Question

what is the image point of $(-4,2)$ after the transformation $d_{3}circ t_{3,-2}$?

Explanation:

Step1: Apply translation \(T_{3,-2}\)

For a point \((x,y)\), the translation \(T_{a,b}\) gives \((x + a,y + b)\).
For the point \((-4,2)\) and \(a = 3,b=-2\), we have \((-4+3,2+( - 2))=(-1,0)\).

Step2: Apply dilation \(D_3\)

For a point \((x,y)\), the dilation \(D_k\) (where \(k\) is the scale factor) gives \((kx,ky)\).
Since \(k = 3\) and the point after translation is \((-1,0)\), we get \((3\times(-1),3\times0)=(-3,0)\).

Answer:

\((-3,0)\)