1. **State the problem:** We are given three functions: $$y = -x + 20$$ $$y = x^2 - 4$$ $$y = 2x + 1$$ We want to analyze these functions and understand their graphs and intersections. 2. **Identify the types of functions:** - $$y = -x + 20$$ is a linear function with slope $$-1$$ and y-intercept $$20$$. - $$y = x^2 - 4$$ is a quadratic function (parabola) opening upwards with vertex at $$(0, -4)$$. - $$y = 2x + 1$$ is a linear function with slope $$2$$ and y-intercept $$1$$. 3. **Find intersections between the parabola and each line:** **Between $$y = x^2 - 4$$ and $$y = -x + 20$$:** Set equal: $$x^2 - 4 = -x + 20$$ Rearranged: $$x^2 + x - 24 = 0$$ Factor or use quadratic formula: $$x = \frac{-1 \pm \sqrt{1 + 96}}{2} = \frac{-1 \pm \sqrt{97}}{2}$$ **Between $$y = x^2 - 4$$ and $$y = 2x + 1$$:** Set equal: $$x^2 - 4 = 2x + 1$$ Rearranged: $$x^2 - 2x - 5 = 0$$ Use quadratic formula: $$x = \frac{2 \pm \sqrt{4 + 20}}{2} = \frac{2 \pm \sqrt{24}}{2} = 1 \pm \sqrt{6}$$ 4. **Find intersection between the two lines:** Set equal: $$-x + 20 = 2x + 1$$ Rearranged: $$-x - 2x = 1 - 20$$ $$-3x = -19$$ $$x = \frac{-19}{-3} = \frac{19}{3}$$ Substitute back: $$y = 2 \times \frac{19}{3} + 1 = \frac{38}{3} + 1 = \frac{41}{3}$$ 5. **Summary of intersections:** - Parabola and $$y = -x + 20$$ intersect at $$x = \frac{-1 \pm \sqrt{97}}{2}$$. - Parabola and $$y = 2x + 1$$ intersect at $$x = 1 \pm \sqrt{6}$$. - Lines intersect at $$\left(\frac{19}{3}, \frac{41}{3}\right)$$. These points can be plotted to visualize the relationships. **Final answers:** - Intersection points parabola and $$y = -x + 20$$: $$\left(\frac{-1 + \sqrt{97}}{2}, \frac{1 - \sqrt{97}}{2} + 20\right)$$ and $$\left(\frac{-1 - \sqrt{97}}{2}, \frac{1 + \sqrt{97}}{2} + 20\right)$$. - Intersection points parabola and $$y = 2x + 1$$: $$\left(1 + \sqrt{6}, 2(1 + \sqrt{6}) + 1\right)$$ and $$\left(1 - \sqrt{6}, 2(1 - \sqrt{6}) + 1\right)$$. - Intersection point of lines: $$\left(\frac{19}{3}, \frac{41}{3}\right)$$.