Chapter 2.4, Problem 32E

To determine

Determine whether the data are linear, quadratic, or neither. Explain. If linear or quadratic, write an equation that fits the data.

Answer to Problem 32E

  $y(x)=-16x^2-x+320.$

Explanation of Solution

Given information: The given data is.

  

Calculation :

$y(0)$		$y(1)$		$y(2)$		$y(3)$		$y(4)$
320		303		254		173		60
	-17		-49		-81		-113
		-32		-32		-32

Because second differences are constant so we can modal the data with a quadratic function .

  $y(x)=a{x}^2+bx+c.$......(1)

  $\begin{array}{l}$ y(0)=a\times {(0)}^2+b\times 0+c.$320=0+0+c.\\ c=320.\\ $ y(1)=a\times {(1)}^2+b\times 1+c.$303=a+b+320.\\ a+b=-\mathrm{17.........}(2)\\ $ y(2)=a\times {(2)}^2+b\times 2+c.$254=4a+2b+320.\\ 4a+2b=-\mathrm{66.......}(3)\\ \text{solving}\text{(2)}\text{and}\text{(3)}.\\ a=-16,b=-1,c=320.\\ \text{Putting}a=-16,b=-1,c=320\text{in}\text{equation}(1)\\ y(x)=-16x^2-x+{320.}^{}\end{array}$