A new 2014 luxury sedan that sold for $38,880 has a book value V of $24,319 after t = 2 years.

Question

(a) Find a linear model for the value V of the sedan.

V(t) =

(b) Find an exponential model for the value V of the sedan. Round the numbers in the model to four decimal places.

V(t) =

(c) Use a graphing utility to graph the two models in the same viewing window.

(d) Which model represents a greater depreciation rate in the first year?

1. linear model
2. exponential model

(e) For what years is the value of the sedan greater using the linear model? the exponential model? (Round your answers to the nearest integer. Enter your answers using interval notation.)

linear model:

exponential model:

Answer 1

Use the given points in the problem to start the set-up. For example the beginning value of the car in 2014 is 38880. That means after 0 years the care is worth 38880. The second point comes from the value of the car after 2 years. After 2 years the value of the care is 24,319. So, the given points are (0,38,880) and (2, 24,319). Find the slope: (24,319-38,880)/(2-0) = -7280.5. Now, use the point-slope form to find the linear equation. Point-slope equation is : y-y1 = m(x-x1).

Part b requires us to use the exponential function Pe^(rt) where we need to find the rate first. So, the set-up after two years is 24,319 = 38,880e^(-r(2)). Divide both sides by 38,880 and you have: .625488. So the equation looks like this so far:

.625488 = e^(-2r)

NOw take the natural log of each side. ln(.625488) = -2r, then solve for r. r = .235.

Our exponential equation becomes A = Pe^(rt) or y = 38,880e^(-.235t)