I have these for points, (1,-2), (4,3), (12,3) and (15,-3), and I was wondering how do I make a model out of this, and how would this relate to polynomials and Pascal's triangle?
I have this formula: $$B(t) = (1-t)^3 P_0 + 3(1-t)^2t P_1 + 3(1-t)t^2 P_2 + t^3 P_3$$ But I am still pretty confused
Thanks for helping.
Simply, convert your expression into parametric equations, a first one dealing with the abscissas, the other with the ordinates
$$x=\color{red}{1}*(1-t)^3 *(1) + \color{red}{3}*(1-t)^2*t*(4) + \color{red}{3}*(1-t)*t^2*(12)+ \color{red}{1}*t^3*(15)$$
$$y=\color{red}{1}*(1-t)^3 *(-2) + \color{red}{3}*(1-t)^2*t*(3) + \color{red}{3}*(1-t)*t^2*(3)+ \color{red}{1}*t^3*(-3).$$
Then use a $plot(x,y)$ for $0 \leq t \leq 1$ or an equivalent in your favorite programming language. See figure below.
Remarks :
1) Coefficients $\color{red}{1,3,3,1}$ constitute the third line of Pascal's triangle.
2) Imagine you had to draw a Bezier curve with endpoints $P_0$ and $P_4$ with three control points ("handles") $P_1,P_2,P_3$, you would use coefficients from the fourth line of Pascal's triangle in the folowing expression :
$$B(t) = \color{red}{1}(1-t)^4 t^0 P_0 + \color{red}{4}(1-t)^3 t^1 P_1 + \color{red}{6}(1-t)^2 t^2 P_2 + \color{red}{4}(1-t) t^3 P_3 + \color{red}{1}(1-t)^0 t^4 P_4$$