Chapter 6.1, Problem 59PPSE

Solution

To write a statement of equations to find cubic polynomial.

  $y=0.35x^3-1.85x^2+3.3x+1$

Given:

The cubic polynomial passes through the following points:

  $\left(-3,-35\right),\left(0,1\right),\left(2,3\right),\left(4,7\right)$

Calculation:

Let the given cubic polynomial be:

  $y=a{x}^3+b{x}^2+cx+d$

The point $\left(-3,-35\right)$ lie on the cubic polynomial, then

  $\begin{array}{l}-35=a{\left(-3\right)}^3+b{\left(-3\right)}^2+c\left(-3\right)+d\\ \Rightarrow -35=-27a+9b-3c+d\dots \left(1\right)\end{array}$

The point $\left(0,1\right)$ lie on the cubic polynomial, then

  $\begin{array}{l}1=a{\left(0\right)}^3+b{\left(0\right)}^3+c\left(0\right)+d\\ \Rightarrow d=1\dots \left(2\right)\end{array}$

The point $\left(2,3\right)$ lie on the cubic polynomial, then

  $\begin{array}{l}3=a{\left(2\right)}^3+b{\left(2\right)}^2+c\left(2\right)+d\\ \Rightarrow 3=8a+4b+2c+d\dots \left(3\right)\end{array}$

The point $\left(4,7\right)$ lie on the cubic polynomial, then

  $\begin{array}{l}7=a{\left(4\right)}^3+b{\left(4\right)}^2+c\left(4\right)+d\\ \Rightarrow 7=64a+16b+4c+d\dots \left(4\right)\end{array}$

Now put $d=1$ from equation $\left(2\right)$ to equation $\left(1\right),\left(3\right),\left(4\right)$ as:

  $\begin{array}{l}-27a+9b-3c+1=-35\\ \Rightarrow -27a+9b-3c=-36\dots \left(5\right)\end{array}$

  $\begin{array}{l}8a+4b+2c+1=3\\ \Rightarrow 8a+4b+2c=2\dots \left(6\right)\end{array}$

  $\begin{array}{l}64a+16b+4c+1=7\\ \Rightarrow 6a+16b+4c=6\dots \left(7\right)\end{array}$

Now, multiply the equation $\left(6\right)$ by $1.5$ :

  $\begin{array}{l}8a\times 1.5+4b\times 1.5+2c\times 1.5=2\times 1.5\\ \Rightarrow 12a+6b+3c=3\end{array}$

Adding the above equation to equation $\left(5\right)$ as:

  $\begin{array}{l}-27a+12a+9b+6b-3c+3c=36+3\\ \Rightarrow -15a+15b=-33\dots \left(8\right)\end{array}$

Now, multiply the equation $\left(6\right)$ by $-2$ as:

  $\begin{array}{l}\Rightarrow 8a\times \left(-2\right)+4b\times \left(-2\right)+2c\times \left(-2\right)=2\times \left(-2\right)\\ \Rightarrow -16a-8b-4c=-4\end{array}$

Now, adding the above equation to $\left(7\right)$ as:

  $\begin{array}{l}64a-16a+16b-8b+4c-4c=6-4\\ \Rightarrow 48a+8b=2\dots \left(9\right)\end{array}$

Now, from equation $\left(8\right)$ , the value of $b$ is:

  $b=\frac{-33+15a}{15}$

Now, put the value of $b$ in equation $\left(9\right)$ as:

  $\begin{array}{l}48a+8\left(\frac{-33+15a}{15}\right)=2\\ \Rightarrow 720a-264+120a=30\\ \Rightarrow 840a=30+264\\ \Rightarrow 840a=294\\ \Rightarrow a=\frac{294}{840}\\ \Rightarrow a=0.35\end{array}$

Next, put the value of $a$ in equation $\left(8\right)$ as:

  $\begin{array}{l}-15a+15b=-33\\ \Rightarrow -15\times 0.35+15b=-33\\ \Rightarrow 15b=-33+5.25\\ \Rightarrow b=\frac{-27.75}{15}\\ \Rightarrow b=-1.85\end{array}$

Now, put the value of $a,b$ in equation $\left(6\right)$ as:

  $\begin{array}{l}8a+4b+2c=2\\ \Rightarrow 8\times 0.35+4\times \left(-1.85\right)+2c=2\\ \Rightarrow 2.8-7.4+2c=2\\ \Rightarrow 2c=2+4.6\\ \Rightarrow 2c=6.6\\ \Rightarrow c=\frac{6.6}{2}\\ \Rightarrow c=3.3\end{array}$

Now, substitute the value of $a,b,c,d$ in the cubic polynomial $y=a{x}^3+b{x}^2+cx+d$ as:

  $y=0.35x^3-1.85x^2+3.3x+1$

Hence, the required cubic polynomial is:

  $y=0.35x^3-1.85x^2+3.3x+1$