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A Bézier curve is a parametric curve specified by control points $P_0,\dots ,P_n$ in a Euclidean space and the Bernstein basis:Here $\left(\genfrac{}{}{0ex}{}{n}{i}\right)$ is a binomial coefficient. Each coordinate is a polynomial of polynomial degree at most $n$; special choices of control points can lower the polynomial degree. The ordered control points form the control polygon. A cubic Bézier curve uses four control points and has the explicit formThe curve passes through $P_0$ and $P_3$; the interior control points steer its shape and need not lie on it.^[ref]

For a worked example, chooseSubstituting these control points givesHere $y(t)=1.5-6(t-1/2{)}^2\le 1.5$, so neither interior control point, both at height $2$, lies on the curve. De Casteljau's algorithm computes the same point using only successive linear interpolations.

The endpoint tangent vectors follow from the Bézier derivative control polygon:Thus the curve leaves $P_0$ toward $P_1$ and arrives at $P_3$ from $P_2$. These nonzero derivatives determine the endpoint tangent lines; if an endpoint derivative vanishes, that formula alone does not determine its tangent direction. Endpoint handles are useful when arranging a parametric first-derivative join of Bézier curves.

The parameter $t$ is usually not arc length. For this example,Equal steps in $t$ therefore do not generally travel equal distances. Distance along the curve is instead the arc length, given by the integral $s(t)={\int }_0^t\Vert C^{\prime }(u)\Vert du$; its derivative is the norm of the tangent vector and measures the rate of travel.

Two geometric facts make Bézier curves especially convenient for design. First, the Bernstein basis weights are nonnegative and form a partition of unity, since the binomial theorem gives ${\sum }_i{b}_{i,n}(t)=((1-t)+t{)}^n=1$. Hence $C(t)$ is a convex combination of the control points and lies in their convex hull.^[ref] The convex hull is the smallest convex set containing the control points, not their axis-aligned bounding box. Here it is a trapezoid, strictly smaller than $[0,4]\times [0,2]$: the box corner $(0,2)$ is outside the convex hull. The curve need not fill or touch every edge of either enclosure. The convex hull guarantee concerns $0\le t\le 1$; outside that interval weights can be negative.

Second, applying an affine map to every control point gives exactly the same result as applying it to the entire curve. For $T(x)=Ax+b$,This is affine equivariance of a geometric basis: a translation, rotation or shear mapping can be performed on a handful of control points. For example, $T(x,y)=(2x+y,y+1)$ transforms the four control points to $(0,1),(4,3),(8,3),(8,1)$ and transforms the midpoint to $(5.5,2.5)$, also obtained from the same weights $(1,3,3,1)/8$.

These editable handles explain the usefulness of Bézier curves in computer aided geometric design, vector graphics paths and outline fonts: a small collection of points controls an entire smooth stroke, and resizing the design preserves its mathematical shape. Cambridge's computer-graphics notes describe their use in graphic design, typography and PostScript outlines. The industrial context includes Pierre Bézier's work at Renault and Paul de Casteljau's work at Citroën.^[ref]

// Target: analysis.bigb

The degree-$n$ Bernstein basis isFor $0\le t\le 1$, every basis function is nonnegative, and the binomial theorem givesTherefore $P(t)={\sum }_i{P}_i{b}_i^n(t)$ is a convex combination of its control points. By the definition of their convex hull,This includes the endpoints because $P(0)=P_0$ and $P(1)=P_n$. It depends on the specified parameter interval: outside it the weights need not be nonnegative.

For $n\ge 1$, differentiating the basis and using the elementary binomial identities giveswhere out-of-range basis indices are zero. ThusHence the Bézier derivative control polygon has points $n(P_{i+1}-P_i)$, andIn particular the endpoint tangent vectors are $n(P_1-P_0)$ and $n(P_n-P_{n-1})$. For $n=0$ the curve is constant and its derivative is zero.

Applying the preceding derivative identity repeatedly yields the Bézier derivative control polygon for every order:where the forward finite difference is $\Delta P_j=P_{j+1}-P_j$ andThe induction step multiplies by the remaining degree $n-r$ and takes one more finite difference. Derivatives above order $n$ vanish. In particular,If the actual parameter interval has length $h$ and $t$ is its normalized coordinate, derivatives in the original parameter acquire the additional factor $h^{-r}$. These formulas allow derivative evaluation and join tests using only the control polygon.

Write $B_j^n(t)=\left(\genfrac{}{}{0ex}{}{n}{j}\right)t^j(1-t{)}^{n-j}$ for the Bernstein basis, and set $B_j^n=0$ when $j$ is outside its index range. Differentiating and using $j\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{j-1}\right)$ and $(n-j)\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{j}\right)$ givesHence, after shifting the first sum by one index,Thus the derivative is a degree-at-most-$(n-1)$ Bézier curve whose control points are $n(P_{j+1}-P_j)$. They are vectors rather than necessarily positions on the original Bézier curve. This is the Bézier derivative control polygon.