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Chapter 2: Problem 14

A new \(\$ 25,000\) car depreciates in value by \(\$ 5000\) per year. Construct a linear function for the value \(V\) of the car after years.

Short Answer

Expert verified
V(t) = -5000t + 25000

Step by step solution

01

Identify the variables

Let the initial value of the car be denoted by the constant term and let the annual depreciation rate be denoted by the coefficient of the variable.
02

Define the Linear Function Format

A linear function has the form: V(t) = mt + b
03

Determine the Slope (Depreciation Rate)

The car depreciates by \$5000 per year. Hence, the slope \(m\) is \ -5000.
04

Identify the Initial Value

The initial value of the car \(V(0)\) is \ $25,000, which means \(b\) in the linear function is 25000.
05

Construct the Linear Function

Combining the slope and the initial value, the linear function that represents the car's value after \(t\) years is: V(t) = -5000t + 25000

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

depreciation
Depreciation is the process by which an asset loses value over time. For items like cars, this is a common phenomenon. In our example, a new car valued at \(\$25,000\) depreciates by \(\$5000\) each year. This means that after one year, the car's value will decrease to \(\$20,000\), after two years it will drop to \(\$15,000\), and so on. Depreciation helps to understand how the value of an asset changes as it gets older and is used.
slope-intercept form
The slope-intercept form is a way of writing linear equations. It is often expressed as \(y = mx + b\). Here, \(m\) represents the slope, and \(b\) represents the y-intercept or initial value. In our exercise, the linear function is \(V(t) = -5000t + 25000\). The slope \(-5000\) shows the rate of depreciation per year, while the y-intercept \(25000\) is the car's initial purchase price. By understanding this form, you can easily determine the value of an asset at any given time.
initial value
Initial value is the starting value of an asset before any changes occur. For the car, this is the purchase price of \(\$25,000\). In the linear function \(V(t) = -5000t + 25000\), the \(b \) term represents this initial value. It is the point at which the function starts when \(t = 0\), meaning the car has just been bought and not yet depreciated. Understanding the initial value helps in predicting the future value of an asset by applying the rate of change (slope).

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Most popular questions from this chapter

On the same axes, graph (and label with the correct equation) three lines that go through the point (0,2) and have the following slopes: a. \(m=-\frac{1}{2}\) b. \(m=-2\) c. \(m=-\frac{5}{6}\)

Assume two individuals, \(A\) and \(B\), are traveling by car and initially are 400 miles apart. They travel toward each other, pass and then continue on. a. If \(A\) is traveling at 60 miles per hour, and \(B\) is traveling at \(40 \mathrm{mph},\) write two functions, \(d_{\mathrm{A}}(t)\) and \(d_{\mathrm{B}}(t),\) that describe the distance (in miles) that \(\mathrm{A}\) and \(\mathrm{B}\) each has traveled over time \(t\) (in hours). b. Now construct a function for the distance \(D_{\mathrm{AB}}(t)\) between A and \(B\) at time \(t\) (in hours). Graph the function for \(0 \leq t \leq 8\) hours.

The greatest integer function \(y=[x]\) is defined as the greatest integer \(\leq x\) (i.e., it rounds \(x\) down to the nearest integer below it). a. What is [2]\(?[2.5] ?[2.9999999] ?\) b. Sketch a graph of the greatest integer function for \(0 \leq x<5 .\) Be sure to indicate whether each endpoint is included or excluded. [Note: A bank employee embezzled hundreds of thousands of dollars by inserting software to round down transactions (such as generating interest on an account) to the nearest cent, and siphoning the round-off differences into his account. He was eventually caught.

Complete the table for each of the linear functions, and then sketch a graph of each function. Make sure to choose an appropriate scale and label the axes. a. $$ \begin{array}{rl} \hline x & f(x)=0.10 x+10 \\ \hline-100 & \\ 0 & \\ 100 & \\ \hline \end{array} $$ b. $$ \begin{array}{ll} \hline x & h(x)=50 x+100 \\ \hline-0.5 & \\ 0 & \\ 0.5 & \end{array} $$

The following problems represent calculations of the slopes of different lines. Solve for the variable in each equation. a. \(\frac{150-75}{20-10}=m\) c. \(\frac{182-150}{28-x}=4\) b. \(\frac{70-y}{0-8}=0.5\) d. \(\frac{6-0}{x-10}=0.6\)
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