Let f : {(1,1), (2,3), (0,-1), (-1,-3)} be a function from z to z defined by f(x) = ax+b, for some integers a, b then (a, b) =
(-1,2)
(2,-1)
(3,-2)
(0,3)
fx=ax+b f1=1⇒a+b=1 f2=3⇒2a+b=3 solving a=2 b=-1