Chapter 9.9, Problem 28PPS

(a)

To determine

The graph and which kind of function models the data

Answer to Problem 28PPS

The ordered pairs represent an exponential function

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:

  $\begin{array}{c}\text{First differences:}\\ d=y_2-y_1\\ =15910-18500\\ =-2590\\ d=y_3-y_2\\ =13682.60-15910\\ =-2227.4\\ d=y_4-y_3\\ =11767.04-13682.60\\ d=-1915.56\\ \text{Second differences:}\\ d=-2227.4-\left(-2590\right)\\ d=362.6\\ d=1915.56-2227.4\\ d=311.84\\ \text{Ratio of y-values:}\\ r_1=\frac{a_2}{a_1}\\ =\frac{15910}{18500}=0.86\\ r_2=\frac{a_3}{a_2}\\ =\frac{13682.6}{15910}=0.86\end{array}$

Thus, the first difference of y-values is not equal but the second difference is also no equal, so the given data represents an exponential model

General equation of exponential function is given by $v=a{b}^x$

(b)

To determine

The equation for the function that models the data

Answer to Problem 28PPS

The equation is $v=18500{\left(0.86\right)}^t$

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 the equation the ratio is 0.86 $b=0.86$ a is the initial value of car $a=18500$

Thus, the equation is given by $v=18500{\left(0.86\right)}^t$

(c)

To determine

How much the car is worth after 7 years.

Answer to Problem 28PPS

The cost is $\$6436.66$ .

Explanation of Solution

Calculation:

In order to find the cost of car in 7 years shown below:

  $\begin{array}{c}v=18500{\left(0.86\right)}^t\\ t=7\\ v=18500{\left(0.86\right)}^7\\ =18500\left(0.3479\right)\\ v=6436.66\end{array}$

Thus, the cost is $\$6436.66$ .