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

Question

Value, v(x) (dollars) 11,000, 9,350, 7,948, 6,755
Which exponential function models this situation?
a. y=85\left(11000\right)^{x}
b. y=11000\left(0.85\right)^{x}
c. y=11000\left(1.15\right)^{x}
d. y=1.15\left(11000\right)^{x}

Answer 1

To find the exponential function that models the value of the car over time, we need to determine the decay factor based on the values given in the table.

1. Initial Value (when x=0):

   • The car is worth $11,000 at the start (x=0).

2. Value after 1 Year (y=1):

   • At x=1, the value is $9,350.
   • To find the decay factor, we can set up the following equation based on the general model \( v(x) = v(0) \cdot r^x \): \[ 9,350 = 11,000 \cdot r^1 \]
   • Solving for \( r \): \[ r = \frac{9,350}{11,000} = 0.850 \]

3. Value after 2 Years (y=2):

   • At x=2, the value is $7,948.
   • We can verify our decay factor: \[ v(2) = 11,000 \cdot (0.85)^2 \] \[ v(2) = 11,000 \cdot 0.7225 = 7,947.5 \quad \text{(which is approximately 7,948)} \]

4. Value after 3 Years (y=3):

   • At x=3, the value is $6,755.
   • Verify again: \[ v(3) = 11,000 \cdot (0.85)^3 \] \[ v(3) = 11,000 \cdot 0.614125 = 6,755 \quad \text{(exact)} \]

5. Exponential Function:

   • We can conclude that the values are modeled by the function: \[ v(x) = 11,000 \cdot (0.85)^x \]
   • This matches option b: \( y = 11000 \cdot (0.85)^x \).

Thus, the correct answer is b. \( y = 11000 \cdot (0.85)^x \).