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)