1. **State the problem:** We need to find the image of triangle $\triangle STU$ after reflecting it over the vertical line $x = -3$. 2. **Reflection rule:** When reflecting a point $(x,y)$ over the line $x = a$, the reflected point $(x',y')$ has the same $y$-coordinate and its $x'$ is given by: $$x' = 2a - x$$ 3. **Apply the reflection to each vertex:** - Let the coordinates of $S, T, U$ be $(x_S,y_S), (x_T,y_T), (x_U,y_U)$. - The reflected points are: $$S' = (2(-3) - x_S, y_S) = (-6 - x_S, y_S)$$ $$T' = (-6 - x_T, y_T)$$ $$U' = (-6 - x_U, y_U)$$ 4. **Explanation:** This formula ensures each point is mirrored across the line $x = -3$ by measuring the horizontal distance from the point to the line and placing the image the same distance on the opposite side. 5. **Final answer:** The image $\triangle S'T'U'$ has vertices at $S'(-6 - x_S, y_S)$, $T'(-6 - x_T, y_T)$, and $U'(-6 - x_U, y_U)$.