what is the pre - image of vertex a if the rule that created the image is $r_{y - axis}(x,y)\to(-x,y)$?\na(-4…

what is the pre - image of vertex a if the rule that created the image is $r_{y - axis}(x,y)\to(-x,y)$?\na(-4,2)\na(-2,-4)\na(2,4)\na(4,-2)

what is the pre - image of vertex a if the rule that created the image is $r_{y - axis}(x,y)\to(-x,y)$?\na(-4,2)\na(-2,-4)\na(2,4)\na(4,-2)

Answer

Answer:

A. $A(-4,2)$

Explanation:

Step1: Identify coordinates of $A'$

From the graph, $A'=(4,2)$.

Step2: Apply inverse transformation

The rule for reflection over $y -$axis is $(x,y)\to(-x,y)$. To find pre - image, we reverse it. If $(x,y)$ is pre - image and $(-x,y)$ is image, given $A'=(4,2)$ (image), then pre - image has $x=-4,y = 2$. So pre - image of $A'$ is $(-4,2)$.