Question

write the coordinates of the vertices after a reflection over the y - axis.
q((□,□))
r((□,□))
s((□,□))
t((□,□))

Explanation:

Step1: Find original coordinates

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

• \( Q \) is at \( (6, 0) \)
• \( R \) is at \( (9, 0) \) (Wait, looking at the graph, \( R \) is at \( (9, 0) \)? Wait, the x-axis: Q is at x=6, y=0; R is at x=9, y=0? Wait, no, looking at the grid, Q is at (6,0), T is at (6,2), S is at (9,2), R is at (9,0). Wait, the x-axis labels: 6, 8, 10. So Q is (6,0), T is (6,2), S is (9,2), R is (9,0).

Step2: Apply reflection over y-axis rule

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

For \( Q(6, 0) \):

Apply the rule: \( x = 6 \to -6 \), \( y = 0 \) remains. So \( Q'(-6, 0) \).

For \( R(9, 0) \):

Apply the rule: \( x = 9 \to -9 \), \( y = 0 \) remains. So \( R'(-9, 0) \).

For \( S(9, 2) \):

Apply the rule: \( x = 9 \to -9 \), \( y = 2 \) remains. So \( S'(-9, 2) \).

For \( T(6, 2) \):

Apply the rule: \( x = 6 \to -6 \), \( y = 2 \) remains. So \( T'(-6, 2) \).

Answer:

\( Q'(-6, 0) \)
\( R'(-9, 0) \)
\( S'(-9, 2) \)
\( T'(-6, 2) \)