# How do you find the domain and range of #sqrt(3t+12)#?

##### 1 Answer

May 1, 2016

#### Answer:

Domain:

Range:

#### Explanation:

Assuming we are talking about Real square roots, we have:

- The square root
#sqrt("expression")# is only defined if#"expression" >= 0# - The notation
#sqrt("expression")# denotes the principal, non-negative square root so is always#>= 0#

So we require

Dividing through by

#t >= -4#

So the domain is

If

#sqrt(3t+12) = y#

So the range includes

Since

So