Question

what is the range of the function $y = \sqrt{x + 5}$?
options:
$y \geq 5$
$y \geq \sqrt{5}$
$y \geq -5$
$y \geq 0$

Explanation:

Step1: Recall square root property

The square root function \( \sqrt{x} \) has a range of \( y \geq 0 \) because the square root of a number (in real numbers) is always non - negative.

Step2: Analyze the given function

For the function \( y=\sqrt{x + 5}\), let \( u=x + 5\). Then \( y = \sqrt{u}\). The expression inside the square root, \(u=x + 5\), affects the domain, but for the range, we focus on the square root part. Since the square root of any real number \(u\) (where \(u\geq0\) for the function to be real - valued) is non - negative, \(y=\sqrt{u}\geq0\). So the range of \(y = \sqrt{x+5}\) is \(y\geq0\).

Answer:

\( y\geq0 \) (the option with \( y\geq0 \))