# Finding a function for the parabola

1. Apr 28, 2010

### Atran

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.

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Thanks...

2. Apr 28, 2010

### rock.freak667

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

3. Apr 28, 2010

### 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...

Last edited: Apr 28, 2010
4. Apr 28, 2010

### 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.

5. Apr 29, 2010

### HallsofIvy

Staff Emeritus
??? An integer can't "approach" 0!

6. Apr 29, 2010

### Atran

Sorry, I just meant a positive rational number.