It has been suggested in the forums as well as in the developer docs that to get a quadratic bezier from Processing's cubic bezier all you have to do is take the two control points to be the same. This is not correct.
To get a quadratic bezier with anchor points (x1,y1) and (x2,y2) and control point (cx,cy), you have to construct the cubic bezier with anchor points (x1,y1) and (x2,y2) and control points (cx1,cy1) = ( (x1 + 2*cx)/3 , (y1 + 2*cy)/3 ) and (cx2,cy2) = ( (2*cx + x2)/3 , (2*cy + y2)/3).
Since this is my first message, I am not allowed to post the link to where I got this information.
But the parametric equation of the quadratic bezier is this:
p(t) = (1-t)²p₀ + 2t(1-t)p₁ + t²p₂, where p₀ and p₂ are the anchor points and p₁ is the control point.
For a  cubic bezier the equation is
p(t) = (1-t)³ p₀ + 3 t (1-t)² p₁ + 3 t²(1-t) p₂ + t³p₃, where p₀ and p₃ are the anchor points and the other two the control points.

With the above subtitution one can verify that the cubic bezier does give the quadratic bezier.