Asked by kelvin calin
On a coordinate plane, a parabola opens up. It starts in quadrant 1, crosses the x axis where the vertex is in quadrant 4, and crosses the x-axis back into quadrant 1.
Which function could be represented by the graph on the coordinate plane?
f(x) = (x – 8)2 + 6
f(x) = (x + 8)2 + 6
f(x) = (x + 8)2 – 6
f(x) = (x – 8)2 – 6
All Answers 1
Answered by
GPT-5 mini
AI
f(x) = (x − 8)^2 − 6
Reason: The parabola opens up (leading coefficient > 0). Its vertex is at (8, −6), which is in quadrant IV (x>0, y<0). The x-intercepts satisfy (x−8)^2 = 6 so x = 8 ± √6, both positive, so the graph crosses the x-axis on the right side (between quadrants I and IV) as described.
Reason: The parabola opens up (leading coefficient > 0). Its vertex is at (8, −6), which is in quadrant IV (x>0, y<0). The x-intercepts satisfy (x−8)^2 = 6 so x = 8 ± √6, both positive, so the graph crosses the x-axis on the right side (between quadrants I and IV) as described.
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