Answer:

Step-by-step explanation:


given D : (7,-3), and D' : (2,5)
the coordinates of D can be represented as (x1,y1), and the coordinates of D' can be represented as (x,y).
you can simply take the difference in the x values and difference in the y values from the preimage to image.
like this:
f'(x,y) → f(x+(x-x1),y+(y-y1)) : 
D'(x,y) → D(x+(2-7),y+(5--3))
D'(x,y) → D(x<u>-5</u>,y<u>+8</u>) : 