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 the line $$y = -2x + 10$$ at the point $(3,4)$. 2. **Understand the conditions:** Since the curve touches the lines at given points, the curve passes through these points and has the same slope as the lines at those points (tangency condition). 3. **Write down the conditions:** - At $(0,1)$, the curve passes through the point: $$a(0)^3 + b(0)^2 + c(0) + d = 1 \implies d = 1$$ - The curve's slope at $x=0$ equals the slope of the line $y = x + 1$, which is 1: $$y' = 3ax^2 + 2bx + c$$ At $x=0$: $$y'(0) = c = 1$$ - At $(3,4)$, the curve passes through the point: $$a(3)^3 + b(3)^2 + c(3) + d = 4$$ $$27a + 9b + 3c + d = 4$$ - The curve's slope at $x=3$ equals the slope of the line $y = -2x + 10$, which is $-2$: $$y'(3) = 3a(3)^2 + 2b(3) + c = 27a + 6b + c = -2$$ 4. **Substitute known values $c=1$ and $d=1$ into the equations:** - From the point $(3,4)$: $$27a + 9b + 3(1) + 1 = 4 \implies 27a + 9b + 4 = 4 \implies 27a + 9b = 0$$ - From the slope at $x=3$: $$27a + 6b + 1 = -2 \implies 27a + 6b = -3$$ 5. **Solve the system:** From the first equation: $$27a + 9b = 0 \implies 3a + b = 0 \implies b = -3a$$ Substitute $b = -3a$ into the second equation: $$27a + 6(-3a) = -3 \implies 27a - 18a = -3 \implies 9a = -3 \implies a = -\frac{1}{3}$$ Then, $$b = -3\left(-\frac{1}{3}\right) = 1$$ 6. **Final coefficients:** $$a = -\frac{1}{3}, \quad b = 1, \quad c = 1, \quad d = 1$$