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
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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
Could you possibly show me an example using the points above.
Thanks for your help.
Or would it make more sense to have the vertex at x=0.
say
5, 0
0, 2
5, 0
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 downopening 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 . . .
Any chance of an example
The quadratic is probably the easiest.
$\displaystyle
\text{At}\; (0,0)\;\; 0 = 0 a + 0 b + c
$
$\displaystyle
\text{At}\; (5,2)\;\; 2 = 25a + 5b + c
$
$\displaystyle
\text{At}\; (10,0) \;\; 0 = 100 a + 10 b + c
$
Solving gives $\displaystyle a =  \frac{2}{25},\;\; b = \frac{4}{5},\;\; c= 0$
and so we have
$\displaystyle
y = \frac{2}{25} x^2 + \frac{4}{5} x.
$
or
$\displaystyle
y = \frac{2}{25} x (10x)
$
Other possibilities
$\displaystyle y = \frac{2}{5^{p+q}} x^p(10x)^q$ where $\displaystyle p, q \in \mathbb{R^+}$.
A trig function
$\displaystyle
y = 2 \sin \frac{\pi x}{10}
$
An ellipse
$\displaystyle
\frac{(x5)^2}{25} + \frac{y^2}{4} = 1
$
Here's a circle
$\displaystyle
(x5)^2 + \left(y + \frac{21}{4} \right)^2 = \left(\frac{29}{4}\right)^2
$.
Like Soroban said  the list is exhaustive.
Thanks everyone. If i wanted to expand that to be given 5 points would a quadratic equation still be the best (simplest) option.
You have to go up in order. So for 5 points try
$\displaystyle
y = ax^4 + bx^3 + cx^2 + dx + e
$
However, you might get lucky and find a polynomial less the degree 4 that works.
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.
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(X2X1) Y3  Y1 = A(X3^2  X1^2) + B(X3X1)
Multiplying the first equation by (X1X3) and the second by (X2X1) and adding them gives:
(Y2Y1)(X1X3) = (X1X3)A(X2^2  X1^2) + B(X2X1)(X1X3) (Y3Y1)(X2X1) = (X2X1)A(X3^2  X1^2) + B(X3X1)(X2X1)
Or
(Y2Y1)(X1X3) + (Y3Y1)(X2X1) = A[(X1X3)(X2^2X1^2) + (X2X1)(X3^2X1^2)]
Solving for A gives:
A = [(Y2Y1)(X1X3) + (Y3Y1)(X2X1)]/[(X1X3)(X2^2X1^2) + (X2X1)(X3^2X1^2)]
Plugging into the previous equation gives:
B = [(Y2  Y1)  A(X2^2  X1^2)] / (X2X1)
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.
Thanks for your time.
The same applies but obviously, more complicated. You could also try Cramer's Rule
Cramer's rule  Wikipedia, the free encyclopedia