(1, 2) (3, 8) (-3, 6)

(1, 2) (8, 3) (-3, 6)

(-1, -2) (-3, -8) (3, -6)

(2, 1) (8, 3) (6, -3)

yes

no

no way to tell

y = x

y =1

y = 0

y = x + 1

 $f^{-1}\left(x\right)=\frac{1}{x+1}$  

 $f^{-1}\left(x\right)=\frac{1}{x-1}$  

 $f^{-1}\left(x\right)=\frac{x}{x+1}$