# What is the domain and range of y = sqrt(x^2 + 2x + 3)?

##### 1 Answer
Apr 25, 2016

With radical functions the argument under the root-sign and the outcome are always non-negative (in real numbers).

#### Explanation:

Domain :
The argument under the root sign must be non-negative:
We 'translate' by completing the square:

${x}^{2} + 2 x + 3 = \left({x}^{2} + 2 x + 1\right) + 2 = {\left(x + 1\right)}^{2} + 2$

Which is always $\ge 2$ for every value of $x$

So there are no restrictions to $x$:
$x \in \left(- \infty , + \infty\right)$

Range:
Since the lowest value the argument can take is $2$, the lowest value of $y = \sqrt{2}$, so:
$y \in \left[\sqrt{2} , + \infty\right)$