Question

harley 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?
left of the ( y )-axis
quadrant i
quadrant ii
above the ( x )-axis

Explanation:

Step1: Analyze rotation

When a figure in Quadrant I is rotated \(90^{\circ}\) clockwise about the origin, it moves to Quadrant IV.

Step2: Analyze reflection

Reflecting a figure in Quadrant IV over the \(x -\)axis moves it to Quadrant I.

Step3: Analyze translation

Translating 3 units to the left moves the figure from Quadrant I (original \(x>0\)) to \(x> - 3\) (still in the right - hand side of the \(y -\)axis if original \(x\geq3\), but more generally, translation 3 units left just changes the \(x -\)coordinate: \(x'=x - 3\)). Translating 3 units up changes the \(y -\)coordinate (\(y'=y + 3\)). The figure is still above the \(x -\)axis.

Answer:

above the \(x -\)axis