# How do you find the domain and range for y = -sqrt ( x + 3)?

May 7, 2018

The domain is $\left\{x \setminus \in \setminus m a t h \boldsymbol{R} : x \setminus \ge q - 3\right\}$

The range is $\left(- \infty , 0\right]$

#### Explanation:

The domain is $\left\{x \setminus \in \setminus m a t h \boldsymbol{R} : x \setminus \ge q - 3\right\}$

In fact, an even root exists if and only if its content is non negative. So, we must impose

$x + 3 \setminus \ge q 0 \setminus \iff x \setminus \ge q - 3$

As for the range, we can observe a some easy things:

• A square root is always positive, when it exists
• A square root is zero only when its content is zero. In our case, this happens only when $x = - 3$
• A square root grows infinitely as its content grows infinitely

So, we know that $\sqrt{x + 3}$ starts at zero for $x = - 3$, and then grows indefinitely, since $x + 3$ grows indefinitely. Its range is thus $\left[0 , \setminus \infty\right)$. But since we have a minus sign in front of the square root, we must reflect the range, obtaining $\left(- \infty , 0\right]$