In 2003, Troy bought a new car for $33,500 . The car was worth $21,000 in the year 2011. If Troy knows that the value of the car depreciated linearly,
Let x be the number of years and y be the cost of the car
Lets assume x=0 for 2003
for 2003, x=0 . so for year 2011 , x= 2011- 2003 = 8
When x=0, y = 33,500
When x= 8, y= 21,000
So two points are (0, 33500) and (8,21000)
To find annual rate of change we find slope because the car depreciated linearly.



Slope = -1562.50
The rate of change of car's value = -1,562.50