College

A car is purchased for the price of [tex]\$7,000[/tex]. After two years, the value of the car has decreased to [tex]\$3,937.50[/tex]. Which exponential model represents the value of this car [tex]x[/tex] years after purchase?

A. [tex]P(x) = 7000 \times 0.75^x[/tex]
B. [tex]P(x) = 7000 \times 0.75^{2r}[/tex]
C. [tex]P(x) = 7000 \times 0.5625^{2x}[/tex]
D. [tex]P(x) = 3937.5 \times 0.5625^x[/tex]

Answer :

We start with the general exponential depreciation model for the car’s value:

[tex]$$
P(x) = 7000 \times r^x,
$$[/tex]

where [tex]\( P(x) \)[/tex] is the value of the car after [tex]\( x \)[/tex] years, and [tex]\( r \)[/tex] is the annual depreciation factor.

After 2 years the car’s value is given as:

[tex]$$
P(2) = 7000 \times r^2 = 3937.50.
$$[/tex]

To find [tex]\( r \)[/tex], we solve for [tex]\( r^2 \)[/tex]:

[tex]$$
r^2 = \frac{3937.50}{7000} = 0.5625.
$$[/tex]

Taking the square root of both sides gives:

[tex]$$
r = \sqrt{0.5625} = 0.75.
$$[/tex]

Thus, the exponential model that represents the value of the car [tex]\( x \)[/tex] years after purchase is:

[tex]$$
P(x) = 7000 \times (0.75)^x.
$$[/tex]

This matches the first option:

[tex]$$
P(x) = 7000 \times 0.75^x.
$$[/tex]