1. **State the problem:** We have triangle \(\triangle ABC\) with vertices \(A(-2, 2)\), \(B(-8, 7)\), and \(C(-7, -2)\). We apply a sequence of transformations: rotate 90° about the origin, dilate by scale factor 3 about the origin, reflect across the y-axis, then translate by \((x, y) \to (x - 4, y + 8)\). We want the coordinates of the transformed vertices \(A'\), \(B'\), and \(C'\). 2. **Rotation 90° about the origin:** The formula for rotating a point \((x, y)\) 90° counterclockwise about the origin is \((x, y) \to (-y, x)\). 3. **Apply rotation to each vertex:** - \(A(-2, 2) \to A_1(-2, -2)\) because \((-y, x) = (-2, -2)\) - \(B(-8, 7) \to B_1(-7, -8)\) - \(C(-7, -2) \to C_1(2, -7)\) 4. **Dilation by scale factor 3 about the origin:** The formula is \((x, y) \to (3x, 3y)\). 5. **Apply dilation:** - \(A_1(-2, -2) \to A_2(-6, -6)\) - \(B_1(-7, -8) \to B_2(-21, -24)\) - \(C_1(2, -7) \to C_2(6, -21)\) 6. **Reflection across the y-axis:** The formula is \((x, y) \to (-x, y)\). 7. **Apply reflection:** - \(A_2(-6, -6) \to A_3(6, -6)\) - \(B_2(-21, -24) \to B_3(21, -24)\) - \(C_2(6, -21) \to C_3(-6, -21)\) 8. **Translation by \((x, y) \to (x - 4, y + 8)\):** 9. **Apply translation:** - \(A_3(6, -6) \to A'(6 - 4, -6 + 8) = (2, 2)\) - \(B_3(21, -24) \to B'(21 - 4, -24 + 8) = (17, -16)\) - \(C_3(-6, -21) \to C'(-6 - 4, -21 + 8) = (-10, -13)\) **Final answer:** \[ A' = (2, 2), \quad B' = (17, -16), \quad C' = (-10, -13) \] This completes the transformation sequence and gives the new vertices of \(\triangle A'B'C'\).