1. **State the problem:** We need to find the image of trapezoid DEFG after reflecting it over the vertical line $x = -3$. 2. **Reflection formula:** When reflecting a point $(x,y)$ over the line $x = a$, the image point is $(2a - x, y)$. 3. **Apply the formula to each vertex:** - For $D(-10,-10)$: $$D' = (2(-3) - (-10), -10) = (-6 + 10, -10) = (4, -10)$$ - For $E(-10,-2)$: $$E' = (2(-3) - (-10), -2) = (4, -2)$$ - For $F(-6,-2)$: $$F' = (2(-3) - (-6), -2) = (-6 + 6, -2) = (0, -2)$$ - For $G(-6,-8)$: $$G' = (0, -8)$$ 4. **Summary:** The reflected trapezoid D'E'F'G' has vertices at $D'(4,-10)$, $E'(4,-2)$, $F'(0,-2)$, and $G'(0,-8)$. This completes the reflection over the line $x = -3$.