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Terrence buys a new car for [tex]\$20,000[/tex]. The value of the car depreciates by [tex]15\%[/tex] each year. If [tex]f(x)[/tex] represents the value of the car after [tex]x[/tex] years, which function represents the car's value?

A. [tex]f(x) = 20,000(0.85)^x[/tex]
B. [tex]f(x) = 20,000(0.15)^x[/tex]
C. [tex]f(x) = 20,000(1.15)^x[/tex]
D. [tex]f(x) = 20,000(1.85)^x[/tex]

Sagot :

Certainly! Let's go through the problem step-by-step.

1. Initial Purchase Value:
- Terrence buys a new car for \[tex]$20,000. 2. Depreciation Rate: - The value of the car depreciates by 15% every year. 3. Understand Depreciation: - Depreciation means a decrease in value. - A 15% depreciation each year means that each year the car retains 85% of its value. This is calculated as \(100\% - 15\% = 85\%\). 4. Express Remaining Value as a Decimal: - 85% in decimal form is 0.85. 5. Setting Up the Depreciation Function: - Let \(f(x)\) represent the value of the car after \(x\) years. - Initially, at \(x = 0\) years, the value is \$[/tex]20,000.
- After 1 year, the value is [tex]\(20,000 \times 0.85\)[/tex].
- After 2 years, the value is [tex]\(20,000 \times 0.85 \times 0.85 = 20,000 \times (0.85)^2\)[/tex].
- After 3 years, the value is [tex]\(20,000 \times 0.85 \times 0.85 \times 0.85 = 20,000 \times (0.85)^3\)[/tex].
- Generally, after [tex]\(x\)[/tex] years, the value will be [tex]\(20,000 \times (0.85)^x\)[/tex].

6. Conclusion:
- The function [tex]\(f(x)\)[/tex] that represents the value of the car after [tex]\(x\)[/tex] years is:
[tex]\[ f(x) = 20,000 \times (0.85)^x \][/tex]

Among the given options:
- [tex]\(f(x) = 20,000(0.85)^x\)[/tex]
- [tex]\(f(x) = 20,000(0.15)^x\)[/tex]
- [tex]\(f(x) = 20,000(1.15)^x\)[/tex]
- [tex]\(f(x) = 20,000(1.85)^x\)[/tex]

The correct function is:
[tex]\[ f(x) = 20,000(0.85)^x \][/tex]