Question

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

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 $y = \sqrt{x + 5}$

For the function $y=\sqrt{x + 5}$, the expression inside the square root, $x + 5$, determines the domain, but for the range, we look at the output of the square root operation. Since the square root of any real number (when the expression inside is non - negative) is non - negative, the value of $y=\sqrt{x + 5}$ will be greater than or equal to 0. So the range of the function $y = \sqrt{x+5}$ is $y\geq0$.

Answer:

The correct option is the one with $y \geq 0$. (If we assume the options are labeled as follows: Let's say the first option (top - left) is A. $y\geq5$, the second option (top - middle) is B. $y\geq0$, the third option (top - right) is C. $y\geq\sqrt{5}$, and the fourth option (bottom - left) is D. $y\geq - 5$. Then the answer is B. $y\geq0$)