1. **State the problem:** Find the equation of the tangent line to the curve $y = x^2 - 2x^3$ at the point $(1, -1)$. 2. **Formula and rules:** The slope of the tangent line at a point is given by the derivative $y' = \frac{dy}{dx}$ evaluated at that point. 3. **Find the derivative:** $$y = x^2 - 2x^3$$ $$y' = \frac{d}{dx}(x^2) - \frac{d}{dx}(2x^3) = 2x - 6x^2$$ 4. **Evaluate the derivative at $x=1$ to find the slope $m$:** $$m = y'(1) = 2(1) - 6(1)^2 = 2 - 6 = -4$$ 5. **Use point-slope form of the line:** $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1) = (1, -1)$ and $m = -4$. 6. **Substitute values:** $$y - (-1) = -4(x - 1)$$ $$y + 1 = -4x + 4$$ 7. **Simplify to slope-intercept form:** $$y = -4x + 4 - 1$$ $$y = -4x + 3$$ **Final answer:** The equation of the tangent line is $$y = -4x + 3$$