Question

the table shows 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(11000)^x
b y=11000(0.85)^x
c y=11000(1.15)^x
d y=1.15(11000)^x

Answer 1

To determine which exponential function models the given situation, we need to analyze the values provided for the car's worth over the years.

The values are as follows:
- At \(x = 0\): \(v(0) = 11,000\)
- At \(x = 1\): \(v(1) = 9,350\)
- At \(x = 2\): \(v(2) = 7,948\)
- At \(x = 3\): \(v(3) = 6,755\)

The value of the car is decreasing, indicating that we are dealing with a decay model.

Looking for a pattern in the values:

From \(v(0)\) to \(v(1)\):
\[
\frac{9,350}{11,000} \approx 0.85
\]

From \(v(1)\) to \(v(2)\):
\[
\frac{7,948}{9,350} \approx 0.85
\]

From \(v(2)\) to \(v(3)\):
\[
\frac{6,755}{7,948} \approx 0.85
\]

This consistent ratio of approximately 0.85 suggests that the car's value decreases by about 15% each year (since \(1 - 0.85 = 0.15\)).

Thus, the value can be modeled by the equation:
\[
v(x) = 11,000 \cdot (0.85)^x
\]

This matches option **b**: \(y = 11,000(0.85)^x\).

So the correct answer is:
**b** \(y = 11000(0.85)^x\).