# What is the domain and range of  y= sqrt(x) -2?

Jan 23, 2017

$\text{Domain=} {\mathbb{R}}^{=} \cup \left\{0\right\} = \left[0 , \infty\right) .$

$\text{Range=} \left[- 2 , \infty\right) .$

#### Explanation:

We will restrict our discussion in $\mathbb{R} .$

As we can not find the square root of $x < 0 , x \ge 0$

So, the Domain is the set of all non-negative reals, i.e.,

${\mathbb{R}}^{+} \cup \left\{0\right\} = \left[0 , \infty\right) .$

Also, $\forall x \in {\mathbb{R}}^{+} \cup \left\{0\right\} , \sqrt{x} \ge 0 \Rightarrow y = \sqrt{x} - 2 \ge - 2.$

Hence, the Range is $\left[- 2 , \infty\right) .$

Enjoy maths.!