find y for x on a Cubic Curve segment

CubicCurveUtils.swift

import Foundation

class CubicCurveUtils {
    /**
     * Returns Y given X on a CubicPath segment
     * PARAM: p0: prev end point
     * PARAM: c0: controllpoint 1 for prev end point
     * PARAM: c1: control point 2 for end point
     * PARAM: p1: end point
     */
    static func point(_ p0:CGPoint,_ c0:CGPoint,_ c1:CGPoint,_ p1:CGPoint,_ x:CGFloat)->CGPoint{
        let a:CGFloat = p0.x
        let b:CGFloat = c0.x
        let c:CGFloat = c1.x
        let d:CGFloat = p1.x
        /**/
        let A:CGFloat = d - 3*c + 3*b - a
        let B:CGFloat = 3*c - 6*b + 3*a
        let C:CGFloat = 3*b - 3*a
        let D:CGFloat = a-x
        /*So we need to solve At³ + Bt² + Ct + D = 0*/
        let t:CGFloat =  Utils.cubic(A,B,C,D);
        /*Replace the t on Bezier function and get x,y */
        let p:CGPoint = Utils.point(p0,c0,c1,p1,t)
        return p
    }
}
private class Utils{
    /**
     * Returns point on a CubicCurve for PARAM: t (0-1)
     */
    static func point(_ a:CGPoint,_ b:CGPoint,_ c:CGPoint,_ d:CGPoint,_ t:CGFloat)->CGPoint{
        var point:CGPoint = CGPoint(x:0,y:0)
        let mt:CGFloat  = 1-t
        let mt2:CGFloat = mt*mt
        let mt3:CGFloat = mt2*mt
        /*fx(t) = x1 * (1-t)³ + x2 * 3 * (1-t)²t + x3 * 3 * (1-t)t² + x4 * t³*/
        point.x = a.x*mt3 + b.x*3*mt2*t + c.x*3*mt*t*t + d.x*t*t*t
        /*fy(t) = y1 * (1-t)³ + y2 * 3 * (1-t)²t + y3 * 3 * (1-t)t² + y4 * t³*/
        point.y = a.y*mt3 + b.y*3*mt2*t + c.y*3*mt*t*t + d.y*t*t*t
        return point
    }
    /**
     * Cubic Equation Calculator
     * refer to http://www.1728.org/cubic.htm
     * FORMULA: ax³ + bx² + cx + d = 0
     * PARAMS: a,b,c,d
     * RETURNS: x
     */
    static func cubic(_ a:CGFloat,_ b:CGFloat,_ c:CGFloat,_ d:CGFloat)->CGFloat{
        var d:CGFloat = d//may not work, could be inout?!? seems to work regardless
        var m:CGFloat = 0
        var m2:CGFloat = 0
        var k:CGFloat = 0
        var n:CGFloat = 0
        var n2:CGFloat = 0
        var x:CGFloat = 0
        var r:CGFloat = 0
        var rc:CGFloat = 0
        var theta:CGFloat = 0
        var sign:CGFloat = 0
        var dans:CGFloat = 0
        /**/
        let f:CGFloat = (((3*c)/a) - (((b*b)/(a*a))))/3;
        /**/
        let b3:CGFloat = b*b*b
        let a3:CGFloat = a*a*a
        let a2:CGFloat = a*a
        /**/
        let g:CGFloat = ((2*((b3)/(a3))-(9*b*c/(a2)) + ((27*(d/a)))))/27
        /**/
        let h:CGFloat = (((g*g)/4) + ((f*f*f)/27));
        /**/
        if (h > 0) {
            m = (-(g/2) + (sqrt(h)))
            k = m < 0 ? -1:1
            m2 = (pow((m*k),(1/3)))
            m2 = m2*k
            n = (-(g/2) - (sqrt(h)))
            k = n<0 ? -1:1
            n2 = (pow((n*k),(1/3)))
            n2 = n2*k
            x =  ((m2 + n2) - (b/(3*a)))
        }else {
            r = ((sqrt((g*g/4)-h)))
            k = r<0 ? -1:1
            rc = pow((r*k),(1/3))*k
            theta = acos((-g/(2*r)))
            x = (2*(rc*cos(theta/3))-(b/(3*a)))
            x=x*1E+14
            x=round(x)
            x=(x/1E+14)
        }
        if ((f+g+h)==0){
            if (d<0) {
                sign = sign - 1
            }
            if (d>=0) {
                sign = 1
            }
            if (sign>0){
                dans = pow(d/a,1/3)
                dans = dans * -1
            }
            if (sign<0){
                d = d * -1
                dans = pow(d/a,1/3)
            }
            x = dans
        }
        return x
    }
}

More discussion and original JS code here: http://stackoverflow.com/questions/5634460/quadratic-bezier-curve-calculate-point