Question

which could be the function graphed below?

\\(f(x) = \sqrt{x} - 2\\)
\\(f(x) = \sqrt{x - 3} + 1\\)
\\(f(x) = \sqrt{2x + 4}\\)
\\(f(x) = \sqrt{x + 1} + 8\\)

Explanation:

Identify the starting point of the graphed function

The graph represents a square root function of the form \(f(x) = a\sqrt{x-h} + k\).
The starting point (vertex) of the curve is located on the y-axis below the origin, corresponding to the coordinates \((0, -2)\).

Match the starting point to the function options

For a general square root function \(f(x) = \sqrt{x-h} + k\), the starting point is \((h, k)\).
Using \((h, k) = (0, -2)\):

$$ f(x) = \sqrt{x-0} - 2 = \sqrt{x} - 2 $$

Verify the domain and range of the matched function

The domain of \(f(x) = \sqrt{x} - 2\) is \(x \ge 0\), which matches the graph starting at the y-axis and extending to the right.
The range is \(y \ge -2\), which matches the graph starting at \(y = -2\) and rising gradually.

Answer:

• (A) \(f(x) = \sqrt{x} - 2\) (Correct answer)
• (B) \(f(x) = \sqrt{x-3} + 1\)
• (C) \(f(x) = \sqrt{2x} + 4\)
• (D) \(f(x) = \sqrt{x+1} + 8\)