MedVision ad

Proving three points define a unique parabola (1 Viewer)

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
Is it true that n points on a cartesian plane define a unique polynomial of degree n-1? How can I prove this?
More specifically I'm looking to prove that three points defines a unique parabola.

Consider y = ax^2 + bx + c

Subbing three points (x1, y1), (x2, y2), (x3, y3) would give:

a(x1)^2 + b(x1) + c = y1
a(x2)^2 + b(x2) + c = y2
a(x3)^2 + b(x3) + c = y3

Since there are three unknowns a,b,c and three equations, the system is solvable and thus the three points define a parabola.
But how do I know only these three points define a UNIQUE parabola?
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
There's a theorem in Linear Algebra that states that if you write those equations in matrix form, the solution (found by finding the inverse of the matrix) is in fact unique.
 

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
Is it as simple as this:

Solving would give a,b,c in terms of x1,y2 ,x2,y2, x3,y3. Thus, the parabola is unique to those three points?

@asianese, damn I have no idea what matrix is lol
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Its just a way of writing those exact equations you had above. When you solve it, the solution is said to be unique which can be proven easily.

There's probably an HSC answer but I don't see it just yet
 

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
n points on a cartesian plane define a unique polynomial of degree at most n-1

For the proof for a quadratic, see Cambridge Year 11 - the intro to exercise 8I.
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,390
Gender
Male
HSC
2006
Suppose there exists another parabola say
y = dx2 + ex + f
for which those three points also satisfy. Show that
a = d
b = e
c = f
must hold simultaneously as the only way this can work. If you can do that you've shown that the parabola must be unique.
 

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
Yep I got it. Thanks everyone
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Here is an outline that is high school level:

Suppose the graphs of two polynomials p and q of degree < n pass through the same n points (no two of which lie on the same vertical line). Then p-q is a poly of degree < n which has at least n roots.

The only such polynomial is the zero polynomial. We can prove this rigorously by using the factor theorem and induction. (Use the factor theorem to reduce the number of roots and the degree of our poly by 1).

Important:
This only establishes that if there exists such a polynomial, it must be unique. We must separately prove that such a polynomial exists, which can be done by explicit construction. This construction is easy enough, and is commonly referred to as Lagrange interpolation.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top