Finding a function for the parabola

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    Function Parabola
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Homework Help Overview

The discussion revolves around finding a quadratic function for a parabola that passes through the points (1, 1), (2, 2), and (3, 3). The original poster expresses uncertainty about the feasibility of the question, noting that the points appear linear rather than quadratic.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Some participants suggest substituting the points into the standard quadratic form to derive equations for the coefficients. Others express concerns about the results being nearly linear, questioning the validity of fitting a quadratic function to the given points.

Discussion Status

The discussion includes various interpretations of the problem, with some participants acknowledging the difficulty of fitting a quadratic function to points that seem to align linearly. There is no explicit consensus, but participants are exploring the implications of the results obtained from their calculations.

Contextual Notes

Participants note that the points are very close together, which complicates the fitting of a quadratic function. There is also a mention of a misunderstanding regarding the approach of using limits with integers, indicating a need for clarity in mathematical terminology.

Atran
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Note: I don't need any answer, all I want to know is whether this question is possible.

Homework Statement


"What is the function of the parabola which has the points (1, 1) (2, 2) and (3, 3)?"
I just asked my teacher to get the question, It's not stated in my textbook.


Homework Equations





The Attempt at a Solution


No idea! I don't get more than a linear function.

- - - - -

Thanks...
 
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While the points look linear, sub the points into y=ax2+bx+c. You'll get three equations with three unknowns.
 
I did this and it turns out kinda ugly. I got 0 for the coefficient of the squared term (well, 1.5x 10^{-14} )I got 1 for the coefficient of the x term; and I got 3x10^{-14} for the constant. Looks pretty linear there...although not EXACTLY linear...but I may have made a mistake.

EDIT: I started this by hand and it got ugly quick. I used my TI89 with the rref function and that's what it got. Those points are in a line...it's difficult to force the quadratic on it because they are so close together. One could plug those coefficients into the Ax^2+Bx+C=0
formula and complete the square...It's REALLY close to linear...and the thing could open up or down...those points are too close together and linear to really make sense of...as far as I can tell...
 
Last edited:
At first it seemed very complicated, but when I uncovered the trick It was so easy.
Yes, I know that it's not possible to write a second-degree equation for a linear graph, therefore I had to use 'limit' as a positive integer 'n' approaching 0.
Thanks for the answers.
 
? An integer can't "approach" 0!
 
HallsofIvy said:
? An integer can't "approach" 0!

Sorry, I just meant a positive rational number.
 

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