To determine at what ordered pair the function \( f(x) = \sqrt{x + 2} - 2 \) will begin on the coordinate plane, we need to find where the function is defined and the value of the function at that point.
1. Identify the domain of the function:
- The function involves a square root, \(\sqrt{x + 2}\). The expression under the square root must be non-negative for the function to be defined.
- This implies \( x + 2 \geq 0 \).
- Solving for \( x \), we get \( x \geq -2 \).
2. Determine the value of the function where it starts:
- We know from the domain analysis that the smallest value \( x \) can take is \( x = -2 \).
3. Calculate \( f(x) \) at \( x = -2 \):
- Substitute \( x = -2 \) into the function \( f(x) \):
[tex]\[
f(-2) = \sqrt{-2 + 2} - 2 = \sqrt{0} - 2 = 0 - 2 = -2
\][/tex]
Therefore, the ordered pair at which the function [tex]\( f(x) = \sqrt{x + 2} - 2 \)[/tex] begins on the coordinate plane is [tex]\((-2, -2)\)[/tex].
