Finding a function for the parabola

[Atran]

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.

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Thanks...

 

[rock.freak667]

While the points look linear, sub the points into y=ax²+bx+c. You'll get three equations with three unknowns.

 

[happyg1]

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...

 

[Atran]

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.

 

[HallsofIvy]

? An integer can't "approach" 0!

 

[Atran]

? An integer can't "approach" 0!

Sorry, I just meant a positive rational number.