Warley graphs a polygon that is located entirely inside quadrant I . He rotates the figure clockwise \( 90 \circ \) about the origin and then reflects the rotated figure over the \( x \)-axis. He translates the resulting figure 3 units to the left and 3 units up Which best describes the location of the final image? quadrant I quadrant II above the \( x \)-axis left of the \( y \)-axis Mark this and return

To determine the location of the final image after Warley's transformations, let's analyze each step: 1. Warley graphs a polygon that is located entirely inside quadrant I. 2. He rotates the figure clockwise \( 90 \circ \) about the origin. 3. Then, he reflects the rotated figure over the \( x \)-axis. 4. Finally, he translates the resulting figure 3 units to the left and 3 units up. After the rotation, the polygon will be in quadrant IV. After the reflection over the \( x \)-axis, the polygon will be in quadrant III. After the translation 3 units to the left and 3 units up, the polygon will be in quadrant II. Therefore, the final image will be located in quadrant II, above the \( x \)-axis, and left of the \( y \)-axis. So, the best description of the location of the final image is quadrant II, above the \( x \)-axis, and left of the \( y \)-axis.