Question

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

$r(\square,\square)$

$s(\square,\square)$

$t(\square,\square)$

$u(\square,\square)$

Explanation:

Step1: Recall the reflection rule over the x - axis

When a point \((x,y)\) is reflected over the \(x\) - axis, the rule is \((x,y)\to(x, - y)\).

Step2: Find the coordinates of \(R\)

From the graph, the coordinates of \(R\) are \((-4,-9)\). Using the reflection rule \((x,y)\to(x, - y)\), for \(R(-4,-9)\), we get \(R'(-4,9)\) since \(y=-9\) and \(-y = 9\).

Step3: Find the coordinates of \(S\)

From the graph, the coordinates of \(S\) are \((5,-9)\). Using the reflection rule \((x,y)\to(x, - y)\), for \(S(5,-9)\), we get \(S'(5,9)\) since \(y = - 9\) and \(-y=9\).

Step4: Find the coordinates of \(T\)

From the graph, the coordinates of \(T\) are \((5,0)\). Using the reflection rule \((x,y)\to(x, - y)\), for \(T(5,0)\), we get \(T'(5,0)\) since \(y = 0\) and \(-y = 0\).

Step5: Find the coordinates of \(U\)

From the graph, the coordinates of \(U\) are \((-4,0)\). Using the reflection rule \((x,y)\to(x, - y)\), for \(U(-4,0)\), we get \(U'(-4,0)\) since \(y = 0\) and \(-y=0\).

Answer:

\(R'(-4,9)\)
\(S'(5,9)\)
\(T'(5,0)\)
\(U'(-4,0)\)