# How do you find the restricted value for (y^2-16)/(y^2+16)?

The domain for $y$ is $\left(- \infty , + \infty\right)$ Any value is possible, as the denominator will never be $0$, actually never $< 16$.
The range of the function will be $\left[- 1 , + 1\right)$ where $\left(0 , - 1\right)$ is the minimum and $y = 1$ is the horizontal asymptote.