A new car worth $21,000 is depreciating in value by $3000 per year?

Question

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.

Answer 1

An important first step is to understand the problem.

A new car worth $21,000 is depreciating in value by $3000 per year?

To write a linear equation, we need to know the initial value and the rate of change. In this case, the initial value is $21,000 for the starting price of the car. The rate of change is $3000 per year (per x), and it will be a negative because it is depreciating or decreasing/going down.

a. y = mx+ b (where m is the rate of change and b is the initial value) y = -3000x + 21000

b. Substitute 12,000 for y (value).

12000 = -3000x + 21000

-21000 - 21000 [subtract the initial value from both sides]

-9000 = -3000x

/-3000 /-3000 [divide by the rate of change]

3 = x

x represents years, so in 3 years, the value of the car will be $12,000.

c. It is difficult to show a graph on a text-based response platform, but I suggest you use desmos graphing calculator to give you a hint. If we worked online together, I'd be able to show you how to draw the graph.

Answer 2

A) The formula in slope-intercept form would be y=-3000x+21,000. The "change" in value is the slope and losing value equals negative slope. The y-intercept is the car's initial value.

B) Plug in 12000 for y as it is the future value and solve for x.

12000=-3000x +21000

-9000=-3000x

3=x

It would take 3 years for the car's value to depreciate to $12,000.

C) Graph y=-3000x+21000 and y=12000 and show the intersection point (3,12000). Desmos.com is a great graphing option.