Question

c)

\\( f(x) = \sqrt{x + 2} - 2 \\)

\\( f(x) = \sqrt{x + 2} \\)

\\( f(x) = -\sqrt{x + 2} + 2 \\)

\\( f(x) = -\sqrt{x + 2} - 2 \\)

Explanation:

Step1: Identify domain start

The graph starts at $x=-2$, so the radicand must be $x+2$ (since $x+2\geq0 \implies x\geq-2$). All options satisfy this.

Step2: Check vertex point

The vertex of the graph is at $(-2, -2)$. Substitute $x=-2$ into each option:

• For $f(x)=\sqrt{x+2}-2$: $f(-2)=\sqrt{-2+2}-2=0-2=-2$, matches.
• For $f(x)=\sqrt{x+2}$: $f(-2)=\sqrt{-2+2}=0

eq-2$, eliminates.

• For $f(x)=-\sqrt{x+2}+2$: $f(-2)=-\sqrt{-2+2}+2=0+2=2

eq-2$, eliminates.

• For $f(x)=-\sqrt{x+2}-2$: $f(-2)=-\sqrt{-2+2}-2=0-2=-2$, vertex matches, but check direction.

Step3: Check increasing trend

The graph rises as $x$ increases. For $f(x)=\sqrt{x+2}-2$, the square root term is positive, so the function increases as $x$ increases. For $f(x)=-\sqrt{x+2}-2$, the negative square root term makes the function decrease as $x$ increases, which does not match the graph.

Answer:

$\boldsymbol{f(x) = \sqrt{x+2} - 2}$