A cubic Bzier curve is a parametric polynomial curve given by: X(t) = (1 t) b...

  

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A cubic Bézier curve is a parametric polynomial curve given by: X(t) = (1 t) ³b + 3(1 – t)²tb₁ + 3(1 – t)t²b₂ + t³b3 where b; are the control points. Suppose the control points are bo = (-1,0), b₁ = (0, 1), b₂ = (0,−1), and b3 = (1, 0). Use de Casteljau's algorithm to find the coordinates of X(0.25), and check it with the polynomial equation above. X(0.25) = A cubic Bézier curve is a parametric polynomial curve given by: X(t) = (1 t) ³b + 3(1 – t)²tb₁ + 3(1 – t)t²b₂ + t³b3 where b; are the control points. Suppose the control points are bo = (-1,0), b₁ = (0, 1), b₂ = (0,−1), and b3 = (1, 0). Use de Casteljau's algorithm to find the coordinates of X(0.25), and check it with the polynomial equation above. X(0.25) = A cubic Bézier curve is a parametric polynomial curve given by: X(t) = (1 t) ³b + 3(1 – t)²tb₁ + 3(1 – t)t²b₂ + t³b3 where b; are the control points. Suppose the control points are bo = (-1,0), b₁ = (0, 1), b₂ = (0,−1), and b3 = (1, 0). Use de Casteljau's algorithm to find the coordinates of X(0.25), and check it with the polynomial equation above. X(0.25) = A cubic Bézier curve is a parametric polynomial curve given by: X(t) = (1 t) ³b + 3(1 – t)²tb₁ + 3(1 – t)t²b₂ + t³b3 where b; are the control points. Suppose the control points are bo = (-1,0), b₁ = (0, 1), b₂ = (0,−1), and b3 = (1, 0). Use de Casteljau's algorithm to find the coordinates of X(0.25), and check it with the polynomial equation above. X(0.25) =