## Precalculus (6th Edition) Blitzer

RECALL: The point $(x, y)$ has: $x$ = directed distance of the point from the y-axis (positive when to the right of the y-axis, negative when to the left) $y$ = directed distance of the point from the x-axis (positive when above the x-axis, negative when below) Thus, the given point is: 4 units to the left of the y-axis 0 units from the x-axis (which means the point is on the x-axis) (refer to the attached image in the answer portion for the graph)