Question
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A home's value increases at an average rate of \( 5.5 \% \) each year. The current value is \( \$ 120,000 \). What function can be used to find the value of the home after \( x \) years? \( f(x)=120,000(1.055 x) \) \( f(x)=120,000(0.055)^{x} \) \( f(x)=120,000(1.055)^{x} \) \( f(x)=[(120,000)(1.055)]^{x} \)

Ask by Haynes Lowe. in the United States
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The function to find the home's value after \( x \) years is \( f(x) = 120,000(1.055)^x \).

Solution

To determine the function that models the value of the home after \( x \) years with an annual increase of \( 5.5\% \), we can use the formula for compound interest: \[ f(x) = \text{Initial Value} \times (1 + \text{Growth Rate})^x \] Given: - **Initial Value** = \$120,000 - **Growth Rate** = \( 5.5\% = 0.055 \) Plugging these values into the formula: \[ f(x) = 120,000 \times (1 + 0.055)^x = 120,000 \times (1.055)^x \] This corresponds to: \[ f(x) = 120,000(1.055)^x \] **Answer:** \( f(x)=120,000(1.055)^{x} \)

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The Deep Dive

To model the increase in a home's value over time, you would use an exponential growth function. The correct choice for the value of the home after \( x \) years is \( f(x) = 120,000(1.055)^{x} \). This equation represents the current value multiplied by the growth factor \( 1.055 \) raised to the power of \( x \), which is the number of years. It's worth noting that the other options provided don't represent the correct exponential growth model. For instance, the option \( f(x)=120,000(0.055)^{x} \) incorrectly applies the percentage, making it decrease instead of grow, while \( f(x)=[(120,000)(1.055)]^{x} \) implies a compounding of both the initial value and growth, which isn’t how exponential growth works in this scenario. Always remember the proper use of the base in growth functions for accurate modeling!

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