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Homework Help: Finding a function for the parabola

  1. Apr 28, 2010 #1
    Note: I don't need any answer, all I want to know is whether this question is possible.

    1. The problem statement, all variables and given/known data
    "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 text book.

    2. Relevant equations

    3. The attempt at a solution
    No idea! I don't get more than a linear function.

    - - - - -

  2. jcsd
  3. Apr 28, 2010 #2


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    Homework Helper

    While the points look linear, sub the points into y=ax2+bx+c. You'll get three equations with three unknowns.
  4. Apr 28, 2010 #3
    I did this and it turns out kinda ugly. I got 0 for the coefficient of the squared term (well, [tex]1.5x 10^{-14}[/tex] )I got 1 for the coefficient of the x term; and I got [tex]3x10^{-14}[/tex] 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 [tex]Ax^2+Bx+C=0[/tex]
    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: Apr 28, 2010
  5. Apr 28, 2010 #4
    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.
  6. Apr 29, 2010 #5


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    ??? An integer can't "approach" 0!
  7. Apr 29, 2010 #6
    Sorry, I just meant a positive rational number.
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