## Intermediate Algebra (6th Edition)

We are given that $f={(-5,5),(0,4),(13,5),(11,-6)}$. A function is one-to-one if each x-value (input) corresponds to only one y-value (output) and also each y-value (output) corresponds to only one x-value (input). Therefore, this function is not one-to-one, because the y-value 5 corresponds to more than one x-value (-5 and 13).