Question

for your turn #2 - 3, write the coordinate rule that maps each preimage to its image. then identify the transformation. record the preimage and image on the graph. #2 is partly done for you.

1. (\triangle abc) maps to (\triangle abc).

preimage
(a(2,6))
(b(-2,6))
(c(2,-3))
image
(a(4,2))
(b(-4,2))
(c(4,-1))
the coordinate rule is:
((x,y)\to(2x,\frac{1}{3}y))
the transformation is:
dilation with a horizontal stretch by a factor of 2
and a vertical compression by a factor of (\frac{1}{3})

Explanation:

Step1: Analyze the x - coordinate transformation

For point \(A(2,6)\) to \(A'(4,2)\), the \(x\) - coordinate changes from \(x = 2\) to \(x'=4\). The formula for the \(x\) - coordinate transformation is \(x'=2x\). For point \(B(- 2,6)\) to \(B'(-4,2)\), when \(x=-2\), \(x' = 2\times(-2)=-4\). For point \(C(2,-3)\) to \(C'(4,-1)\), when \(x = 2\), \(x'=2\times2 = 4\).

Step2: Analyze the y - coordinate transformation

For point \(A(2,6)\) to \(A'(4,2)\), the \(y\) - coordinate changes from \(y = 6\) to \(y'=2\). The formula for the \(y\) - coordinate transformation is \(y'=\frac{1}{3}y\). For point \(B(-2,6)\) to \(B'(-4,2)\), when \(y = 6\), \(y'=\frac{1}{3}\times6 = 2\). For point \(C(2,-3)\) to \(C'(4,-1)\), when \(y=-3\), \(y'=\frac{1}{3}\times(-3)=-1\).

Answer:

The coordinate rule is \((x,y)\to(2x,\frac{1}{3}y)\). The transformation is a dilation with a horizontal stretch factor of \(2\) and a vertical compression factor of \(\frac{1}{3}\).