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

Aug 12, 2015

Domain is $x \ge - 8$

Range is $y \ge 0$

#### 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.

$x + 8 \ge 0$

subtract 8 from both sides to get,

$x \ge - 8$ , 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:

$y \ge 0$

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 $x = \pm 2$