Finding a function for the parabola

  • Thread starter Thread starter Atran
  • Start date Start date
  • Tags Tags
    Function Parabola
AI Thread Summary
The discussion revolves around the possibility of finding a quadratic function for the points (1, 1), (2, 2), and (3, 3). Despite attempting to fit these points into the quadratic equation y = ax^2 + bx + c, the results indicate that the points are nearly linear, yielding coefficients that suggest a linear relationship. The user notes that the calculations become complex, and the coefficients derived are very close to those of a linear function. The conversation highlights the challenge of fitting a quadratic function to points that are closely aligned in a linear fashion. Ultimately, it concludes that a true quadratic function cannot be accurately defined for these points.
Atran
Messages
93
Reaction score
1
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...
 
Physics news on Phys.org
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.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Similar threads

Identifying the equation of a parabola
Replies
2
Views
2K
Intersection of a circle and a parabola
Replies
5
Views
4K
Graphing to find the intersections of lines and a parabola in this limit
Replies
2
Views
2K
Upside down parabola equation -- needed for a quiz
Replies
17
Views
8K
Finding the focus of a parabola given an equation
Replies
4
Views
2K
Arriving at parabola formula via distance formula
Replies
4
Views
2K
Triangle formed by tangents to a parabola
Replies
3
Views
3K
Solve for a When Parabola y=x^2-a(x+1)+3 Intersects x-Axis Once
Replies
9
Views
3K
Is ##f(x)=2^{x}-1## considered an exponential function?
Replies
12
Views
2K
Downward parabola: value of p = 1/a?
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top