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.<br /><br />Let's start with the first point in the domain: $(1,-2)$.<br />Substituting $x=1$ and $y=-2$ into the function, we get:<br />$f(1,-2) = (2(1)-3, 2(-2)+5) = (-1,1)$<br /><br />Now let's move on to the second point: $(3,-1)$.<br />Substituting $x=3$ and $y=-1$ into the function, we get:<br />$f(3,-1) = (2(3)-3, 2(-1)+5) = (3,3)$<br /><br />Finally, let's apply the function to the third point: $(4,-4)$.<br />Substituting $x=4$ and $y=-4$ into the function, we get:<br />$f(4,-4) = (2(4)-3, 2(-4)+5) = (5,-3)$<br /><br />So the range of the function is $(-1,1),(3,3),(5,-3)$.<br /><br />Therefore, the correct answer is B. $(-1,1),(3,3),(5,-3)$.