Problem 22 - College Algebra and Calculus: An Applied Approach

Question

You are given the 2005 value of a product and the rate at which the value is expected to change during the next 5 years. Use this information to write a linear equation that gives the dollar value of the product in terms of the year. (Let \(t=5\) represent 2005.) 2005 Value \(\quad\) Rate $$156 $$quad $$ 4.50$$ increase per year

Step-by-Step Solution

Answer

The linear equation representing the dollar value of the product in any given year is \(y = 4.5t + 141.5\)

Step 1: Understand the problem and organize the given data

Given: Initial value of the product in 2005 (\(t=5\)) is \(y = \$156\). The value of the product increases at a rate of \$4.50 per year. Let \(t\) be the number of years since 2000, and let \(y\) be the dollar value of the product.

Step 2: Identify the slope and the y-intercept

Since the product value increases $4.50 each year, this is the slope (m) of our linear function. The y-intercept (b) is the initial value of the product in the year 2005, or when \(t = 5\). Thus, \(m = 4.5\) and \(b = 156\). However, because we adjusted our time scale so that t = 5 represents 2005, we need to adjust our y-intercept accordingly, i.e., \(b = 156 - 4.5 * 5 = 141.5\).

Step 3: Formulate the linear equation

A linear equation has the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. Substituting the identified values, we get \(y = 4.5t + 141.5\). This equation represents the dollar value of the product in any given year.

Key Concepts

Slope

In the context of a linear equation, the slope represents how steep the line is. It indicates how much the y-value will change for each unit change in the x-value. In simpler words, it's the rate of how something is increasing or decreasing.
For example, in the problem given, the value of the product increases by \(4.50 every year.

• This increase of \)4.50 is the slope of the line.
• It shows that for every year that passes, the value of the product goes up by $4.50.

In the equation of a line, represented as \(y = mx + b\), \(m\) is the slope.
Here, the slope is denoted by 4.5, since the product value increases consistently by that amount each year.
Remember: Slope is a keystone concept in understanding how changes occur over time in linear relationships.

Y-intercept

The y-intercept is the point where the line crosses the y-axis. It signifies the value of \(y\) when \(x\) is zero. In practical terms, it represents the starting point of a line within a graph.
In the given problem, the y-intercept needs special attention because of the time frame adjustment.
Initially, in 2005, the value is \(156 with \(t=5\), but for the line to accurately depict values starting from 2000 (\(t=0\)), we must adjust that initial value.

• This adjustment corrects the y-intercept to \)141.50, accounting for the five years leading up to 2005 (2000 to 2005) with an increase of \(4.50 per year.

In the linear equation formula \(y = mx + b\), \(b\) represents the y-intercept.
Thus, the y-intercept of \)141.50 reflects the adjusted starting point of value for year zero, in this adjusted scale.
Understanding the y-intercept ensures clarity on where the line "begins" in terms of value before changes account for future increments.

Rate of Change

Rate of change is a fundamental concept when discussing linear equations. It tells us how much one variable changes in relation to another.
For linear relationships, this rate is consistent and steady across the entire scope of the equation.

• In this specific problem, the rate of change is the $4.50 increase per year.
• This tells us that, no matter the year, the product's value steadily increases by this fixed amount annually.

This constant rate is valuable for predictions, as it allows one to use the linear equation \(y = 4.5t + 141.5\) to compute the expected value of the product in any year since you know it grows by this same amount each time period.
An easily grasped rate of change helps when interpreting and using linear equations in real-world applications. It relays a straight forward prediction method, emphasizing why lines on graphs stay straight – the change is constant.