Question

write the coordinates of the vertices after a reflection over the y-axis.
graph with vertices s, t, u, r on a coordinate grid, and boxes for r(□, □), s(□, □), t(□, □), u(□, □)

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of each vertex from the graph:

• \( R(-9, -4) \)
• \( S(-9, 1) \) (Wait, looking at the graph, S is at (-9, 1)? Wait, no, the y-axis: S is at y=1? Wait, the grid: let's check again. The x-axis: S is at x=-9, y=1? Wait, no, the blue square: S is at (-9, 1)? Wait, no, the y-coordinate: the horizontal line for S and T is at y=1? Wait, the grid lines: each square is 1 unit. So S is at (-9, 1)? Wait, no, the graph shows S at (-9, 1)? Wait, no, the y-axis: the line for S and T is at y=1? Wait, the vertical axis: 0 is the x-axis, so above x-axis is positive y. So S is (-9, 1), T is (-4, 1), U is (-4, -4), R is (-9, -4). Wait, let's confirm:

Looking at the graph:

• S: x=-9, y=1 (since it's 9 units left of y-axis, 1 unit above x-axis)
• T: x=-4, y=1 (4 units left of y-axis, 1 unit above x-axis)
• U: x=-4, y=-4 (4 units left of y-axis, 4 units below x-axis)
• R: x=-9, y=-4 (9 units left of y-axis, 4 units below x-axis)

Step2: Apply reflection over y-axis

The rule for reflection over the y-axis is \( (x, y) \to (-x, y) \). So we change the sign of the x-coordinate, keep y the same.

For \( R(-9, -4) \):
New x: \( -(-9) = 9 \), y remains -4. So \( R'(9, -4) \)

For \( S(-9, 1) \):
New x: \( -(-9) = 9 \), y remains 1. So \( S'(9, 1) \)

For \( T(-4, 1) \):
New x: \( -(-4) = 4 \), y remains 1. So \( T'(4, 1) \)

For \( U(-4, -4) \):
New x: \( -(-4) = 4 \), y remains -4. So \( U'(4, -4) \)

Wait, wait, maybe I made a mistake in original coordinates. Let's recheck the graph. The x-axis: from -10 to 10, y-axis from -10 to 10. The blue square: S is at x=-9, y=1? Wait, no, the horizontal line for S and T is at y=1? Wait, the y-coordinate: the grid lines, each square is 1 unit. So S is at (-9, 1)? Wait, no, the y-axis: 0 is the x-axis, so S is at (-9, 1), T at (-4, 1), U at (-4, -4), R at (-9, -4). Yes, that's correct.

So applying reflection over y-axis (x becomes -x, y same):

• R: (-9, -4) → (9, -4)
• S: (-9, 1) → (9, 1)
• T: (-4, 1) → (4, 1)
• U: (-4, -4) → (4, -4)

Answer:

\( R'(9, -4) \)
\( S'(9, 1) \)
\( T'(4, 1) \)
\( U'(4, -4) \)