1. **State the problem:** Find the equation of the normal line to the curve $y = 3x^3 - 2x$ at the point $(1,1)$. 2. **Recall the formula:** The slope of the tangent line to the curve at a point is given by the derivative $y' = \frac{dy}{dx}$ evaluated at that point. 3. **Find the derivative:** $$y = 3x^3 - 2x \implies y' = 9x^2 - 2$$ 4. **Evaluate the slope of the tangent at $x=1$:** $$m_{tangent} = 9(1)^2 - 2 = 9 - 2 = 7$$ 5. **Find the slope of the normal line:** The normal line is perpendicular to the tangent, so its slope is the negative reciprocal: $$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{7}$$ 6. **Use point-slope form to find the equation of the normal line:** Point-slope form: $$y - y_1 = m(x - x_1)$$ Substitute $m = -\frac{1}{7}$ and point $(1,1)$: $$y - 1 = -\frac{1}{7}(x - 1)$$ 7. **Simplify the equation:** $$y - 1 = -\frac{1}{7}x + \frac{1}{7}$$ $$y = -\frac{1}{7}x + \frac{1}{7} + 1 = -\frac{1}{7}x + \frac{8}{7}$$ 8. **Rewrite in standard form:** Multiply both sides by 7: $$7y = -x + 8$$ **Final answer:** The equation of the normal line is $$7y = -x + 8$$. Note: The user mentioned $79y = -x + 78$, which is not correct for the given curve and point.