Question

show how each vertex of \\( \triangle efg \\) maps to its image. describe in words how to map \\( \triangle efg \\) to its image \\( \triangle efg \\) using a reflection.
complete the table with the coordinates of the image points \\( e \\), \\( f \\), and \\( g \\)
\\( \

$$\begin{array} { | c | c | c | } \\hline & \\triangle efg & \\triangle efg \\\\ \\hline e & ( - 3,4 ) & e \\\\ \\hline f & ( - 2,3 ) & f \\\\ \\hline g & ( - 4,2 ) & g \\\\ \\hline \\end{array}$$

\\)
(type ordered pairs.)

Explanation:

Step1: Determine the reflection rule

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

Step2: Apply the rule to point \(E\)

For point \(E(-3,4)\), using the rule \((x,y)\to(x, - y)\), we get \(E'(-3,-4)\).

Step3: Apply the rule to point \(F\)

For point \(F(-2,3)\), using the rule \((x,y)\to(x, - y)\), we get \(F'(-2,-3)\).

Step4: Apply the rule to point \(G\)

For point \(G(-4,2)\), using the rule \((x,y)\to(x, - y)\), we get \(G'(-4,-2)\).

Answer:

\(\triangle EFG\)	\(\triangle E'F'G'\)
\(F(-2,3)\)	\(F'(-2,-3)\)
\(G(-4,2)\)	\(G'(-4,-2)\)

To map \(\triangle EFG\) to \(\triangle E'F'G'\), reflect \(\triangle EFG\) over the \(x -\)axis.