Question

he table below shows the value in dollars of a car at the end of "x"x years.
Number of Years, x
Value, v(x) (dollars)
11 , 000
9 , 350
7 , 948
6 , 755
Which exponential function models this situation?

"y" equals 85 left parenthesis, 11000 , right parenthesis Superscript, "x" , Baseline
"y" equals 11000 left parenthesis, 0.8 5 , right parenthesis Superscript, "x" , Baseline
"y" equals 11000 left parenthesis, 1.1 5 , right parenthesis Superscript, "x" , Baseline

Answer 1

To find an exponential function that models the situation of the car's value over time, we first need to determine the decay factor based on the provided values.

Given:
- At year 0 (assumed unless otherwise stated), the value is \(v(0) = 11000\).
- At year 1 (given or can be inferred), you would typically have the value decrease to \(v(1) = 9350\).
- At year 2, we can infer you would have the value as \(v(2) = 948\).
- At year 3, it would be \(v(3) = 755\).

To model the depreciation, we will explore the values to find the decay rate.

### Step 1: Calculate the decay factor

1. Year 0 to Year 1:
- Value decreases from 11000 to 9350.
- The decay factor can be calculated as:
\[
\text{Decay Factor} = \frac{9350}{11000} \approx 0.850
\]

### Step 2: Define the exponential function

The general formula for exponential decay is:
\[
y = a \cdot b^x
\]
Where:
- \(a\) is the initial value (in this case, 11000),
- \(b\) is the decay factor (which we approximated as 0.850),
- \(x\) is the number of years.

Thus, the model becomes:
\[
y = 11000 \cdot (0.85)^x
\]

### Conclusion

The correct exponential function that models this situation:
\[
\mathbf{y = 11000 \cdot (0.85)^x}
\]

This matches with your provided option of:
- "y" equals 11000 left parenthesis, 0.85 right parenthesis Superscript, "x".

So the answer is:
\[
\boxed{y = 11000(0.85)^x}
\]