## Intermediate Algebra (6th Edition)

One-to-one, $h^{-1}={(14,-9),(8,6),(12,-11),(15,15)}$
We are given that $h={(-9,14),(6,8),(-11,12),(15,15)}$. 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 one-to-one. To find the elements of its inverse, we interchange the coordinates of each point. For example, $(-9,14)$ becomes $(14,-9)$. Therefore, $h^{-1}={(14,-9),(8,6),(12,-11),(15,15)}$.