Casey recently purchased a sedan and a pickup truck at about the same time for a new business. The value of the sedan S
, in dollars, as a function of the number of years t
after the purchase can be represented by the equation S(t)=24,400(0.82)t. The equation P(t)=35,900(0.71)t2 represents the value of the pickup truck P
P
, in dollars, t
t
years after the purchase. Analyze the functions S(t) and P(t) to interpret the parameters of each function, including the coefficient and the base. Then use the interpretations to make a comparison on how the value of the sedan and the value of the pickup truck change over time.
To analyze the functions S(t) and P(t), let's first interpret the parameters of each function.
For the sedan's value function, S(t) = 24,400(0.82)t:
- The coefficient 24,400 represents the initial value of the sedan, i.e., its value at t=0 (the time of purchase).
- The base 0.82 represents the rate of depreciation per year. Since it is less than 1, it indicates that the value of the sedan is decreasing over time.
For the pickup truck's value function, P(t) = 35,900(0.71)t^2:
- The coefficient 35,900 represents the initial value of the pickup truck, i.e., its value at t=0 (the time of purchase).
- The base 0.71 represents the rate of depreciation per year squared. Since it is less than 1, it indicates that the value of the pickup truck is also decreasing over time but at an even faster rate compared to the sedan.
Now, let's compare how the value of the sedan and the value of the pickup truck change over time.
Based on the functions, as the number of years (t) increases:
- The value of the sedan, S(t), will decrease at a rate of 0.82 per year, representing a constant annual depreciation.
- The value of the pickup truck, P(t), will also decrease, but at an increasing rate since t is squared. The depreciation per year will be 0.71^2 = 0.5041, which means the value decreases faster compared to the sedan.
In summary, both the sedan and the pickup truck lose value over time, but the pickup truck depreciates at a faster pace due to the squared term in its equation.
the second one if P(t)=35,900(0.71)^t/2
means it depreciates by 29% every half year
You have a typo
if S(t) = 24400(.82)t, then the value would actually increase which is nonsense
you must have meant: S(t) = 24400(.82)^t
which tells me that it depreciates 18% each year.
P(t)=35,900(0.71)t2 , also makes no sense, please fix it.
S(t)=24,400(0.82)^t
P(t)=35,900(0.71)^t/2