What is the domain of #y=sqrt(x+16)#?

1 Answer
Jun 18, 2018

#x in [-16, infty)#

Explanation:

The domain is restricted by where the quantity #x+16 >=0#

This means that #x >= -16#

There is no restriction on how large #x# can be, as the quantity is always positive.

So the domain is

#x in [-16, infty)#