o $f(x)=sqrt{x + 3}$\no $f(x)=-sqrt{x + 3}-3$\no $f(x)=-sqrt{x + 3}+3$\no $f(x)=sqrt{x + 3}-3$

Answer

Explanation:

Step1: Analyze domain

For a square - root function $y = \sqrt{u}$, the expression inside the square - root $u\geq0$. For the function $y=\sqrt{x + 3}$, the domain is $x+3\geq0$ or $x\geq - 3$. The graph starts at $x=-3$.

Step2: Analyze y - intercept

Set $x = 0$. For $y=\sqrt{x + 3}$, when $x = 0$, $y=\sqrt{0 + 3}=\sqrt{3}\approx1.73$.

Step3: Check each option

Option 1: $f(x)=\sqrt{x + 3}$: The domain is $x\geq - 3$. When $x=-3$, $y = 0$; when $x = 0$, $y=\sqrt{3}\approx1.73$. This function matches the graph.
Option 2: $f(x)=-\sqrt{x + 3}-3$. The negative sign in front of the square - root flips the graph of $y = \sqrt{x+3}$ over the $x$ - axis and the $-3$ shifts it down 3 units, which does not match the given graph.
Option 3: $f(x)=-\sqrt{x + 3}+3$. The negative sign flips the graph of $y=\sqrt{x + 3}$ over the $x$ - axis and the $+3$ shifts it up 3 units, which does not match the given graph.
Option 4: $f(x)=\sqrt{x + 3}-3$ shifts the graph of $y=\sqrt{x + 3}$ down 3 units, which does not match the given graph.

Answer:

$f(x)=\sqrt{x + 3}$