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graph of a function with a grid, axes from -10 to 10 on both x and y. the graph has two branches, one on the left (x < -3) and one on the right (x > 3), opening upwards, with vertices at (-3, 0) and (3, 0), and passing through (-5, 4) and (5, 4) approximately.

select one:
○ a. $f(x) = \sqrt{x^2 - 9}$
○ b. $f(x) = \sqrt{x} + 9$
○ c. $f(x) = \sqrt{x + 9}$
○ d. $f(x) = \sqrt{x^2 + 9}$

Explanation:

Step1: Analyze Domain from Graph

The graph has two branches, one for \( x \leq -3 \) (since the left end starts near \( x=-3 \)) and one for \( x \geq 3 \) (right end starts near \( x = 3 \)). So the domain is \( x \leq -3 \) or \( x \geq 3 \), which implies the expression inside the square root must be non - negative when \( x^2-9\geq0\) (i.e., \( x^2\geq9\) or \( |x|\geq3 \)).

Step2: Analyze Each Option

• Option a: \( f(x)=\sqrt{x^{2}-9} \). The domain of this function is \( x^{2}-9\geq0\), which is \( x\leq - 3\) or \( x\geq3 \), matching the graph's domain. Let's check a point. When \( x = 5 \), \( f(5)=\sqrt{25 - 9}=\sqrt{16}=4\)? Wait, no, wait the graph at \( x = 5 \) seems to have \( y = 4\)? Wait, no, looking at the graph, when \( x = 5 \), the \( y \) - value: let's recalculate. Wait \( x = 5 \), \( x^{2}-9=25 - 9 = 16\), \( \sqrt{16}=4 \)? But the graph at \( x = 5 \) is at \( y = 4 \)? Wait, maybe my initial look was wrong. Wait, let's check another point. When \( x = 3 \), \( f(3)=\sqrt{9 - 9}=0 \), which matches the graph (the vertex at \( x = 3 \) and \( x=-3 \) is on the x - axis).
• Option b: \( f(x)=\sqrt{x}+9 \). The domain of this function is \( x\geq0 \), but the graph has a left branch for \( x<0 \), so this is incorrect.
• Option c: \( f(x)=\sqrt{x + 9} \). The domain is \( x+9\geq0\) or \( x\geq - 9 \). But when \( x = 0 \), \( f(0)=\sqrt{9}=3 \), but the graph at \( x = 0 \) has no point (the graph is symmetric about the y - axis and has no point at \( x = 0 \) between \( - 3\) and \( 3 \)). Also, the shape is not a square - root of a linear function. The graph of \( y=\sqrt{x + 9}\) is a square - root function shifted left 9 units, which is a single - branch function (increasing), but our graph has two branches. So this is incorrect.
• Option d: \( f(x)=\sqrt{x^{2}+9} \). The domain of this function is all real numbers (\( x^{2}+9\geq0\) for all real \( x \)), but the graph has a gap between \( - 3\) and \( 3 \), so this is incorrect.

Answer:

a. \( f(x)=\sqrt{x^{2}-9} \)