Question 1008550
year 2000 car is bought for 26,000.
year 2005, value of car is 15,000.


let x = 0 represent the year 2000
then x = 5 represents the year 2005
then x = 8 represents the year 2008


y = mx + b is the slope intercdept form of a straight line.


y = value of the car.
x = year
m is the slope of the change in the value of y divided by the change in the value of x.
b is the y-intercept which is the value of y when the value of x is 0.


when x = 0, y = 26,000
when x = 5, y = 15,000


the slope is equal to (y2-y1)/(x2-x1)


y2 = 15,000
y1 = 26,000
x2 = 5
x1 = 0


slope = (15,000 - 26,000) / (5 - 0) = -11,000 / 5 = -2,200.


this means that the car is losing value by 2,2000 per year.


y = mx + b becomes y = -2,200 * x + b


b is the value of y when x is equal to 0 which is the value of the car when it was brand new.


b is therefore equal to 26,000 and the equation is therefore equal to:


y = -2,200 * x + 26,000


in year 0, the car is worth 26,000.
in year 5, the car is worth 26,000 - 2,200 * 5 = 15,000.
in year 8, the car is worth 26,000 - 2,200 * 8 = 8,400.


the offer of 7,000 is 1,400 dollar below the remaining value of the car based on straight line depreciation.


you can see this graphically below:


the first picture shows x = 0 and x = 5
the value at x = 0 is 26,000
the vlaue at x = 5 is 15,000
this is based on loss of value of 2,200 per year based on straight line depreciation.


<img src = "http://theo.x10hosting.com/2015/122703.jpg" alt="$$$" </>


the second picture shows x = 5 and x = 8.
the value at x = 5 is 15,000
the value at x = 8 is 8,400
you can also see the offer of 7,000 at x = 8.
it is below the value of 8,400.
the difference is 1,400.


<img src = "http://theo.x10hosting.com/2015/122704.jpg" alt="$$$" </>


the red line is the value of the car in year x.
the blue line is the value of the offer in year x.


the actual year is 2000 + x.


when x is 0, the year is 2000
when x is 5, the year is 2005
when x is 8, the year is 2008