Asked by chris
If you scan the y‑values, something jumps out:
The values increase from 15 → 21 → 26
Then they decrease from 26 → 20 → 14
That’s a rise and fall pattern — the hallmark of a quadratic relationship, not a linear one.
All Answers 1
Answered by
GPT-5 mini
AI
Exactly — a rise-then-fall shows a turning point, which a straight line cannot have. That pattern is characteristic of a parabola (a quadratic), but it isn’t proof by itself — other nonlinear shapes can do the same.
If the x-values are evenly spaced (say x = 1,2,3,4,5), you can check for a true quadratic by looking at finite differences:
- y: 15, 21, 26, 20, 14
- 1st differences: +6, +5, −6, −6 (not constant → not linear)
- 2nd differences: −1, −11, 0 (not constant → the five points do not lie exactly on a single quadratic)
So the data are non‑linear and have a clear local maximum near the third point, which is consistent with a quadratic vertex, but the second differences show they don’t lie exactly on one parabola. If you want, I can:
- fit the best quadratic (least-squares) to these five points and show the fit and residuals, or
- find the exact quadratic that passes through any three chosen points (e.g., the middle three to capture the peak). Which would you prefer?
If the x-values are evenly spaced (say x = 1,2,3,4,5), you can check for a true quadratic by looking at finite differences:
- y: 15, 21, 26, 20, 14
- 1st differences: +6, +5, −6, −6 (not constant → not linear)
- 2nd differences: −1, −11, 0 (not constant → the five points do not lie exactly on a single quadratic)
So the data are non‑linear and have a clear local maximum near the third point, which is consistent with a quadratic vertex, but the second differences show they don’t lie exactly on one parabola. If you want, I can:
- fit the best quadratic (least-squares) to these five points and show the fit and residuals, or
- find the exact quadratic that passes through any three chosen points (e.g., the middle three to capture the peak). Which would you prefer?
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