Step-by-step explanation:
Given the table of values where it represents the relationship between the change in value of a car and number of years.
<h3>Slope:</h3>
Before we can establish the linear equation, we need to determine its slope by using two ordered pairs from the table.
Let (x₁, y₁) = (1, 84)
(x₂, y₂) = (3, 64)
Substitute these values into the following <u>slope formula:</u>
m = (y₂ - y₁)/(x₂ - x₁)
m = (64 - 84)/(3 - 1)
m = -20/2
m = -10
Therefore, the slope of the linear function is -10.
<h3>1. Enter a function to represent the car's value over time, assuming that the car's value is linear for the first 5 years.</h3>
Assuming that the car's depreciation in value is linear for the first 5 years, then we can establish the following function:
Next, using one of the given ordered pairs from the table, (1, 84), and the slope, m = -10, substitute these values into the slope-intercept form to solve for the value of the car at x = 0 (the y-intercept, <em>b </em>):
84 = -10(1) + b
84 = -10 + b
Add 10 to both sides to isolate b:
84 + 10 = -10 + 10 + b
94 = b
Therefore, the linear function is f(x) = -10x + 94.
To verify whether this is the correct function to represent the given linear model, substitute the values for year 5, where the car's value is at 44%:
f(5) = -10(5) + 94
f(5) = -50 + 94
f(5) = 44 ⇒ This matches the value from the given table, where it implies that at year 5, the value of the car is 44%.
<h3>2. According to your model, how much did the car's value drop the day it was purchased and driven off the lot?</h3>
Since the car's value is expressed in percentage (%), then it means that at year = 0, the car's value is 100%. The value of the car in year 1 is 84%.
Therefore, the car's value will decrease by 16% from the day it was purchased (at year = 0) <u>through year 1</u>.
To determine the value of the car <u>the day it was purchased</u> and driven off the lot:
Let x = 1 out of 365 days = 1/365 = 0.0027
f(0.0027) = -10(0.0027) + 94
f(0.0027) = -10(0.0027) + 94
f(0.0027) = -0.027 + 94
f(0.0027) = 93.973
The car's value at <u>Day 1</u> is 93.97%, which means that <u>it decreased by 6%</u>.
<h3>3. Do you think the linear model would still be useful after 10 years? Explain.</h3>
In order to determine whether the linear model would still be useful after 10 years, let x = 10:
f(x) = -10x + 94
f(10) = -10(10) + 94
f(10) = -100 + 94
f(10) = -100 + 94
f(10) = -6
The value of the car by the 10th year will be -6%, implying that it is no longer useful by year 10. The linear model is <u>only useful up to the 9th year</u>:
f(9) = -10(9) + 94
f(9) = -90 + 94
f(9) = 4 ⇒ The car's value by the 9th year is 4%.
<h3>4. Suppose you used months instead of years to write a function. How would your model change?</h3>
Since in the original function, x represents the number of years, and there are 12 months in a year:
Let 12x = 12 months × 1 year
Substitute 12x as an input into the original function:
f(x) = -10x + 94
f(x) = -10(12x) + 94
f(x) = -120x + 94 ⇒ This represents the function that models linear relationship between the percentage value of a car relative to the number of months that the car is driven.
