Question

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 #

Answers

Answered by GPT-4o mini
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.**
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