Question

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

Answer 1

The domain for `y` is `(-oo,+oo)` Any value is possible, as the denominator will never be `0`, actually never `<16`.

Explanation:

The range of the function will be `[-1,+1)` where `(0,-1)` is the minimum and `y=1` is the horizontal asymptote.
graph{(x^2-16)/(x^2+16) [-11.25, 11.25, -5.625, 5.625]}