1. The problem asks to identify the algebraic rule that transforms parallelogram GHJK to G'H'J'K'. 2. The given points for GHJK are approximately G(-4,-6), H(-8,-4), J(-6,-3), K(-2,-5). 3. The image points G'H'J'K' are approximately G'(6,-4), H'(4,-8), J'(3,-6), K'(5,-2). 4. To find the transformation, we analyze the coordinates and their changes. 5. Notice that the original parallelogram is in the lower-left quadrant (negative x-values), and the image is in the lower-right quadrant (positive x-values). 6. The transformation appears to reflect points across the y-axis, changing the sign of the x-coordinate while keeping the y-coordinate the same. 7. The algebraic rule for reflection across the y-axis is: $$ (x,y) \to (-x,y) $$ 8. Checking a point: G(-4,-6) reflected across y-axis becomes G'(4,-6), but the image point is G'(6,-4), so this is not a simple reflection. 9. Next, consider rotation. The problem suggests options including rotations $R_{0,90^\circ}$ and $R_{0,270^\circ}$. 10. Rotation $R_{0,90^\circ}$ about the origin transforms $(x,y)$ to $(-y,x)$. 11. Applying $R_{0,90^\circ}$ to G(-4,-6): $$ (-(-6), -4) = (6, -4) $$ which matches G'(6,-4). 12. Similarly, H(-8,-4) under $R_{0,90^\circ}$: $$ (-(-4), -8) = (4, -8) $$ matches H'(4,-8). 13. J(-6,-3) under $R_{0,90^\circ}$: $$ (-(-3), -6) = (3, -6) $$ matches J'(3,-6). 14. K(-2,-5) under $R_{0,90^\circ}$: $$ (-(-5), -2) = (5, -2) $$ matches K'(5,-2). 15. Therefore, the transformation is a rotation of 90 degrees about the origin. 16. The algebraic rule describing this transformation is: $$ (x,y) \to (-y,x) $$ Final answer: The transformation is $R_{0,90^\circ}$, a rotation of 90 degrees about the origin.