The value of a certain car model decreases by 12\% each year. If the initial value of the … — SAT

The value of a certain car model decreases by $12\%$ each year. If the initial value of the car is $\$25,000$ , which of the following functions models the value of the car, $V(t)$ , in dollars, $t$ years after it was purchased?

A
$V(t) = 25,000 - 0.12t$
B
$V(t) = 25,000(0.12)^t$
$V(t) = 25,000(0.88)^t$ Answer
D
$V(t) = 25,000(1.12)^t$

Answer

The function

V(t) = 25,000(0.88)^t

The initial value of the car is

25,000

. An annual decrease of

12\%

means that the car retains

100\% - 12\% = 88\%

of its value from the previous year. This results in a decay factor of

1 - 0.12 = 0.88

. Substituting these values into the exponential model

V(t) = a(b)^t

yields

V(t) = 25,000(0.88)^t

Step-by-Step Solution

Identify the initial value of the car.

The initial value is

25,000

This is the value of the car at

t = 0

, which corresponds to the coefficient

a

in the general exponential model

V(t) = a(b)^t

Determine the growth or decay factor base,

b

The base is

1 - 0.12 = 0.88

Since the car's value decreases by

12\%

each year, it retains

100\% - 12\% = 88\%

of its value. This is represented as a decay factor of

0.88

Formulate the exponential decay equation.

The equation is

V(t) = 25,000(0.88)^t

By substituting the initial value

a = 25,000

and the decay base

b = 0.88

into

V(t) = a(b)^t

Key Concept

Exponential decay models and interpretation of growth/decay factors

Alternative Method

Calculate the value of the car after one year (

t = 1

). A

12\%

decrease on a

\$25,000

car means its value drops by

\$3,000

\$22,000

. Substituting

t=1

into the correct function should yield

22,000

. Calculating for

V(t) = 25,000(0.88)^t

t = 1

gives

25,000(0.88)^1 = 22,000

, confirming the choice.

Estimated Time:45s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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