Chapter 9.9, Problem 12CYU

To determine

Which kind of model best describe the data and equation

Answer to Problem 12CYU

The equation is $y=\frac{1}{4}x-1$

Explanation of Solution

Given:

  

Concept Used:

If first difference of y-values is equal, then model be linear

If second difference be equal, then the model be quadratic

If first and second differences are not equal then the model will be exponential

Calculation:

In order to check whether the given data is linear, quadratic or exponential, find the first difference as shown below:

  $\begin{array}{c}\text{First differences:}\\ d=y_2-y_1\\ =-1-\left(-1.25\right)\\ =0.25\\ d=y_3-y_2\\ =-0.75-\left(-1\right)\\ =0.25\\ d=y_4-y_3\\ =-0.5-\left(-0.75\right)\\ d=0.25\end{array}$

Thus, the first difference is equal, so the given data represents a linear model of the form of $y=mx+b$

The equation is for points $\left(0,-1\right)$

  $\begin{array}{l}\left(0,-1\right)\\ y=mx+b\\ -1=\frac{1}{4}\left(0\right)+b\\ -1=0+b\\ \Rightarrow b=-1\end{array}$

Thus, the equation is $y=\frac{1}{4}x-1$