To determine which exponential function models the depreciation of the car's value, we can analyze the given values at the end of \(x\) years.
The values given are:
- At \(x = 0\), \(v(0) = 11,000\)
- At \(x = 1\), \(v(1) = 9,350\)
- At \(x = 2\), \(v(2) = 7,948\)
- At \(x = 3\), \(v(3) = 6,755\)
An exponential decay function can generally be modeled in the form: \[ v(x) = v_0 \cdot (r)^x \] where \(v_0\) is the initial value and \(r\) is the decay factor (a number between 0 and 1).
Starting with the initial value, we see \(v_0 = 11,000\).
Next, we need to calculate the decay factor \(r\). We can find it by looking at the values over one year.
From \(x = 0\) to \(x = 1\): \[ \frac{v(1)}{v(0)} = \frac{9,350}{11,000} \approx 0.850 \]
From \(x = 1\) to \(x = 2\): \[ \frac{v(2)}{v(1)} = \frac{7,948}{9,350} \approx 0.851 \]
From \(x = 2\) to \(x = 3\): \[ \frac{v(3)}{v(2)} = \frac{6,755}{7,948} \approx 0.849 \]
These calculations show that the decay factor \(r\) is roughly \(0.85\).
Now, substituting this information back into our exponential decay function form: \[ v(x) = 11,000(0.85)^x \]
This aligns with option B from the list provided.
Thus, the correct function that models the value of the car over time is: B: \( y = 11,000(0.85)^x \).