1. **State the problem:** We are given the quadratic equation $$y = (x + 2)(x - 8)$$ and asked to sketch its graph by understanding its key features. 2. **Rewrite the equation:** Expand the product to standard form: $$y = (x + 2)(x - 8) = x^2 - 8x + 2x - 16 = x^2 - 6x - 16$$ 3. **Identify the roots (x-intercepts):** Set $$y=0$$: $$0 = (x + 2)(x - 8)$$ So, $$x = -2$$ or $$x = 8$$. These are the points where the graph crosses the x-axis. 4. **Find the vertex:** The vertex of a parabola $$y = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. Here, $$a=1$$ and $$b=-6$$, so: $$x = -\frac{-6}{2 \times 1} = \frac{6}{2} = 3$$ 5. **Calculate the y-coordinate of the vertex:** Substitute $$x=3$$ into the equation: $$y = (3 + 2)(3 - 8) = 5 \times (-5) = -25$$ So the vertex is at $$(3, -25)$$. 6. **Graph shape:** Since $$a=1 > 0$$, the parabola opens upward, not downward as described. The description in the problem seems inconsistent with the equation. 7. **Movable points:** The points given are approximately $$(-2, 0)$$, $$(0, -16)$$, and $$(2, 0)$$. - $$(-2, 0)$$ and $$(8, 0)$$ are roots. - At $$x=0$$, $$y = (0+2)(0-8) = 2 \times (-8) = -16$$. - At $$x=2$$, $$y = (2+2)(2-8) = 4 \times (-6) = -24$$, so the point $$(2, 0)$$ is not on the graph; likely a typo or misinterpretation. 8. **Summary:** The parabola opens upward, crosses the x-axis at $$x=-2$$ and $$x=8$$, and has a vertex at $$(3, -25)$$. **Final answer:** The graph of $$y = (x + 2)(x - 8)$$ is a parabola opening upward with roots at $$x=-2$$ and $$x=8$$ and vertex at $$(3, -25)$$.