1. **State the problem:** We need to find the image of trapezoid JKLM after reflecting it over the vertical line $x = -1$. 2. **Reflection formula:** When reflecting a point $(x,y)$ over the line $x = a$, the image point $(x',y')$ is given by: $$x' = 2a - x, \quad y' = y$$ 3. **Apply the formula to each vertex:** - For $J(-8, 2)$: $$x'_J = 2(-1) - (-8) = -2 + 8 = 6, \quad y'_J = 2$$ - For $K(-6, 2)$: $$x'_K = 2(-1) - (-6) = -2 + 6 = 4, \quad y'_K = 2$$ - For $L(-3, 4)$: $$x'_L = 2(-1) - (-3) = -2 + 3 = 1, \quad y'_L = 4$$ - For $M(-9, 4)$: $$x'_M = 2(-1) - (-9) = -2 + 9 = 7, \quad y'_M = 4$$ 4. **Final image vertices:** $$J' = (6, 2), \quad K' = (4, 2), \quad L' = (1, 4), \quad M' = (7, 4)$$ 5. **Explanation:** Each point is flipped horizontally across the line $x = -1$, so the $y$-coordinates remain the same while the $x$-coordinates are adjusted according to the reflection formula. This completes the reflection of trapezoid JKLM over the line $x = -1$.