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 & (-2,4) & f \hline g & (-5,2) & g \hlineend{array}) (type ordered pairs.)

Explanation:

Step1: Identify the reflection rule

Looking at the graph, the reflection is over the \(x -\)axis. The rule for reflecting a point \((x,y)\) 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 substitute \(x=-3\) and \(y = 4\). So \(E'(-3,-4)\).

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

For point \(F(-2,4)\), using the rule \((x,y)\to(x, - y)\), we substitute \(x=-2\) and \(y = 4\). So \(F'(-2,-4)\).

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

For point \(G(-5,2)\), using the rule \((x,y)\to(x, - y)\), we substitute \(x=-5\) and \(y = 2\). So \(G'(-5,-2)\).

Answer:

\(\triangle EFG\)	\(E(-3,4)\)	\(E'(-3,-4)\)
	\(G(-5,2)\)	\(G'(-5,-2)\)