Polynomials model many variable relationships, including volume and surface area.

Cubic polynomials are often used in economics to model cost functions. The cost of producing \(x\) items is an increasing function of \(x\text{,}\) but its rate of increase is usually not constant.

In

Example 8.1.16c, we solved a cubic equation graphically. There is a cubic formula, analogous to the quadratic formula, that allows us to solve cubic equations algebraically, but it is complicated and not often used.

Cubic polynomials are also used to model smooth curves connecting given points. Such a curve is called a cubic spline.

####
Investigation 8.1.1. Bézier Curves.

A Bézier curve is actually a sequence of short curves pieced together. Each piece has two endpoints and (for nonlinear curves) at least one control point. The control points do not lie on the curve itself, but they determine its shape. Two polynomials define the curve, one for the \(x\)-coordinate and one for the \(y\)-coordinate.

*A. Linear Bézier Curves*

The linear Bézier curve for two endpoints, \((x_1, y_1)\) and \((x_2, y_2)\text{,}\) is the straight line segment joining those two points. The curve is defined by the two functions

\begin{align*}
x \amp= f(t) = x_1\cdot (1 - t) + x_2\cdot t \\
y \amp = g(t) = y_1 \cdot (1 - t) + y_2 \cdot t
\end{align*}

for \(0\le t\le 1\text{.}\)

Find the functions \(f\) and \(g\) defining the linear Bézier curve joining the two points \((-4,7)\) and \((20)\text{.}\) Simplify the formulas defining each function.

Fill in the table of values and plot the curve.

\(t\) |
\(0\) |
\(0.25\) |
\(0.5\) |
\(0.75\) |
\(1\) |

\(x\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |

\(y\) |
\(\) |
\(\) |
\(\) |
\(\) |
\(\) |

*B. Quadratic Bézier Curves: Drawing a Simple Numeral 7*

The quadratic Bézier curve is defined by two endpoints, \((x_1, y_1)\) and \((x_3, y_3)\text{,}\) and a control point, \((x_2, y_2)\text{.}\)

\begin{align*}
x \amp= f(t)=x_1\cdot(1 - t)^2+2x_2\cdot t (1 - t) + x_3\cdot t^2\\
y \amp = g(t)= y_1\cdot(1 - t)^2 +2y_2\cdot t (1 - t) + y_3 \cdot t^2
\end{align*}

for \(0\le t\le 1\text{.}\)

Find the functions \(f\) and \(g\) for the quadratic Bézier curve defined by the endpoints \((-4, 7)\) and \((2, 0)\text{,}\) and the control point \((0, 5)\text{.}\) Simplify the formulas defining each function.

Fill in the table of values and plot the curve.

\(t\) |
\(0\) |
\(0.25\) |
\(0.5\) |
\(0.75\) |
\(1\) |

\(x\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |

\(y\) |
\(\) |
\(\) |
\(\) |
\(\) |
\(\) |

Draw a line segment from \((-4,7)\) to \((4,7)\) on the grid above to complete the numeral 7.

We can adjust the curvature of the diagonal stroke of the 7 by moving the control point. Find the functions \(f\) and \(g\) for the quadratic Bézier curve defined by the endpoints \((4,7)\) and \((0,-7)\text{,}\) and the control point \((0,-3)\text{.}\) Simplify the formulas defining each function.

Fill in the table of values and plot the curve.

\(t\) |
\(0\) |
\(0.25\) |
\(0.5\) |
\(0.75\) |
\(1\) |

\(x\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |

\(y\) |
\(\) |
\(\) |
\(\) |
\(\) |
\(\) |

Draw a line segment from \((-4, 7)\) to \((4, 7)\) on the grid above to complete the numeral 7.

On your graphs in steps (5) and (8), plot the three points that defined the curved section of the numeral 7, then connect them in order with line segments. How does the position of the control point change the curve?

*C. Cubic Bézier Curves: Drawing a Letter y*

A cubic Bézier curve is defined by two endpoints, \((x_1,y_1)\) and \((x_4,y_4)\text{,}\) and two control points, \((x_2,y_2)\) and \((x_3,y_3)\text{.}\)

\begin{align*}
x \amp= f(t) = x_1\cdot (1-t)^2 +3x_2\cdot t(1-t)^2 +3x_3\cdot t^2(1-t) + x_4\cdot t^3\\
y \amp = g(t) = y_1\cdot (1-t)^2 +3y_2\cdot t(1-t)^2 +3y_3\cdot t^2(1-t) + y_4 \cdot t^3
\end{align*}

for \(0\le t \le 1\text{.}\)

Find the functions \(f\) and \(g\) for the cubic Bézier curve defined by the endpoints \((4,7)\) and \((-4,-5)\text{,}\) and the control points \((3,3)\) and \((0,-8)\text{.}\) Simplify the formulas defining each function.

Fill in the table of values and plot the curve.

\(t\) |
\(0\) |
\(0.25\) |
\(0.5\) |
\(0.75\) |
\(1\) |

\(x\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |

\(y\) |
\(\) |
\(\) |
\(\) |
\(\) |
\(\) |

Connect the four given points in order using three line segments. How does the position of the control points affect the curve? Finish the letter y by including the linear Bézier curve you drew for step (2).