Which of the following is NOT a blending function of a quadratic Bezier curve?

The blending (basis) functions of a Bézier curve of degree n are the Bernstein polynomials Bi,n(u) = C(n, i) · ui · (1 − u)n−i for i = 0, 1, …, n, where C(n, i) is the binomial coefficient. These n + 1 polynomials always sum to exactly 1 for every value of u, and each coefficient is fixed by the binomial expansion — never an arbitrary constant.

1. For a quadratic Bézier curve the degree is n = 2, so the index i takes the values 0, 1, and 2, giving exactly three blending functions.

2. Term i = 0: B0,2(u) = C(2, 0) · u0 · (1 − u)2 = (1 − u)2, since C(2, 0) = 1.

3. Term i = 1: B1,2(u) = C(2, 1) · u1 · (1 − u)1 = 2u(1 − u), since the binomial coefficient C(2, 1) = 2.

4. Term i = 2: B2,2(u) = C(2, 2) · u2 · (1 − u)0 = u2, since C(2, 2) = 1.

5. So the three genuine quadratic Bézier blending functions are (1 − u)2, 2u(1 − u), and u2.

Cross-check: adding all three blending functions gives (1 − u)2 + 2u(1 − u) + u2 = ((1 − u) + u)2 = 12 = 1 for every value of u — the defining partition-of-unity property that a valid set of blending functions must satisfy.

• (1 − u)2 is the i = 0 term, using coefficient C(2, 0) = 1.

• 2u(1 − u) is the i = 1 term, using the binomial coefficient C(2, 1) = 2.

• 3u(1 − u) uses the coefficient 3 in place of the required binomial coefficient C(2, 1) = 2, so it does not match any of the three Bernstein terms for n = 2.

• u2 is the i = 2 term, using coefficient C(2, 2) = 1.

Therefore, among the four given expressions, 3u(1 − u) is the one that is NOT a blending function of the quadratic Bézier curve, since its coefficient does not match any of the three Bernstein basis terms derived above.