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Depreciation is a term used to describe the reduction in the value of an asset over time. For a car, this process usually occurs because of wear and tear, usage, and eventually, obsolescence as newer models become available. In our exercise, the car initially worth \( \\(26,000 \) depreciates at a rate of \( \\)1,500 \) each year. Thus, depreciation is a key factor in understanding how the car's value decreases annually.

• Depreciation is often expressed as a negative value because it represents a loss in worth.
• This decrease is considered linear if the value reduces by a consistent amount each year.
• Knowing depreciation helps owners anticipate the future value of their asset over time.

The importance of understanding depreciation is that it allows us to create a realistic financial plan. For instance, with this car, we can predict that its worth decreases steadily every year, which aids in decision-making about future selling or trading options.

The slope-intercept form is a mathematical equation of a line. A linear function is typically expressed in this format: \( y = mx + b \), where:

• \( m \) is the slope of the line, indicating how the value changes with each unit increase in \( x \).
• \( b \) is the y-intercept, representing the initial starting value before any changes.

In the context of our exercise, we use the function for modeling the car's depreciation: \( V(t) = -1500t + 26000 \).

• The slope, \( m = -1500 \), reflects the yearly decrease in the car's value.
• The y-intercept, \( b = 26000 \), stands for the car's initial purchase price.

Utilizing the slope-intercept form helps visualize and calculate the change in the car's value over time. It's a practical tool for predicting these changes consistently, as long as the depreciation rate remains constant.

Function evaluation involves finding the output of a function for a specific input. In our car depreciation example, evaluating the function \( V(t) = -1500t + 26000 \) at \( t = 4 \) years gives us a way to determine the car's projected value at that point in time.

• To evaluate, substitute the given time (in years) into the linear function where \( t \).
• Then calculate the resulting expression to find the value after that duration.

For \( V(4) \), the calculation steps are as follows: - Replace \( t \) with 4: \( V(4) = -1500(4) + 26000 \) - Compute \( -1500 \times 4 = -6000 \) - Add \( 26000 - 6000 = 20000 \) Thus, the evaluated function, \( V(4) = 20000 \), tells us that after 4 years, the vehicle's estimated value will be \( \$20,000 \). This straightforward process assists in making informed predictions based on the current depreciation model, ensuring accurate planning for future financial aspirations.