Understanding Bézier Curves in Asymptote

I am trying to correct Mathics‘ ability to draw Bézier curves properly in Asymptote, and I have a hard time understanding how to translate Mathematica’s Specification into Asymptote. I haven’t found an easy-to-understand description or tutorial of how to draw Bézier Curves in Asymptote.

So let me try show what I want to be able to do with a basic example:

Mathematica (actually, Mathics):

Graphics[BezierCurve[{{0,0}, {1,1}, {2, -1}, {3,0}}]]

Incorrect translation of first example above:

usepackage("amsmath");
size(5.8333cm, 5.8333cm);
// BezierCurveBox
draw(0,.0.0),  rgb(0, 0, 0)+linewidth(0.66667));
draw((0.,0.)..controls(155.56,213.89) and 
(194.44,194.44)..(233.33,116.67), 
rgb(0, 0, 0)+linewidth(0.66667));

Above, I have given how Mathics currently converts this. One of the bugs seen above seems to have to do with translating control-point coordinates to center the figure in overall image window, while not translating some of the non-control points. And there is what looks like a stray point drawn at the beginning.

I leave this in. But in giving a solution, feel free to totally disregard this and describe something from scratch that has the same shape as what is intended.

So how do I fix this, and how do I think about doing correct translations like the Mathematica-to-Asymptote example above?

(I can also show the correct translations into SVG if that helps, but think that this may just complicate things.)

import graph;

pair Bezier(pair P[], real t)
{ // https://tex.stackexchange.com/a/554290/236162
  pair Bezi;
  for (int k=0; k <= P.length-1; ++k)
  {  
    Bezi=Bezi+choose(P.length-1,k)*(1-t)^(P.length-1-k)*t^k*P[k];
  }
  return Bezi;
}

unitsize(1cm);
pair[] P={(0,0),(1,1),(2,-1)};
pair[] Q={(0,0),(1,1),(2,-1),(3,0)};
pair F(real t){return Bezier(P,t);}
pair G(real t){return Bezier(Q,t);}

draw(graph(F,0,1,350),red);
draw(operator --(... P));
draw(shift(0,-2)*graph(G,0,1,350),red);
draw(shift(0,-2)*operator --(... Q));