Calculus (3rd Edition)

The equation of the Bézier curve is $\left( { - 4{t^3} + 6{t^2} - 1, - 6{t^2} + 6t - 1} \right)$.
We have ${P_0} = \left( {{a_0},{b_0}} \right) = \left( { - 1, - 1} \right)$, ${P_1} = \left( {{a_1},{b_1}} \right) = \left( { - 1,1} \right)$, ${P_2} = \left( {{a_2},{b_2}} \right) = \left( {1,1} \right)$, ${P_3} = \left( {{a_3},{b_3}} \right) = \left( {1, - 1} \right)$. Using Eq. (9) and Eq. (10) we obtain $x\left( t \right) = - {\left( {1 - t} \right)^3} - 3t{\left( {1 - t} \right)^2} + 3{t^2}\left( {1 - t} \right) + {t^3}$, $x\left( t \right) = - 4{t^3} + 6{t^2} - 1$ $y\left( t \right) = - {\left( {1 - t} \right)^3} + 3t{\left( {1 - t} \right)^2} + 3{t^2}\left( {1 - t} \right) - {t^3}$ $y\left( t \right) = - 6{t^2} + 6t - 1$ So, the equation of the Bézier curve is $\left( { - 4{t^3} + 6{t^2} - 1, - 6{t^2} + 6t - 1} \right)$.