1. **State the problem:** Find the equations of the tangent and normal lines to the graph of the function $$F(x) = x^2 + 5x$$ at the point where $$x = -2$$. 2. **Find the point on the curve:** Evaluate $$F(-2)$$. $$F(-2) = (-2)^2 + 5(-2) = 4 - 10 = -6$$ So the point is $$(-2, -6)$$. 3. **Find the derivative:** The derivative $$F'(x)$$ gives the slope of the tangent line. $$F'(x) = \frac{d}{dx}(x^2 + 5x) = 2x + 5$$ 4. **Find the slope of the tangent at $$x = -2$$:** $$F'(-2) = 2(-2) + 5 = -4 + 5 = 1$$ 5. **Equation of the tangent line:** Using point-slope form $$y - y_1 = m(x - x_1)$$ with $$m=1$$ and point $$(-2, -6)$$: $$y - (-6) = 1(x - (-2))$$ $$y + 6 = x + 2$$ $$y = x - 4$$ 6. **Slope of the normal line:** The normal line is perpendicular to the tangent, so its slope is the negative reciprocal: $$m_{normal} = -\frac{1}{1} = -1$$ 7. **Equation of the normal line:** Using point-slope form with slope $$-1$$ and point $$(-2, -6)$$: $$y - (-6) = -1(x - (-2))$$ $$y + 6 = -1(x + 2)$$ $$y + 6 = -x - 2$$ $$y = -x - 8$$ **Final answers:** - Tangent line: $$y = x - 4$$ - Normal line: $$y = -x - 8$$