### Part A: Identifying the Type of Function
To determine the type of function that describes the value of each car over time, we need to analyze how their values change year by year.
Car 1:
- Value in year 1: [tex]$32,000
- Value in year 2: $[/tex]26,000
- Value in year 3: [tex]$20,000
We observe a consistent decrease in value each year. Specifically:
- From year 1 to year 2: $[/tex]32,000 - [tex]$26,000 = $[/tex]6,000
- From year 2 to year 3: [tex]$26,000 - $[/tex]20,000 = [tex]$6,000
Since the value decreases by a fixed amount each year, this is indicative of a linear function. In a linear function, the difference between consecutive values (common difference) remains constant.
Car 2:
- Value in year 1: $[/tex]32,300
- Value in year 2: [tex]$27,455
- Value in year 3: $[/tex]23,336.75
We note a percentage decrease each year. Specifically:
- From year 1 to year 2: [tex]$27,455 / $[/tex]32,300 ≈ 0.85
- From year 2 to year 3: [tex]$23,336.75 / $[/tex]27,455 ≈ 0.85
Since the value decreases by a consistent ratio each year, this matches an exponential function. In an exponential function, the ratio between consecutive values (common ratio) remains constant.
### Part B: Writing the Function for Each Car
Car 1: Linear Function
Given:
- Constant decrease per year: [tex]$6,000
- Initial value (year 0): Let's denote it as $[/tex]a_0[tex]$, which we need to find.
If $[/tex]a_n[tex]$ represents the value of the car at year $[/tex]n[tex]$, then:
1. $[/tex]a_1 = a_0 - 6,000 = 32,000[tex]$, hence $[/tex]a_0 = 38,000[tex]$
2. Linear function form: $[/tex]f(x) = a_0 - d \times x[tex]$ where $[/tex]d[tex]$ is the constant decrease per year.
So for Car 1:
\[ f(x) = 38,000 - 6,000x \]
Car 2: Exponential Function
Given:
- Constant ratio per year: $[/tex]0.85[tex]$
- Initial value: $[/tex]a_0 = 32,300[tex]$ (value at year 1)
Exponential function form: $[/tex]f(x) = a_0 \times r^x[tex]$ where $[/tex]r[tex]$ is the ratio.
So for Car 2:
\[ f(x) = 32,300 \times (0.85)^{x-1} \]
### Comparison after Five Years
Value of Car 1 after 5 years:
\[ f(5) = 38,000 - 6,000 \times 5 \]
\[ f(5) = 38,000 - 30,000 \]
\[ f(5) = 8,000 \]
Value of Car 2 after 5 years:
\[ f(5) = 32,300 \times (0.85)^{5-1} \]
\[ f(5) = 32,300 \times (0.85)^4 \]
\[ f(5) ≈ 32,300 \times 0.52200625 \]
\[ f(5) ≈ 16,846.319 \]
Comparing the values after five years:
- Car 1: \$[/tex]8,000
- Car 2: \[tex]$14,331.68
Therefore, after five years, Car 2 is projected to have a higher value (\$[/tex]14,331.68) compared to Car 1 (\$8,000). This suggests that Car 2 retains its value better over time.
