Quadratic Graph 160Fc2

1. **State the problem:** We need to graph the quadratic function $$y = -x^2 + 6x - 8$$ and plot 5 points including the roots and the vertex. 2. **Formula and important rules:** The quadratic function is in the form $$y = ax^2 + bx + c$$ where $$a = -1$$, $$b = 6$$, and $$c = -8$$. 3. **Find the vertex:** The vertex $$x$$-coordinate is given by $$x = -\frac{b}{2a} = -\frac{6}{2(-1)} = 3$$. 4. **Calculate the vertex $$y$$-coordinate:** $$y = -(3)^2 + 6(3) - 8 = -9 + 18 - 8 = 1$$. 5. **Find the roots:** Solve $$-x^2 + 6x - 8 = 0$$. Multiply both sides by $$-1$$ to simplify: $$\cancel{-}x^2 + \cancel{6}x - \cancel{8} = 0 \Rightarrow x^2 - 6x + 8 = 0$$ 6. **Factor the quadratic:** $$x^2 - 6x + 8 = (x - 2)(x - 4) = 0$$. 7. **Roots are:** $$x = 2$$ and $$x = 4$$. 8. **Calculate additional points:** Choose $$x = 1$$ and $$x = 5$$ to get more points. For $$x=1$$: $$y = -(1)^2 + 6(1) - 8 = -1 + 6 - 8 = -3$$. For $$x=5$$: $$y = -(5)^2 + 6(5) - 8 = -25 + 30 - 8 = -3$$. 9. **Summary of points to plot:** - Roots: $$(2, 0)$$ and $$(4, 0)$$ - Vertex: $$(3, 1)$$ - Additional points: $$(1, -3)$$ and $$(5, -3)$$ These points form a downward opening parabola with vertex at $$(3,1)$$.