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, its coordinates \((x,y)\) change to \((y, -x)\). So it moves to Quadrant IV.

Step2: Analyze reflection

Reflecting a figure in Quadrant IV over the \(x -\)axis. If a point has coordinates \((a,b)\) in Quadrant IV (\(a>0,b <0\)), after reflection over the \(x -\)axis, the coordinates become \((a,-b)\) (now \(a>0,-b>0\)), so it moves to Quadrant I.

Step3: Analyze translation

Translating the figure \(3\) units to the left (\(x=x - 3\)) and \(3\) units up (\(y=y + 3\)). The \(y -\)coordinate of the figure (which was positive after reflection) remains positive (since we add \(3\) to it). The \(x -\)coordinate may still be non - negative depending on the original \(x\) values (but the key is the position relative to the \(x -\)axis). After all the transformations (rotation, reflection, translation), the figure is above the \(x -\)axis.

Answer:

above the \(x -\)axis