https://file.aiquickdraw.com/mathbot/file/1726427717135irn6zsq6.jpg

Let's go through each part of the problem:

### a) How do we know that we should use a linear model?
We should use a linear model because the car’s value is decreasing at a constant rate, which matches the characteristics of a linear relationship.

### b) Fill in the table:
The car's value decreases by $1500 per year. Starting from $19,000 in 2005, we can calculate the missing values.

- For $ x = 4 $ years:
\[
y = 19000 - 1500 \times 4 = 19000 - 6000 = 13000
\]
This value is already filled in the table.

- For $ x = 12.8 $ years:
\[
y = 19000 - 1500 \times 12.8 = 19000 - 19200 = -200
\]
However, the value cannot go below $0, so it is truncated to $ y = 0 $.

The complete table should look like this:
| $ x $ (years) | $ y $ (value of car in dollars) |
|:---------------:|:--------------------------------:|
| 0               | 19000                           |
| 1               | 17500                           |
| 4               | 13000                           |
| 12.8            | 0                               |

### c) Write the function relating $ y $ to $ x $:
The slope of the linear equation is the rate of decrease, which is $1500 per year. The initial value (intercept) is $19,000. Therefore, the linear function is:
\[
y = 19000 - 1500x
\]

### d) Graph the function:
The graph shows the linear decrease of the car's value over time. The y-axis represents the value of the car (in thousands of dollars), and the x-axis represents the number of years since 2005.

You can plot points such as:
- $ (0, 19) $ (starting value $19,000)
- $ (1, 17.5) $
- $ (4, 13) $
- $ (12.8, 0) $

Would you like any details or further explanation?

Here are 5 related questions to expand on this:
1. What would the car's value be after 10 years using the same model?
2. How would the function change if the rate of decrease was $2000 per year instead of $1500?
3. Can the linear model still be applied if the car's value hits zero before the end of the timeline?
4. What does the slope of the graph represent in this scenario?
5. How does the y-intercept of the graph affect the interpretation of the car’s value?

**Tip**: Linear models work best when rates of change are constant, as they ensure a straight-line relationship between the variables.

The value of a certain car was $19,000 in 2005 and has been decreasing at a constant rate of $1500 per year since then. How do we know that we should use a linear model to represent the value of the car? Fill in the table with missing values and write a function relating y (value of the car) to x (number of years since 2005).

Solve a problem where a car's value, starting at $19,000 in 2005, decreases by $1500 per year. Learn how to fill in missing table values, derive the linear equation y = 19000 - 1500x, and graph the function. Suitable for Grades 8-10.

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