Question

which could be the function graphed below?
options:
$f(x)=\sqrt{x} - 2$
$f(x)=\sqrt{x - 3} + 1$
$f(x)=\sqrt{2x + 4}$
$f(x)=\sqrt{x + 1} + 8$ (last option text may have ocr error)

Explanation:

Step1: Analyze the domain and starting point

The graph starts at \( x = 0 \) (or near \( x = 0 \)) and is defined for \( x\geq0 \) (or similar non - negative start). Let's check the domain of each function:

• For \( f(x)=\sqrt{x}-2 \), the domain is \( x\geq0 \), and when \( x = 0 \), \( f(0)=\sqrt{0}-2=-2\). But the graph starts at \( x = 0 \) with \( y\) - value around \( - 2\)? Wait, no, let's check the other functions.
• For \( f(x)=\sqrt{x - 3}+1 \), the domain is \( x\geq3 \), so the graph should start at \( x = 3 \), but our graph starts at \( x = 0 \) (or near \( x = 0 \)), so this is out.
• For \( f(x)=-\sqrt{2x + 4} \), the domain is \( 2x+4\geq0\Rightarrow x\geq - 2 \). When \( x=-2 \), \( f(-2)=-\sqrt{0}=0 \). But the graph we have starts at \( x = 0 \) (or near \( x = 0 \)) with a negative \( y\) - intercept? Wait, no, let's re - evaluate. Wait, the graph in the picture starts at \( x = 0 \) (the vertex is at \( x = 0 \), \( y\) negative). Wait, let's check the first function: \( f(x)=\sqrt{x}-2 \). Domain \( x\geq0 \). When \( x = 0 \), \( f(0)=-2 \). The function \( y = \sqrt{x}\) is a curve that starts at \( (0,0) \) and increases slowly. \( y=\sqrt{x}-2 \) starts at \( (0,-2) \) and increases. Let's check the other options:
• \( f(x)=\sqrt{x - 3}+1 \): domain \( x\geq3 \), so the graph starts at \( x = 3 \), which does not match the given graph (which starts at \( x = 0 \)).
• \( f(x)=-\sqrt{2x + 4} \): domain \( x\geq - 2 \), and it's a reflection over the \( x\) - axis of \( y=\sqrt{2x + 4} \). When \( x = 0 \), \( f(0)=-\sqrt{4}=-2 \). Wait, but the shape: \( y =-\sqrt{2x + 4}\) is a decreasing curve (since the square root is multiplied by - 1), but our graph is increasing. Wait, no, the first function \( f(x)=\sqrt{x}-2 \) is an increasing function (since the derivative \( f^\prime(x)=\frac{1}{2\sqrt{x}}>0 \) for \( x>0 \)), and it starts at \( (0,-2) \), which matches the graph's starting point (at \( x = 0 \), \( y\) is negative and then increases as \( x \) increases).
• The last option seems to have a typo, but among the first three (and considering the typo - free ones), \( f(x)=\sqrt{x}-2 \) (the first option) has domain \( x\geq0 \), starts at \( (0,-2) \) and increases, which matches the graph. Wait, but let's re - check the third option: \( f(x)=-\sqrt{2x + 4} \) is a decreasing function (because of the negative sign), while the given graph is increasing. So the first option \( f(x)=\sqrt{x}-2 \) is an increasing function with domain \( x\geq0 \), starting at \( (0,-2) \), which matches the graph.

Step2: Verify the shape and intercepts

The graph is a square - root - like curve (increasing, starting at \( x = 0 \) with \( y=-2 \)). The function \( f(x)=\sqrt{x}-2 \) has the form \( y=\sqrt{x}+k \) (here \( k=-2 \)), which is a vertical shift of the parent function \( y = \sqrt{x} \) down by 2 units. The parent function \( y=\sqrt{x} \) starts at \( (0,0) \) and increases, and \( y=\sqrt{x}-2 \) starts at \( (0,-2) \) and increases, which matches the shape of the given graph. The other functions either have the wrong domain (like \( f(x)=\sqrt{x - 3}+1 \) with domain \( x\geq3 \)) or the wrong direction (like \( f(x)=-\sqrt{2x + 4} \) which is decreasing).

Answer:

\( f(x)=\sqrt{x}-2 \) (the first option, assuming the options are labeled as: A. \( f(x)=\sqrt{x}-2 \), B. \( f(x)=\sqrt{x - 3}+1 \), C. \( f(x)=-\sqrt{2x + 4} \), D. (typo - ed option)) So the answer is A. \( f(x)=\sqrt{x}-2 \)