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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) ? (1 point) Responses y=0.75(30,000)x y equals 0.75 left parenthesis 30,000 right parenthesis superscript x baseline y=30,000(34)x y equals 30,000 left parenthesis Start Fraction 3 over 4 End Fraction right parenthesis superscript x baseline y=22,500(34)x y equals 22,500 left parenthesis Start Fraction 3 over 4 End Fraction right parenthesis superscript x baseline y=30,000(43)x y equals 30,000 left parenthesis Start Fraction 4 over 3 End Fraction right parenthesis superscript x baseline Skip to navigation

To determine the correct exponential equation for the given graph, we can use the standard form of an exponential decay function:

\[
y = y_0 \cdot b^x
\]

Where:
- \( y_0 \) is the initial value (value of the car at \( x = 0 \)),
- \( b \) is the base representing the decay factor,
- \( x \) is the number of years since purchase, and
- \( y \) is the car's value after \( x \) years.

### Step 1: Identify the given points.
From the graph:
- When \( x = 0 \), \( y = 30,000 \), so the initial value \( y_0 = 30,000 \).
- When \( x = 1 \), \( y = 22,500 \).

### Step 2: Solve for the decay factor \( b \).
We know that after 1 year (\( x = 1 \)), the car's value has dropped to 22,500, so we can substitute these values into the equation:

\[
22,500 = 30,000 \cdot b^1
\]

Solving for \( b \):

\[
b = \frac{22,500}{30,000} = 0.75
\]

### Step 3: Write the exponential equation.
Now that we have \( y_0 = 30,000 \) and \( b = 0.75 \), the equation becomes:

\[
y = 30,000 \cdot (0.75)^x
\]

### Step 4: Verify the correct option.
The correct option from the given responses is:

\[
\boxed{y = 30,000 \left(\frac{3}{4}\right)^x}
\]

This matches the second option in your provided choices.

Would you like more details or have any further questions? Here are 5 related questions:

1. How does exponential decay differ from linear decay in car depreciation models?
2. How can the decay factor be used to predict the car's future value after a certain number of years?
3. What would happen if the decay factor was greater than 1 instead of less than 1?
4. How can you use logarithms to solve for the time it takes for the car's value to reach a specific amount?
5. What is the half-life of the car's value, and how can it be calculated from the decay factor?

Tip: When working with exponential models, always look at the base of the exponent to determine if the function represents growth (greater than 1) or decay (less than 1).

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)?

Understand how exponential decay models the depreciation of a car's value over time using the equation y = 30,000 * (0.75)^x. Learn how to calculate the decay factor from graph data points and write the correct exponential equation for depreciation.

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