1. **Problem Statement:** Write an equation of the line tangent to the graph of $$y = x^3 + 3x^2 + 2$$ at its point of inflection. 2. **Find the point of inflection:** The point of inflection occurs where the second derivative is zero. 3. **First derivative:** $$y' = 3x^2 + 6x$$ 4. **Second derivative:** $$y'' = 6x + 6$$ 5. Set the second derivative equal to zero to find the inflection point: $$6x + 6 = 0$$ $$6x = -6$$ $$x = -1$$ 6. **Find the y-coordinate at $$x = -1$$:** $$y = (-1)^3 + 3(-1)^2 + 2 = -1 + 3 + 2 = 4$$ 7. **Find the slope of the tangent line at $$x = -1$$:** $$m = y'(-1) = 3(-1)^2 + 6(-1) = 3 - 6 = -3$$ 8. **Write the equation of the tangent line using point-slope form:** $$y - y_1 = m(x - x_1)$$ $$y - 4 = -3(x + 1)$$ **Final answer:** $$y - 4 = -3(x + 1)$$