The domain of the coordinate transformation function f(x,y)=(2x-3,2y+5) is (1,-2),(3,-1),(4,-4) What is the range of the function? A. (-2,3),(0,4),(1,1) B (-1,1),(3,3),(5,-3) C. (2,-4),(6,-2),(8,-8) D (-7,7),(-5,11),(-11,13)

To find the range of the coordinate transformation function $$f(x,y)=(2x-3,2y+5)$$ , we need to apply the function to each point in the given domain and see what the resulting coordinates are.

Let's start with the first point in the domain: $$(1,-2)$$ .
Substituting $$x=1$$ and $$y=-2$$ into the function, we get:
$$f(1,-2) = (2(1)-3, 2(-2)+5) = (-1,1)$$

Now let's move on to the second point: $$(3,-1)$$ .
Substituting $$x=3$$ and $$y=-1$$ into the function, we get:
$$f(3,-1) = (2(3)-3, 2(-1)+5) = (3,3)$$

Finally, let's apply the function to the third point: $$(4,-4)$$ .
Substituting $$x=4$$ and $$y=-4$$ into the function, we get:
$$f(4,-4) = (2(4)-3, 2(-4)+5) = (5,-3)$$

So the range of the function is $$(-1,1),(3,3),(5,-3)$$ .

Therefore, the correct answer is B. $$(-1,1),(3,3),(5,-3)$$ .