Question

Depreciation

Answer 1

Answer:

a) $y=-3950\cdot{x}+21500$

b) f(x) $=21500\cdot{(0.7953)^x}$

c) Year 1: Linear $17550, exponential $17099

Year 4: Linear $5700, exponential $8601

d) Exponential model

e) The linear model depreciates the value quicker than exponential model long term around 4 years

Step-by-step explanation:

a) At year 0 the price is 21500 and at year 2 the price is 13600

WE can use points (0,21500) and (2,13600)

We can determine the gradient

$m=(13600-21500)/(2-0)=-3950$

We can use the point slow formula:

$y-y_1=m\cdot{(x-x_1)}$

$y-21500=-3950\cdot{(x-0)}$

$y=-3950\cdot{x}+21500$

b) We can use the following equation:

f(x) $=a\cdot{(1+r)^x}$

f(x) is the depreciation value after amount of time t, a is the new value, r is the rate of depreciation and x is the time.

$13600=21500\cdot{(1+r)^2}$

$r=-0.2047$

The depreciation rate is 20.47% and is negative because it decreases the new value of the car

c) Year 1:

$y=-3950\cdot{1}+21500=17550$

f(x) $=21500\cdot{(0.7953)^1}=17099$

Year 4:

$y=-3950\cdot{4}+21500=5700$

f(x) $=21500\cdot{(0.7953)^4}=8601$

d) Year 2

$y=-3950\cdot{2}+21500=13600$

f(x) $=21500\cdot{(0.7953)^2}=13598$