Question

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

Explanation:

Step1: Identify the starting point

The graph starts at $x=-2$, $f(-2)=0$. Substitute $x=-2$ into each option:

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

eq 0$

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

eq 0$

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

eq 0$

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

Step2: Verify a second point

Take $x=2$. From the graph, $f(2)\approx2$. Substitute $x=2$ into $f(x)=\sqrt{x+2}$:
$f(2)=\sqrt{2+2}=\sqrt{4}=2$, which matches the graph.

Answer:

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