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 formulas:** - The slope of the tangent line to the curve at a point is given by the derivative $y' = \frac{dy}{dx}$. - The slope of the normal line is the negative reciprocal of the tangent slope: if $m_t$ is tangent slope, then $m_n = -\frac{1}{m_t}$. - Equation of a line with slope $m$ passing through $(x_0,y_0)$ is $y - y_0 = m(x - x_0)$. 3. **Find the derivative:** $$y = 3x^3 - 2x$$ $$\frac{dy}{dx} = 9x^2 - 2$$ 4. **Evaluate the slope of the tangent at $x=1$:** $$m_t = 9(1)^2 - 2 = 9 - 2 = 7$$ 5. **Find the slope of the normal line:** $$m_n = -\frac{1}{7}$$ 6. **Write the equation of the normal line passing through $(1,1)$:** $$y - 1 = -\frac{1}{7}(x - 1)$$ 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. **Multiply both sides by 7 to clear denominators:** $$7y = -x + 8$$ **Note:** The problem states the normal line as $79y = -x + 78$, which seems to be a typographical error. The correct normal line equation is: $$7y = -x + 8$$