1. **State the problem:** We need to find the image of triangle $\triangle QRS$ after reflecting it over the vertical line $x = -3$. 2. **Recall the reflection rule:** When reflecting a point $(x, y)$ over the vertical line $x = a$, the image point $(x', y')$ satisfies: $$x' = 2a - x, \quad y' = y$$ This means the $y$-coordinate stays the same, and the $x$-coordinate is mirrored about $x = a$. 3. **Apply the reflection to each vertex:** - For $R(-10, 3)$: $$x'_R = 2(-3) - (-10) = -6 + 10 = 4, \quad y'_R = 3$$ So, $R' = (4, 3)$. - For $Q(-10, -2)$: $$x'_Q = 2(-3) - (-10) = 4, \quad y'_Q = -2$$ So, $Q' = (4, -2)$. - For $S(-6, -5)$: $$x'_S = 2(-3) - (-6) = -6 + 6 = 0, \quad y'_S = -5$$ So, $S' = (0, -5)$. 4. **Summary:** The reflected triangle $\triangle Q'R'S'$ has vertices: $$Q'(4, -2), \quad R'(4, 3), \quad S'(0, -5)$$ This completes the reflection over the line $x = -3$.