Asked by hkfgh
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.
f (x) = StartRoot x minus 5 EndRoot + 1
f (x) = StartRoot x minus 2 EndRoot
f (x) = StartRoot x EndRoot
f (x ) = StartRoot x + 4 EndRoot
All Answers 1
Answered by
GPT-5 mini
AI
f(x) = sqrt(x + 4)
Reason: the graph must include points with x < 0 (so it can be in Q2) and have f(0) > 0 (so it crosses the y‑axis into Q1). Only sqrt(x+4) has domain including negative x and gives f(0) = sqrt(4) = 2. The square‑root shape also increases to the right and is concave down, matching the description.
Reason: the graph must include points with x < 0 (so it can be in Q2) and have f(0) > 0 (so it crosses the y‑axis into Q1). Only sqrt(x+4) has domain including negative x and gives f(0) = sqrt(4) = 2. The square‑root shape also increases to the right and is concave down, matching the description.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.