To answer this question, we can remember that when we reflect a point (x, y) across the x-axis, the point is (x, -y).
1. From the graph, we have that the coordinates of the figure are in Quadrant II. Since the figure will have the same values for x but will change the values from y to -y, then the figure will be reflected on Quadrant III.
2. For instance, if we have the points of the figure PQRS are:
• P(-3, 3)
,
• Q(-2, 3)
,
• R(-1, 1)
,
• S(-3, 2)
3. After a reflection across the x-axis, and following the rule for this transformation, we will have that the image for that pre-image will be:
• P(-3,3) ---> P'(-3, -(3)) ---> P'(-3, -3)
,
• Q(-2, 3) ---> Q'(-2, -3)
,
• R(-1, 1) ---> R'(-1, -1)
,
• S(-3, 2) ---> S'(-3, -2)
We can see Quadrant III above.
And we can see that all of the values for the coordinates pairs are negative, and this is possible in Quadrant III where the values for x and y are negative.
In summary, therefore, we can conclude that the image of figure PQRS will be located in Quadrant III (third option).