Question

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

Explanation:

Step1: Find original coordinates

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

• \( C(-6, -8) \)
• \( D(-3, -8) \) (Wait, looking at the graph, the x-coordinate of D: from the grid, E is at (-3, -1)? Wait no, let's recheck. Wait the graph: F is at (-6, -1)? Wait no, the y-axis: the points are at y=-1? Wait no, the green points: F is at (-6, -1)? Wait no, the vertical lines: F is at (-6, -1)? Wait no, the grid: each square is 1 unit. Let's look again. The points: C is at (-6, -8), D is at (-3, -8), E is at (-3, -1), F is at (-6, -1). Wait, the y-coordinate: when x is -6, y is -8 for C? Wait no, the bottom point C: looking at the y-axis, the bottom is -10, so C is at (-6, -8), D at (-3, -8), E at (-3, -1), F at (-6, -1).

Step2: Apply reflection over y-axis

The rule for reflection over the y-axis is \( (x, y) \to (-x, y) \).

For \( C(-6, -8) \):

Apply the rule: \( x = -6 \to -(-6) = 6 \), \( y = -8 \) remains. So \( C'(6, -8) \).

For \( D(-3, -8) \):

Apply the rule: \( x = -3 \to -(-3) = 3 \), \( y = -8 \) remains. So \( D'(3, -8) \).

For \( E(-3, -1) \):

Apply the rule: \( x = -3 \to -(-3) = 3 \), \( y = -1 \) remains. So \( E'(3, -1) \).

For \( F(-6, -1) \):

Apply the rule: \( x = -6 \to -(-6) = 6 \), \( y = -1 \) remains. So \( F'(6, -1) \).

Wait, maybe I misread the original coordinates. Let's re-express the original points:

Looking at the graph:

• Point F: x=-6, y=-1 (since it's on the line y=-1, x=-6)
• Point E: x=-3, y=-1 (x=-3, y=-1)
• Point D: x=-3, y=-8 (x=-3, y=-8)
• Point C: x=-6, y=-8 (x=-6, y=-8)

Yes, that's correct. So reflection over y-axis: (x,y) → (-x, y).

So:

• \( C(-6, -8) \) → \( C'(6, -8) \)
• \( D(-3, -8) \) → \( D'(3, -8) \)
• \( E(-3, -1) \) → \( E'(3, -1) \)
• \( F(-6, -1) \) → \( F'(6, -1) \)

Answer:

\( C'(6, -8) \)
\( D'(3, -8) \)
\( E'(3, -1) \)
\( F'(6, -1) \)