Asked by David
Which could be the function graphed below?
On a coordinate plane, a curve starts in quadrant 2 and crosses the y-axis and goes into quadrant 1. The curve opens down and to the right.
A)f (x) = StartRoot x minus 5 EndRoot + 1
B)f (x ) = StartRoot x + 4 EndRoot
All Answers 1
Answered by
GPT-5 mini
AI
B) f(x) = sqrt(x + 4)
Reason: sqrt(x+4) has domain x ≥ −4, so it begins at (−4,0) (on the left) and for −4 < x < 0 lies in quadrant II (x<0, y>0). At x = 0 it gives y = 2, so it crosses the y‑axis and then continues into quadrant I, increasing to the right and concave down.
By contrast, sqrt(x−5)+1 has domain x ≥ 5 and starts at (5,1) in quadrant I, so it cannot produce the described left‑starting curve.
Reason: sqrt(x+4) has domain x ≥ −4, so it begins at (−4,0) (on the left) and for −4 < x < 0 lies in quadrant II (x<0, y>0). At x = 0 it gives y = 2, so it crosses the y‑axis and then continues into quadrant I, increasing to the right and concave down.
By contrast, sqrt(x−5)+1 has domain x ≥ 5 and starts at (5,1) in quadrant I, so it cannot produce the described left‑starting curve.
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