There is nothing better than to explain the answer to your question with a simple example.
Say we have two segment of control polygon formed by the three points \(\displaystyle (0,0),(1,2),(2,0)\).
And we have a Bezier curve defined by:
\(\displaystyle \bold{B}(t) = (2t, 4t(1-t)), \ \ \ \ \ t \in [0,1]\)
Suppose we have a line \(\displaystyle y = 1\) and we want to see how many times the control polygon crosses it and how how many times the Bezier curve crosses it.
If you analyze this problem carefully, you will find out that the control polygon crosses \(\displaystyle (y = 1) \ \bold{2}\) times while the Bezier curve \(\displaystyle \bold{1}\) time.
Let us see if this goes with the variation diminishing property.
\(\displaystyle \#(\text{Curve} \cap L) \leq \#(\text{Control Polygon} \cap L)\)
\(\displaystyle \textcolor{red}{\bold{1}} \leq \textcolor{red}{\bold{2}}\)
So this confirms the property. In other words, the Bezier curve does not intersect any straight line more than the control polygon does.
How does this help us? \(\displaystyle \rightarrow\) What are the consequences of this result?
\(\displaystyle \bold{1.}\) The Bezier curve is smoother than the control polygon.
\(\displaystyle \bold{2.}\) Bezier curves do not introduce extra oscillations
\(\displaystyle \bold{3.}\) Useful in curve fitting and approximation
\(\displaystyle \bold{4.}\) Simplifies geometric reasoning
\(\displaystyle \bold{5.}\) Supports recursive subdivision (de Casteljau’s algorithm)
And more\(\displaystyle \cdots\)
What about Bernstein polynomials?
Well the curve \(\displaystyle \bold{B}(t) = (2t, 4t(1-t))\) was defined by:
\(\displaystyle \bold{B}(t) = \sum_{i=0}^{2}\bold{P}_iB_i^2(t)\)
where \(\displaystyle \bold{P}_i\) are the control points and \(\displaystyle B_i^2(t) = \binom{2}{i} (1 - t)^{2 - i} t^i\) is the Bernstein Basis Polynomials.
What is their purpose?
The Bezier curve preserves its properties of smoothness, stability, and geometric control because it is defined using these Bernstein polynomials as basis functions.
