1. The problem asks to find the translation rule that maps the pre-image parallelogram $\square ABCD$ to the image parallelogram $\square A'B'C'D'$.\n\n2. Translation moves every point of a figure the same distance in the same direction. The rule for translation is given by $T_{(h,k)}(x,y) = (x+h, y+k)$, where $h$ is the horizontal shift and $k$ is the vertical shift.\n\n3. From the given points, the pre-image point $A(-4,4)$ moves to $A'(-2,5)$. To find the translation vector, calculate the change in $x$ and $y$:\n$$h = -2 - (-4) = 2$$\n$$k = 5 - 4 = 1$$\n\n4. Therefore, the translation rule is $T_{(2,1)}(x,y) = (x+2, y+1)$.\n\n5. This matches the option $T_{(2,1)}(x,y)$, which means the figure is translated 2 units to the right and 1 unit up.\n\nFinal answer: \n$$T_{(2,1)}(x,y) = (x+2, y+1)$$