Final coefficients:** $$a = -\frac{1}{3}, \quad b = 1, \quad c = 1, \quad d = 1$$