a) enter the equation into a graphing calculator in this case V(t) would be the "y =" part and t would be X
b) put 2 in for t, so you have V(2) = 14000*(3/4)^2
c) To see where the car depreciates most rapidly look at the graph of the equation. Where the slope is the steepest is where the car depreciates most rapidly. It will be realistic if the slope is steepest at the beginning of the graph (for lower values of t) and the steepness will decrease over time and almost flatten out.
b) put 2 in for t, so you have V(2) = 14000*(3/4)^2
c) To see where the car depreciates most rapidly look at the graph of the equation. Where the slope is the steepest is where the car depreciates most rapidly. It will be realistic if the slope is steepest at the beginning of the graph (for lower values of t) and the steepness will decrease over time and almost flatten out.