Question

A new car worth $21,000 is depreciating in value by $3,000 per year. Complete parts (a) through (c) below. a. Write a formula that models the car's value, y, in dollars, after x years. y=

Answer 1

Answer:

Step-by-step explanation:

If the car's value is going down by 3000 a year, this is a linear function.  Most depreciation is exponential, but no matter.  The steady falling indicates linear.  And the steady rate that it is falling, 3000, is the slope of the line.  You could look at it in terms of 2 coordinate points.  (0, 21000) is one point, where x is the time gone by and y is the value of the car.  This point tells us that when 0 time has gone by, the car is brand new, worth 21000.  Another point could be (1, 18000), where after 1 year, the cars value is 18000.  Using those in the slope formula will give you the slope of the line.

$m=\frac{18000-21000}{1-0}=-3000$

Now use that slope along with either one of the coordinates to write the equation of the line.  I am going to use (0, 21000) to get

y - 21000 = -3000(x - 0) and

y - 21000 = -3000x + 0 and

y = -3000x + 21000

That is the linear function that models that particular situation.  I do not see a part b and c, so maybe this equation can help you with those parts.