Question

which of the following is the graph of this square root function? f(x) = \sqrt{x - 2} + 2

Explanation:

Step1: Find the domain of the function

For the square root function \( f(x)=\sqrt{x - 2}+2 \), the expression inside the square root must be non - negative. So we set \( x - 2\geq0 \), which gives \( x\geq2 \). This means the graph of the function should start at \( x = 2 \) or to the right of \( x=2 \).

Step2: Find the y - intercept (when x = 2)

Substitute \( x = 2 \) into the function: \( f(2)=\sqrt{2 - 2}+2=\sqrt{0}+2 = 0 + 2=2 \). So the starting point (the vertex of the square - root - shaped graph) of the function is at the point \( (2,2) \).

Step3: Analyze the options

• The first graph has a starting point with \( x\lt2 \), so it does not satisfy the domain condition.
• The second graph has a starting point at \( x = 2 \) (or near \( x = 2 \)) and when \( x = 2 \), the y - value is 2 (matches our calculation of \( f(2)=2 \)). As \( x \) increases, \( \sqrt{x - 2} \) increases, so \( f(x)=\sqrt{x - 2}+2 \) will also increase, which is consistent with the shape of a square - root function.
• The third graph has a starting point with \( y\lt0 \) when \( x = 2 \) (or near \( x = 2 \)), which does not match our calculation of \( f(2)=2 \).

Answer:

The second graph (the one with the x - axis starting from - 1 and going to 9, and the y - axis starting from - 1 and going to 8, with the curve starting at (2,2) approximately) is the graph of \( f(x)=\sqrt{x - 2}+2 \).