How do I reflect over x=3 ??
This last claim is because x can be thought of as 2+a, where "a" could be negative! Now x needs to go to 2-a, but solveing for "a" in x=2+a gives a=x-2, so the new x is 2-a = 2-(x-2) = 4-x. (-1,1) => (5,1) (0,4) => (4,4) (-3,4) => (7,4)