Example: Let F be an arbitrary polynomial F(x) = ax^{2} + bx
+ c, of degree < 2. We want to
demonstrate how this curve is controlled by 3 points.

Step 1: Blossom of F. From F we can obtain a unique symmetric
biaffine
map f (x_{1}, x_{2}) = ax_{1}x_{2} +
b (x_{1}+x_{2})/2 + c, such that F(x) = f(x, x), for
all x in R. This process is called
the blossom (or polar form) of F.

Step 2: Linear interpolation of f. Let t_{i} =
(1-l_{i})r +
l_{i}s, i=1,2.
Let us compute f(t_{1}, t_{2}) = f(
(1-l_{1})r +
l_{1}s,
(1-l_{2})r +
l_{2}s) .......
(*) Note that f is symmetric and biaffine, so we can use the
properties
of such a map to expand (*) to get the following form, f(
t_{1}, t_{2}) =
(1-l_{1})
(1-l_{2}) f(r, r) +
((1-l_{1})
l_{2} +
l_{1}
(1-l_{2})) f(r, s) +
l_{1}
l_{2} f(s,s) .......Š(**) Solving
l_{i} in
terms of t_{i} we get

l_{i} = (

t_{i} - r
s - r

)

Then (**) becomes

f( t_{1}, t_{2} ) = (

s - t_{1}
s - r

) (

s - t_{2}
s - r

) f(r,r) + [(

s - t_{1}
s - r

) (

t_{2} - r
s - r

) +

(

t_{1} - r
s - r

) (

s - t_{2}
s - r

)] f(r,s) + (

t_{1} - r
s - r

) (

t_{2} - r
s - r

) f(s,s)

The points f(r, r), f(r, s) and
f(s, s) are called the control points (or Bézier control points) of
the curve F.

Step 3: Determination of a curve by control points Now let us see how
to determine any point on the curve by the control points f(r,r), f(r,s),
f(s,s). In practice we often choose r=0, s=1 in ti = (1-li)r + li s.
Thus, ti = li . Now the polynomial function F corresponding to f(t1,
t2) is F(t) = f(t, t) = (Please compare with (**)). The polynomials
are known as Bernstein polynomials of degree 2. Thus, F(t) is determined
by the control points f(0,0), f(0,1), f(1,1) and the Bernstein polynomials.
In fact, we have written F(t) as a linear combination of Bernstein polynomials
of degree 2. It is customary to call such a curve a Bézier Curve (in
terms of Bernstein polynomials).

Remark: Clearly we can turn the above procedure into an algorithm.
The obtained algorithm will be called deCastlejau algorithm.

deCastlejau Algorithm --

(for order n=2) Recall from the previous example what f(r,r), f(r,s)
and f(s,s) are. The deCastlejau algorithm consists of two stages.

Stage1: We compute the 2 points f(r,t) = f(r, (1-l)r + l s) = (1-l)f(r,r)
+ lf(r,s) and f(t,s) = f((1-l)r + l s, s) = (1-l)f(r,s) + lf(s,s), by
linear interpolation, where f(r,t) is computed from the two control
points f(r,r) and f(r,s), and f(t,s) is computed from the two control
points f(r,s) and f(s,s), the ratio of interpolation being .

Stage 2: Since by symmetry, f(r,t) = f(t,r), we compute the point f(t,t)
= f(t, (1-l)r + l s) = (1-l)f(t,r) + lf(t,s), from the points f(t,r)
and f(t,s) computed during the first stage, the ratio of interpolation
being the same as before. Thus, by three linear interpolation steps,
we obtain the point F(t) on the curve. Note that the two control points
f(r,r) = F(r) and f(s,s) = F(s) are on the curve but f(r,s) is not.