More Algebra 2 problems.

__August 2017, Part IV__

The Question in Part IV is worth 6 credits. Work need be shown (or explained or justified) for full credit. Correct numerical answers with no work receive one credit.

*37. The value of a certain small passenger car based on its use in years is modeled by
V(t) = 28482.698(0.684) ^{t}, where V(t) is the value in dollars and t is the time in years. Zach had to take out a loan to purchase the small passenger car. The function Z(t) = 22151.327(0.778)^{t}, where Z(t) is measured in dollars, and t is the time in years, models the unpaid amount of Zach’s loan over time.
*

*Graph V(t) and Z(t) over the interval 0 < t < 5, on the set of axes below.
*

*State when V(t) = Z(t), to the nearest hundredth, and interpret its meaning in the context of the
problem.
*

*Zach takes out an insurance policy that requires him to pay a $3000 deductible in case of a
collision. Zach will cancel the collision policy when the value of his car equals his deductible.
To the nearest year, how long will it take Zach to cancel this policy? Justify your answer.
*

**Answer: **

Create Tables of Values for both V(t) and Z(t). You will see that an interval of 0 to 5 will be sufficient. Next, determine the scale. There is enough room on the y-axis for the scale to be 1 box to equal $2,000. The x-axis will measure the time, and you can use every 2, 3 or 4 boxes for each year (be consistent!) to spread the graph out, for readability.

The plot the points and label the graphs and the axes.

See the images below.

The point where V(t) = Z(t) can be found be finding the point of intersection in your calculator. This happens at about 1.95233, or 1.95, to the nearest hundredth. In the context of the problem, this is the point where the value of the car is equal to the remaining balance of the loan.

For the final portion of the question, since they want the answer rounded the nearest year, you don't need to solve algebraically. You only need to continue the table and find when the car will be worth less than $3000. At year 6, the car's value is V(6) = 28482.698(0.684)^{6} = 2916.87.

It is reasonable after six years, because that is the first time that the car will be worth less than $3000.

Comments and questions welcome.

More Algebra 2 problems.

## No comments:

Post a Comment