One thing to consider in addition to the solution from Jason S. (from Katy, TX, above) is that the assumption made here is that the relationship is linear.

But with the help of a little technology, something interesting comes to be noted.

If we enter the given two points (6, 900) and (8,750) using the STAT - EDIT feature on the TI-83/84 graphing calculator (make sure the DiagnosticOn feature is enabled from the CATALOG menu), we can then use regression models.

First to find the LINEAR equation, use STAT-CALC-4:LinReg(ax+b), and you get the equation:

y = - 75 x + 1350 with r^{2}=1 (ignore r^{2} - not analytically relevant) , but more importantly note that

r = - 1 (this means the equation has an EXACT negative correlation - a "perfect" fit to the given data)

Using this equation, the original amount (when x=0) is $1350, which is from Jason S.'s solution.

Next we look at the EXPONENTIAL equation. Use STAT-CALC-0:ExpReg, and you get the equation:

y = 1555.2 (.9128709292)^x with the SAME r^{2} and r values as for the linear equation.

This means that the exponential equation is ALSO an EXACT negative correlation - a "perfect" to the data.

So for this equation, the original amount (when x=0) is $1555.20, over $200 more than the linear equation yields.

So, it is important to solve this problem in context. If this problem is embedded among other problems involving linear relationships, then it's safe to assume it's linear.

Without any other context, you simply cannot assume the relationship is linear. The fact that BOTH the linear equation and the exponential equation are an exact fit means that technically, unless you have more information (like a third data point), then the original value of the computer could be $1350 if you assume a linear decrease, or it's $1555.20, if you assume an exponential decrease.

If this were a multiple-choice question and only one of the answers were available, then obviously you choose that correct answer. If BOTH the linear and the exponential answers were available, then I would boldly challenge this problem with your teacher.