1. **State the problem:** We need to graph the image of square EFGH after reflecting it over the y-axis. 2. **Reflection rule:** When reflecting a point $(x,y)$ over the y-axis, the image point becomes $(-x,y)$. 3. **Apply the rule to each vertex:** - Let the vertices of square EFGH be $E(x_1,y_1)$, $F(x_2,y_2)$, $G(x_3,y_3)$, and $H(x_4,y_4)$ in the bottom-right quadrant (where $x>0$, $y<0$). - After reflection, the vertices become $E'(-x_1,y_1)$, $F'(-x_2,y_2)$, $G'(-x_3,y_3)$, and $H'(-x_4,y_4)$. 4. **Explanation:** Reflecting over the y-axis changes the sign of the x-coordinate but keeps the y-coordinate the same. 5. **Result:** The reflected square E'F'G'H' will be in the bottom-left quadrant, symmetric to the original square EFGH. Final answer: The image of square EFGH after reflection over the y-axis has vertices $E'(-x_1,y_1)$, $F'(-x_2,y_2)$, $G'(-x_3,y_3)$, and $H'(-x_4,y_4)$, which is the mirror image of the original square across the y-axis.