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A car purchased for [tex]$12,000 depreciates by $[/tex]850 each year under a straight-line method. Which equation best models this depreciation?

A. [tex]\( y = 12000 - 850x \)[/tex]
B. [tex]\( y = 12000 + 850x \)[/tex]
C. [tex]\( y = 12000x + 850 \)[/tex]
D. [tex]\( y = 12000x - 850 \)[/tex]

Sagot :

To determine the best model for the car's depreciation using the straight-line method, let's break down the information given and go through the logical steps:

1. Initial Cost of the Car: The car is purchased for [tex]$12,000. 2. Annual Depreciation: The car loses $[/tex]850 each year.

3. Formulating the Depreciation Equation:
- Let [tex]\( y \)[/tex] represent the value of the car after [tex]\( x \)[/tex] years.
- Initially, when [tex]\( x = 0 \)[/tex], the value of the car [tex]\( y \)[/tex] is its initial cost: [tex]$12,000. - Each year, the car loses $[/tex]850 in value. Therefore, after [tex]\( x \)[/tex] years, the total depreciation would be [tex]\( 850x \)[/tex].

4. Constructing the Equation:
- We start with the initial cost and subtract the total depreciation over [tex]\( x \)[/tex] years.
- Thus, the value of the car after [tex]\( x \)[/tex] years can be modeled by the equation:
[tex]\[ y = 12000 - 850x \][/tex]

5. Evaluating the Given Options:
- Option A: [tex]\( y = 12000 - 850x \)[/tex]
- This matches our derived equation.
- Option B: [tex]\( y = 12000 + 850x \)[/tex]
- This would imply the car's value increases each year by [tex]$850, which contradicts the depreciation. - Option C: \( y = 12000x + 850 \) - This equation incorrectly suggests the value depends on multiplying the number of years by the initial cost and then adding $[/tex]850.
- Option D: [tex]\( y = 12000x - 850 \)[/tex]
- This suggests an initial value of zero and grows based on the formula, which does not match our initial condition.

Thus, the correct equation that models this depreciation is:

[tex]\[ y = 12000 - 850x \][/tex]

So, the answer is:

A. [tex]\( y = 12000 - 850x \)[/tex]