1. **State the problem:** Graph the parabola given by the equation $$y = -x^2 + 2x + 8$$ and plot 5 points including the roots and the vertex. 2. **Formula and rules:** The equation is a quadratic function in standard form $$y = ax^2 + bx + c$$ where $$a = -1$$, $$b = 2$$, and $$c = 8$$. - The parabola opens downward because $$a < 0$$. - The vertex formula is $$x = -\frac{b}{2a}$$. - Roots are found by solving $$y=0$$. 3. **Find the vertex:** $$x = -\frac{2}{2 \times (-1)} = -\frac{2}{-2} = 1$$ Substitute $$x=1$$ into the equation: $$y = -(1)^2 + 2(1) + 8 = -1 + 2 + 8 = 9$$ Vertex is at $$(1, 9)$$. 4. **Find the roots:** Solve $$-x^2 + 2x + 8 = 0$$ Multiply both sides by $$-1$$ to simplify: $$\cancel{-1} \times (-x^2 + 2x + 8) = \cancel{-1} \times 0$$ $$x^2 - 2x - 8 = 0$$ Factor: $$(x - 4)(x + 2) = 0$$ Roots are $$x = 4$$ and $$x = -2$$. 5. **Find symmetric points:** Since the vertex is at $$x=1$$, points equidistant from 1 have the same $$y$$. At $$x=0$$: $$y = -(0)^2 + 2(0) + 8 = 8$$ At $$x=2$$: $$y = -(2)^2 + 2(2) + 8 = -4 + 4 + 8 = 8$$ 6. **Summary of points to plot:** - Roots: $$(-2, 0)$$ and $$(4, 0)$$ - Vertex: $$(1, 9)$$ - Symmetric points: $$(0, 8)$$ and $$(2, 8)$$ These 5 points define the parabola shape. **Final answer:** The parabola opens downward with vertex at $$(1, 9)$$, roots at $$(-2, 0)$$ and $$(4, 0)$$, and symmetric points at $$(0, 8)$$ and $$(2, 8)$$.