How do you find the domain and range of #y = sqrt(x+8)#?

1 Answer
Aug 12, 2015

Answer:

Domain is #x >= -8#

Range is #y >= 0#

Explanation:

Let's take a look at the equation first.

# y = sqrt(x+8)#

For the domain, we're interested in the values of x that give a "valid result".

In other words, we're looking for values of x that won't "break" the equation.

We notice that #x+8# is inside of a square root. We also know that anything inside of a square root must be non-negative, since you can't take the square root of a negative number.

Therefore, we can set this up.

#x+8 >=0#

subtract 8 from both sides to get,

#x >= -8# , which is the domain.

Now, for the range.

We know that the square root of a number cannot be negative. This is the same for function such as #x + 8#. So, we get:

#y >= 0#

It's worth noting here that the square root symbol is used to indicate only the positive root. For example,

#sqrt(4) = 2#

while if #x^2 = 4#, then #x = +-2#