write the coordinates of the vertices after a reflection over the x-axis.

Answer

Explanation:

Step1: Identify original coordinates

First, find the coordinates of the original vertices. From the graph:

( I ): ( (1, 0) )
( U ): ( (-2, -5) ) (Wait, no, looking at the grid, let's recheck. Wait, the y-axis: let's see the grid. Wait, the point ( U ): x=-2, y=-5? Wait, no, the grid lines: each square is 1 unit. Let's check again. Wait, the point ( I ) is at (1, 0)? Wait, no, the origin is (0,0), and ( I ) is at (1, 0)? Wait, no, looking at the graph, ( I ) is at (1, 0)? Wait, no, the x-axis is horizontal, y-axis vertical. Let's list the vertices:
( I ): (1, 0) (on x-axis, x=1, y=0)
( U ): (-2, -5)? Wait, no, the point ( U ) is at x=-2, y=-5? Wait, no, the grid: from x=-2, y=-5? Wait, no, the y-coordinate for ( U ): looking at the vertical lines, the point ( U ) is at x=-2, y=-5? Wait, no, maybe I misread. Wait, the point ( R ) is at (1, -10)? Wait, no, the bottom point ( R ) is at (1, -10)? Wait, no, the y-axis: the bottom arrow is -10, so ( R ) is at (1, -10)? Wait, no, let's check again.

Wait, correct original coordinates:

( I ): (1, 0) (x=1, y=0)
( U ): (-2, -5) – no, wait, the point ( U ) is at x=-2, y=-5? Wait, no, the grid: each square is 1 unit. Let's see the vertical line x=-2, and horizontal line y=-5? Wait, no, the point ( U ) is at x=-2, y=-5? Wait, maybe I made a mistake. Wait, the point ( S ) is at (8, -5)? Wait, no, the right point ( S ) is at (8, -5)? Wait, no, looking at the graph, ( S ) is at (8, -5)? Wait, no, the vertical line x=8, y=-5? Wait, no, the grid: x=8, y=-5? Wait, maybe the original vertices are:
( I ): (1, 0)
( U ): (-2, -5)
( R ): (1, -10)
( S ): (8, -5)

Wait, let's confirm. The figure is a quadrilateral with vertices ( I ), ( U ), ( R ), ( S ).

Now, reflection over x-axis: the rule for reflection over x-axis is ( (x, y) \rightarrow (x, -y) ).

So let's apply the rule:

For ( I ): (1, 0) → (1, -0) = (1, 0) (since -0=0)

For ( U ): Let's recheck original ( U ). Wait, maybe I misread ( U )'s coordinates. Wait, the point ( U ): x=-2, y=-5? No, wait, the grid: the point ( U ) is at x=-2, y=-5? Wait, no, the y-coordinate for ( U ): looking at the horizontal lines, the point ( U ) is at y=-5? Wait, maybe the original coordinates are:

( I ): (1, 0)
( U ): (-2, -5)
( R ): (1, -10)
( S ): (8, -5)

Wait, no, that can't be. Wait, the figure is a quadrilateral, so four vertices: ( I ), ( U ), ( R ), ( S ).

Wait, let's look at the graph again. The top vertex ( I ) is at (1, 0) (on x-axis). Then ( U ) is at (-2, -5) (left side), ( R ) is at (1, -10) (bottom), ( S ) is at (8, -5) (right side). Yes, that makes a quadrilateral.

Now, reflection over x-axis: the rule is ( (x, y) \rightarrow (x, -y) ).

So:

( I ): (1, 0) → (1, -0) = (1, 0)
( U ): (-2, -5) → (-2, -(-5)) = (-2, 5)
( R ): (1, -10) → (1, -(-10)) = (1, 10)
( S ): (8, -5) → (8, -(-5)) = (8, 5)

Wait, that makes sense. So original coordinates:

( I ): (1, 0)
( U ): (-2, -5)
( R ): (1, -10)
( S ): (8, -5)

Now, apply reflection over x-axis (rule: (x, y) → (x, -y)):

Step2: Apply reflection rule

For each vertex:

( I ): (1, 0) → (1, -0) = (1, 0)
( U ): (-2, -5) → (-2, 5) (since -y = -(-5) = 5)
( R ): (1, -10) → (1, 10) (since -y = -(-10) = 10)
( S ): (8, -5) → (8, 5) (since -y = -(-5) = 5)

Wait, but let's confirm the original coordinates again. Maybe I misread ( U ) and ( S ). Let's check the graph again. The point ( U ) is at x=-2, y=-5? Wait, the vertical line x=-2, and the horizontal line y=-5 (since each grid is 1 unit). Then ( S ) is at x=8, y=-5. ( R ) is at x=1, y=-10. ( I ) is at x=1, y=0. Yes, that seems correct.

So after reflection over x-axis, the coordinates are:

( I' ): (1, 0)
( U' ): (-2, 5)
( R' ): (1, 10)
( S' ): (8, 5)

Answer:

The coordinates of the vertices after reflection over the x - axis are ( I(1, 0) ), ( U(-2, 5) ), ( R(1, 10) ), and ( S(8, 5) ).