# Curve equation

• January 15th 2010, 06:04 AM
mrwelcam
Curve equation
Hi, I would like to know if its possible to find a equation for a curve if 3 points were given. Say

1. 0, 0
2. 5, 2
3. 10, 0

If so, how would i go about solving this.

Thanks all
• January 15th 2010, 06:11 AM
Jester
Quote:

Originally Posted by mrwelcam
Hi, I would like to know if its possible to find a equation for a curve if 3 points were given. Say

1. 0, 0
2. 5, 2
3. 10, 0

If so, how would i go about solving this.

Thanks all

Sure - try a quadratic $y = ax^2+bx+c$. Sub. in your points and solve a system of equations for $a, b$ and $c.$
• January 15th 2010, 06:30 AM
mrwelcam
Could you possibly show me an example using the points above.

• January 15th 2010, 06:37 AM
mrwelcam
Or would it make more sense to have the vertex at x=0.

say

-5, 0
0, 2
5, 0
• January 15th 2010, 06:38 AM
Soroban
Hello, mrwelcam!

Quote:

I would like to know if its possible to find a equation for a curve
if 3 points were given, say: .A(0,0), B(5,2), C(10,0)

If so, how would i go about solving this?

Code:

      |       |      B       |      o       |   . . o - - - - - - - o - -       A              C

With only three points, there is an infinite number of equations.

The first idea would be a down-opening parabola.

But there are an infinite number of cubics, quartics, etc. that contain those points
. . as well as zillions of circles, ellipses, sine/cosine functions . . .

• January 15th 2010, 06:44 AM
mrwelcam
Any chance of an example
• January 15th 2010, 07:12 AM
Jester
Quote:

Originally Posted by mrwelcam
Any chance of an example

The quadratic is probably the easiest.

$
\text{At}\; (0,0)\;\; 0 = 0 a + 0 b + c
$

$
\text{At}\; (5,2)\;\; 2 = 25a + 5b + c
$

$
\text{At}\; (10,0) \;\; 0 = 100 a + 10 b + c
$

Solving gives $a = - \frac{2}{25},\;\; b = \frac{4}{5},\;\; c= 0$

and so we have

$
y =- \frac{2}{25} x^2 + \frac{4}{5} x.
$

or

$
y = \frac{2}{25} x (10-x)
$

Other possibilities

$y = \frac{2}{5^{p+q}} x^p(10-x)^q$ where $p, q \in \mathbb{R^+}$.

A trig function

$
y = 2 \sin \frac{\pi x}{10}
$

An ellipse

$
\frac{(x-5)^2}{25} + \frac{y^2}{4} = 1
$

Here's a circle

$
(x-5)^2 + \left(y + \frac{21}{4} \right)^2 = \left(\frac{29}{4}\right)^2
$
.

Like Soroban said - the list is exhaustive.
• January 15th 2010, 08:05 AM
mrwelcam
Thanks everyone. If i wanted to expand that to be given 5 points would a quadratic equation still be the best (simplest) option.
• January 15th 2010, 08:47 AM
Jester
You have to go up in order. So for 5 points try

$
y = ax^4 + bx^3 + cx^2 + dx + e
$

However, you might get lucky and find a polynomial less the degree 4 that works.
• January 15th 2010, 10:39 PM
gyan1010
whatever the degree of the polynomial is I believe you need that many + 1 points to get it exactly.

so 3 points would give you any x^2 degree equation. So the answer is yes for parabolas but no for any curve.
• January 16th 2010, 05:07 AM
mrwelcam
Hi. I'm found this on the web:

A quadratic equation has the form:

Y = A * X^2 + B * X + C
Since we have three points, we have three sets of values for X and Y. That gives us three equations and three unknowns: A, B, and C.
Plugging in the points (X1, Y1), (X2, Y2), and (X3, Y3) gives:
Y1 = AX1^2 + BX1 + C Y2 = AX2^2 + BX2 + C Y3 = AX3^2 + BX3 + C
Subtracting equation 1 from the other two gives:
Y2 - Y1 = A(X2^2 - X1^2) + B(X2-X1) Y3 - Y1 = A(X3^2 - X1^2) + B(X3-X1)
Multiplying the first equation by (X1-X3) and the second by (X2-X1) and adding them gives:
(Y2-Y1)(X1-X3) = (X1-X3)A(X2^2 - X1^2) + B(X2-X1)(X1-X3) (Y3-Y1)(X2-X1) = (X2-X1)A(X3^2 - X1^2) + B(X3-X1)(X2-X1)
Or
(Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1) = A[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
Solving for A gives:
A = [(Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1)]/[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
Plugging into the previous equation gives:
B = [(Y2 - Y1) - A(X2^2 - X1^2)] / (X2-X1)
Plugging into the original equation gives:
C = Y1 - AX1^2 - BX1

This formula seems to be working well for me. How would i go about expanding this formula to allow for 5 points.

• January 16th 2010, 06:23 AM
Jester
Quote:

Originally Posted by mrwelcam
Hi. I'm found this on the web:

A quadratic equation has the form:

Y = A * X^2 + B * X + C
Since we have three points, we have three sets of values for X and Y. That gives us three equations and three unknowns: A, B, and C.
Plugging in the points (X1, Y1), (X2, Y2), and (X3, Y3) gives:
Y1 = AX1^2 + BX1 + C Y2 = AX2^2 + BX2 + C Y3 = AX3^2 + BX3 + C
Subtracting equation 1 from the other two gives:
Y2 - Y1 = A(X2^2 - X1^2) + B(X2-X1) Y3 - Y1 = A(X3^2 - X1^2) + B(X3-X1)
Multiplying the first equation by (X1-X3) and the second by (X2-X1) and adding them gives:
(Y2-Y1)(X1-X3) = (X1-X3)A(X2^2 - X1^2) + B(X2-X1)(X1-X3) (Y3-Y1)(X2-X1) = (X2-X1)A(X3^2 - X1^2) + B(X3-X1)(X2-X1)
Or
(Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1) = A[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
Solving for A gives:
A = [(Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1)]/[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
Plugging into the previous equation gives:
B = [(Y2 - Y1) - A(X2^2 - X1^2)] / (X2-X1)
Plugging into the original equation gives:
C = Y1 - AX1^2 - BX1

This formula seems to be working well for me. How would i go about expanding this formula to allow for 5 points.