Question

Use the image to answer the question.

A descending curve with an arrow on the right end passes through 10 points plotted on a coordinate plane. The x-axis is labeled years since purchase, and the y-axis is labeled value of car.

Given the graph of the exponential equation representing the value of a car since purchase, which option is the correct exponential equation for the graph if the graph goes through the points (0, 30000)
and (1, 22500)
?

Answer 1

To find the correct exponential equation that fits the given points (0, 30000) and (1, 22500), we can use the general form of an exponential decay equation, which is:

\[ V(t) = V_0 \cdot e^{-kt} \]

Where:
- \( V(t) \) is the value of the car at time \( t \) (in years).
- \( V_0 \) is the initial value of the car (value at \( t = 0 \)).
- \( k \) is a decay constant.
- \( e \) is the base of the natural logarithm.
- \( t \) is the number of years since purchase.

Given the point (0, 30000), we start with:
- \( V(0) = 30000 \)

This implies that:
\[ V_0 = 30000 \]

Next, using the point (1, 22500), we have:
\[ V(1) = 30000 \cdot e^{-k} = 22500 \]

To solve for \( k \), we can divide both sides by 30000:
\[ e^{-k} = \frac{22500}{30000} = 0.75 \]

Now, we take the natural logarithm of both sides:
\[ -k = \ln(0.75) \]

Therefore:
\[ k = -\ln(0.75) \]

Now we can express the equation:
\[ V(t) = 30000 \cdot e^{-\ln(0.75) \cdot t} \]

This can also be expressed in a simpler base form using the property of logarithms, since:
\[ e^{-\ln(0.75)} = \frac{1}{0.75} \]

Thus, in terms of decay:
\[ V(t) = 30000 \cdot (0.75)^t \]

So the full exponential equation that represents the value of the car since purchase is:

\[ V(t) = 30000 \cdot (0.75)^t \]