A quartic Bézier curve is determined by five control points,

\({{\bf{p}}_{\bf{o}}}{\bf{,}}\,{\rm{ }}{{\bf{p}}_{\bf{1}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{2}}}\,{\bf{,}}{\rm{ }}{{\bf{p}}_{\bf{3}}}\)and \({{\bf{p}}_4}\):

\({\bf{x}}\left( t \right) = {\left( {1 - t} \right)^4}{{\bf{p}}_0} + 4t{\left( {1 - t} \right)^3}{{\bf{p}}_1} + 6{t^2}{\left( {1 - t} \right)^2}{{\bf{p}}_2} + 4{t^3}\left( {1 - t} \right){{\bf{p}}_3} + {t^4}{{\bf{p}}_4}\)for \(0 \le t \le 1\)

Construct the quartic basis matrix \({M_B}\) for \({\bf{x}}\left( t \right)\).

The weighted vector of \({\rm{x}}\left( t \right)\) is shown below:

\(\left( {\begin{array}{{}}{1 - 4t + 6{t^2} - 4{t^3} + {t^4}}\\{4t - 12{t^2} + 12{t^3} - 4{t^4}}\\{6{t^2} - 12{t^3} + 6{t^4}}\\\begin{array}{l}4{t^3} - 4{t^4}\\\,\,\,\,\,\,\,\,\,\,{t^4}\end{array}\end{array}} \right)\)

Factor the weighted vector as\({M_B}{\bf{u}}\left( t \right)\), where\({\bf{u}}\left( t \right)\)is the column vector involving ascending powers of t as shown below:

\({M_B}u\left( t \right) = \left( {\begin{array}{{}}1&{ - 4}&6&{ - 4}&1\\0&4&{ - 12}&{12}&{ - 4}\\0&0&6&{ - 12}&6\\0&0&0&4&{ - 4}\\0&0&0&0&1\end{array}} \right)\left( \begin{array}{l}1\\t\\{t^2}\\{t^3}\\{t^4}\end{array} \right)\)

Thus, the vector is \({M_B} = \left( {\begin{array}{{}}1&{ - 4}&6&{ - 4}&1\\0&4&{ - 12}&{12}&{ - 4}\\0&0&6&{ - 12}&6\\0&0&0&4&{ - 4}\\0&0&0&0&1\end{array}} \right)\).