Chapter 7.4, Problem 90E

Solution

i.

The cubic equation whose graph passes through the given points.

The cubic equation is $y(t)=\frac{302}{3}{t}^3-1473t^2+\frac{21676}{3}t-8908$ .

Given:

The ordered pair $(t,y(t))$ .The points are $(3,2229),(4,2868),(5,2977),(6,3160)$

Calculation:

The standard cubic expression is,

  $y(t)=a{t}^3+b{t}^2+ct+d$

Substituting the points on the standard form,

  $\begin{array}{l}y(3)=a{3}^3+b{3}^2+c3+d=2229\\ y(4)=a{4}^3+b{4}^2+c4+d=2868\\ y(5)=a{5}^3+b{5}^2+c5+d=2977\\ y(6)=a{6}^3+b{6}^2+c6+d=3160\end{array}$

The system of equation is:

  $\begin{array}{l}27a+9b+3c+d=2229\\ 64a+16b+4c+d=2868\\ 125a+25b+5c+d=2977\\ 216a+36b+6c+d=3160\end{array}$

It can be written in matrix form as,

The augmented matrix is:

  $\left(\begin{array}{ccccc}27& 9& 3& 1& 2229\\ 64& 16& 4& 1& 2868\\ 125& 25& 5& 1& 2977\\ 216& 36& 6& 1& 3160\end{array}\right)$

Solving we get,

  $\begin{array}{l}a=\frac{302}{3}\\ b=-1473\\ c=\frac{21676}{3}\\ d=-8908\end{array}$

Therefore the cubic expression is,

  $y(t)=\frac{302}{3}{t}^3-1473t^2+\frac{21676}{3}t-8908$ .

ii.

The cubic function.

  

Given: $y(t)=\frac{302}{3}{t}^3-1473t^2+\frac{21676}{3}t-8908$ .

Calculation: