Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 7 - Exponential and Logarithmic Functions - Mid-Chapter Quiz - Page 461: 8


Function that models the car's expected value after $x$ years: $\text{FV}=25000(0.85)^x$ Expected value after $5$ years: $\$11,092.63$

Work Step by Step

Recall: Depreciation Formula The future value $\text{FV}$ of a physical property that depreciates in value over time is given by the formula: $$\text{FV}=P(1-r)^x$$ where $P$ = present value $r$ = depreciation rate per year $x$ = time in years Thus, the future value of a car worth $\$25,000$ at the moment and depreciates $r\%$ per year will be modeled by the formula: $$\text{FV} = 25,000\left(1-r\right)^x$$ The value of the car after one year is $\$21,250$. Substitute $1$ to $x$ and $21250$ to $\text{FV}$ to find the value of $r$: \begin{align*} 21250& = 25000\left(1-r\right)^1\\\\ 21250& = 25000\left(1-r\right)\\\\ \frac{21250}{25000}& = \frac{25000\left(1-r\right)}{25000}\\\\ 0.85 &=1-r\\\\ 0.85 - 1 &= -r\\ -0.15&= -r\\ 0.15 &= r \end{align*} Thus, the exponential function that models the car's expected value ($\text{FV}$) after $x$ years is: \begin{align*} \text{FV}&=25,000(1-0.15)^x\\ \text{FV}&=25000(0.85)^x \end{align*} The estimated value of the car after $5$ years can be found by substituting $5$ to $x$: \begin{align*} \text{FV}&=25,000(0.85)^5\\ &\approx 11,092.63 \end{align*}
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