A coordinate plane's axes range from negative 2 to 4, both by 1-unit increments. A solid curve is plotted passing through the first quadrant.

Select the option that matches the graph.

Option #1: The graph of the function matches f(x)=2x−−√3
.

Option #2: The graph of the function matches f(x)=2x−−√
.

(1 point)
Option #

1 answer

To determine which option matches the graph in the first quadrant, let's analyze both functions given in the options:

  1. Option #1: \( f(x) = \sqrt{2x} - 3 \)

    To check this function in the first quadrant (where \( x \) is positive):

    • At \( x = 0 \): \[ f(0) = \sqrt{2(0)} - 3 = -3 \] (not in the first quadrant)
    • At \( x = 1 \): \[ f(1) = \sqrt{2(1)} - 3 = \sqrt{2} - 3 \approx -1.59 \] (not in the first quadrant)
    • At \( x = 2 \): \[ f(2) = \sqrt{2(2)} - 3 = \sqrt{4} - 3 = 2 - 3 = -1 \] (not in the first quadrant)
    • At \( x = 3 \): \[ f(3) = \sqrt{2(3)} - 3 = \sqrt{6} - 3 \approx -0.55 \] (not in the first quadrant)
    • At \( x = 4 \): \[ f(4) = \sqrt{2(4)} - 3 = \sqrt{8} - 3 \approx -0.17 \] (not in the first quadrant)

    This function does not intersect the first quadrant since it remains negative for all tested points.

  2. Option #2: \( f(x) = \sqrt{2x} \)

    To check this function in the first quadrant:

    • At \( x = 0 \): \[ f(0) = \sqrt{2(0)} = 0 \quad (\text{on the boundary of the first quadrant}) \]
    • At \( x = 1 \): \[ f(1) = \sqrt{2(1)} = \sqrt{2} \approx 1.41 \quad (\text{in the first quadrant}) \]
    • At \( x = 2 \): \[ f(2) = \sqrt{2(2)} = \sqrt{4} = 2 \quad (\text{in the first quadrant}) \]
    • At \( x = 3 \): \[ f(3) = \sqrt{2(3)} = \sqrt{6} \approx 2.45 \quad (\text{in the first quadrant}) \]
    • At \( x = 4 \): \[ f(4) = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \quad (\text{in the first quadrant}) \]

    This function intersects the first quadrant for positive values of \( x \).

Based on the analysis above, Option #2: \( f(x) = \sqrt{2x} \) matches the graph in the first quadrant.

The answer is: Option #2.