A new car worth \$36,000 is depreciating in value by \$4000 per year. a. Write a formula that models the car's value, y, in dollars, after x years. b. Use the formula from part (a) to determine after how many years the car's value will be \$12,000. c. Graph the formula from part (a) in the first quadrant of a rectangular coordinate system. Then show your solution to part (b) on the graph.

Verified step by step guidance

1. For part (a), recognize that the car's value decreases by a fixed amount each year, which is a linear relationship. The general form of a linear equation modeling depreciation is \(y = mx + b\), where \(m\) is the rate of change (slope) and \(b\) is the initial value (y-intercept).
2. Identify the initial value \(b\) as \$36,000\( (the car's worth at year 0) and the rate of depreciation \)m\( as \)-4000\( (since the value decreases by \)4000\( each year). So, the formula becomes \)y = -4000x + 36000$.
3. For part (b), use the formula \(y = -4000x + 36000\) and set \(y\) equal to \$12,000\( to find the number of years \)x\( when the car's value reaches \)12,000\(. This gives the equation \)12000 = -4000x + 36000$.
4. Solve the equation for \(x\) by isolating the variable: subtract \$36000\( from both sides to get \)12000 - 36000 = -4000x\(, then simplify and divide both sides by \)-4000\( to find \)x$.
5. For part (c), plot the linear equation \(y = -4000x + 36000\) on a coordinate plane with \(x\) representing years and \(y\) representing the car's value. The graph will be a straight line starting at \((0, 36000)\) and decreasing by \$4000\( for each increase of 1 in \)x\(. Mark the point where \)y = 12000$ on the graph to visually show the solution from part (b).

Key Concepts

Linear Functions and Modeling

A linear function represents a relationship with a constant rate of change, often written as y = mx + b. In this problem, the car's value decreases by a fixed amount each year, making it a linear model where the slope (m) is negative, reflecting depreciation.

Solving Linear Equations

To find when the car's value reaches a certain amount, you set the linear equation equal to that value and solve for the variable. This involves isolating the variable using algebraic operations, which helps determine the time (x) when the car's value is $12,000.

Graphing Linear Equations

Graphing a linear equation involves plotting points that satisfy the equation on a coordinate plane. The first quadrant is used here since time and value are non-negative. Marking the solution on the graph visually shows the point where the car's value equals $12,000.