1. **Problem:** Find the equations of the tangent and normal lines to the curve $$y = 3x^4 - 5x^3 + 6x + 8$$ at the point where $$x=1$$. 2. **Formula and rules:** - The slope of the tangent line at a point is given by the derivative $$\frac{dy}{dx}$$ evaluated at that point. - The equation of the tangent line at $$x=a$$ is $$y - y(a) = m_t (x - a)$$ where $$m_t = \left. \frac{dy}{dx} \right|_{x=a}$$. - The slope of the normal line is the negative reciprocal of the tangent slope: $$m_n = -\frac{1}{m_t}$$. - The equation of the normal line is $$y - y(a) = m_n (x - a)$$. 3. **Find $$y(1)$$:** $$y(1) = 3(1)^4 - 5(1)^3 + 6(1) + 8 = 3 - 5 + 6 + 8 = 12$$. 4. **Find the derivative $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = 12x^3 - 15x^2 + 6$$. 5. **Evaluate the derivative at $$x=1$$:** $$m_t = 12(1)^3 - 15(1)^2 + 6 = 12 - 15 + 6 = 3$$. 6. **Equation of the tangent line:** Using point-slope form: $$y - 12 = 3(x - 1)$$ Simplify: $$y = 3x - 3 + 12 = 3x + 9$$. 7. **Slope of the normal line:** $$m_n = -\frac{1}{3}$$. 8. **Equation of the normal line:** $$y - 12 = -\frac{1}{3}(x - 1)$$ Simplify: $$y = -\frac{1}{3}x + \frac{1}{3} + 12 = -\frac{1}{3}x + \frac{37}{3}$$. **Final answers:** - Tangent line: $$y = 3x + 9$$ - Normal line: $$y = -\frac{1}{3}x + \frac{37}{3}$$