Chapter 2, Problem 4CT

To determine

To calculate: To explain why a quadratic function models the data. Further, use a linear system to find the model.

Answer to Problem 4CT

The equation of the parabola is $y=-x^2-\frac{1}{2}x+5$ .

Explanation of Solution

Given information: Following data is given,

x	y
2	0
4	-13
6	-34
8	-63
10	-100

Formula used: nth function models the data if nth difference of the model is constant.

Calculation:

Find the first and second differences,

x	y	First difference	Second difference
2	0
4	-13	$-13-0=-13$
6	-34	$\begin{array}{c}-34-\left(-13\right)=-34+13\\ =-21\end{array}$	$\begin{array}{c}-21-\left(-13\right)=-21+13\\ =-8\end{array}$
8	-63	$\begin{array}{c}-63-\left(-34\right)=-63+34\\ =-29\end{array}$	$\begin{array}{c}-29-\left(-21\right)=-29+21\\ =-8\end{array}$
10	-100	$\begin{array}{c}-100-\left(-63\right)=-100+63\\ =-37\end{array}$	$\begin{array}{c}-37-\left(-29\right)=-37+29\\ =-8\end{array}$

Clearly, it can be observed that the second differences are constant and therefore the quadratic function models the data.

That is, the data can be modeled by the equation,

  $y=a{x}^2+bx+c$

Since three unknowns are to be determined, substitute three values from the model data and simplify for a, b and c

  $\begin{array}{c}y=a{x}^2+bx+c\\ 0=a{\left(2\right)}^2+b\left(2\right)+c\left(x=2,y=0\right)\\ 4a+2b+c=0\mathrm{......}\left(1\right)\\ -13=a{\left(4\right)}^2+b\left(4\right)+c\left(x=4,y=-13\right)\\ 16a+4b+c=-13\mathrm{......}\left(2\right)\\ -34=a{\left(6\right)}^2+b\left(6\right)+c\left(x=6,y=-34\right)\\ 36a+6b+c=-34\mathrm{......}\left(3\right)\end{array}$

On simplifying equations (1), (2) and (3),

  $\begin{array}{c}a=-1\\ b=-\frac{1}{2}\\ c=5\end{array}$

Hence, the quadratic equation that models the data is,

  $y=-x^2-\frac{1}{2}x+5$