Question

The table below shows the value in dollars of a car at the end of x years.
Number of Years, x: 0, 1, 2, 3
Value, v(x) (dollars): 11,000, 9,350, 7,948, 6,755

Which exponential function models this situation?

A: y = 85(11,000)^x

B: y= 11,000(0.85)^x

C: y = 11,000(1.15)^x

D: y = 1.15 ( 11,000) ^x

Answer 1

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: \( y = 11,000(0.85)^x \)**.