Problem 25 - College Algebra Essentials

Question

A new car worth \(\$ 24,000\) is depreciating in value by \(\$ 3000\) 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 \(\$ 9000\). 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.

Step-by-Step Solution

Answer

a. The car's value after \(x\) years, \(y\), is given by the formula \(y = 24000 - 3000x\). b. It will take 5 years for the car's value to depreciate to \$9000. c. Graphing this equation, we get a straight line starting from (0,24000) declining at a slope of -3000; the point corresponding to the car's value at \$9000 is shown at (5,9000).

Step 1: Formulate the linear equation

We know that the car price is depreciating linearly, at a constant rate per year. This leads to the following linear model:\[y = 24000 - 3000x\]Where \(y\) is the car's value after \(x\) years.

Step 2: Determine the car's value after a certain time

We know that we need to find out when the car's value will be \$9000. This can be solved by setting \(y\) to 9000 in our equation developed in step 1. So the equation becomes:\[9000 = 24000 - 3000x\]We then solve for \(x\), which yields \(x = 5\), meaning it will take 5 years for the car's value to depreciate to \$9000.

Step 3: Graph the linear depreciation model

First, label the axes appropriately: `x` for years and `y` for car value. The y-intercept of the graph is 24000, representing the initial money invested. The graph is a straight line with a slope of -3000, representing the annual depreciation. To show the solution to part (b), place a point at the coordinates (5,9000) on the graph.