Chapter 4.8B, Problem 1GCE

To determine

A cubic polynomial function whose graph passes through the given points.

Explanation of Solution

Given information:

The given pints are $(-2,-23),(-1,-21),(0,15)\&(1,1)$ .

Calculations:

To get the desired function, let us consider the function be,

  $y=a{x}^3+b{x}^2+cx+d$

Now putting the value of each point in the function, we

find the value of the constraints.

  $\begin{array}{l}\text{For the first point,}(-2,-23)\text{, we take }x=-2\text{ and }y=-23\text{ and putting it in the equation,}\\ a{x}^3+b{x}^2+cx+d=y\\ $ a{(-2)}^3+b{(-2)}^2+c(-2)+d=-23$-8a+4b-2c+d=-\mathrm{23..............}(1)\\ \text{For the second point,}(-1,21)\\ a{x}^3+b{x}^2+cx+d=y\\ $ a{(-1)}^3+b{(-1)}^2+c(-1)+d=21$-a+b-c+d=\mathrm{21..............}(2)\\ \text{For the third point,}(0,15)\\ a{x}^3+b{x}^2+cx+d=y\\ $ a{(0)}^3+b{(0)}^2+c(0)+d=15$d=\mathrm{15..............}(3)\\ \text{For the fourth point,}(1,1)\\ a{x}^3+b{x}^2+cx+d=y\\ $ a{(1)}^3+b{(1)}^2+c(1)+d=1$a+b+c+d=\mathrm{1..............}(4)\end{array}$

Now we solve the four equations and find the constrains,

  $\begin{array}{l}-8a+4b-2c+d=-\mathrm{23..............}(1)\\ -a+b-c+d=\mathrm{21..............}(2)\\ d=\mathrm{15..............}(3)\\ a+b+c+d=\mathrm{1..............}(4)\\ \text{Putting the value of equation(3) in equation (2) we get,}\\ -a+b-c+15=21\\ -a+b-c=\mathrm{6.....................}(5)\\ Adding(4)+(5)\\ a+b+c+d-a+b-c=1+6\\ 2b+d=7\\ 2b=7-15\\ 2b=-8\\ b=-4\\ \text{Multiplying equation (2) with 2 we get,}\\ -2a+2b-2c+2d=21\times \mathrm{2......}(6)\\ \text{Subtracting equation, (2)from equation(1),we get, }\\ -8a+4b-2c+d+2a-2b+2c-2d=-23-42\\ -6a+2b-d=-65\\ -6a=-65+d-2b\\ -6a=-65+15-(2\times (-4))\\ -6a=-42\\ a=7\\ \text{Now from equation(5),}\\ -a+b-c=6\\ -7-4-c=6\\ c=-17\\ \text{Thus we get,}\\ a=7,\\ b=-4,\\ c=-17\\ d=15\\ \text{Therefore the desired equation is,}\\ y=7x^3-4x^2-17x+15\end{array}$