Video Transcript
The original value of a car is 28000 dollars. If it depreciates by 15 percent each year, which of the following functions 𝑓 models the value of the car, in dollars, 𝑡 years later? Is it option A) 𝑓 of 𝑡 is equal to 28000 multiplied by 0.15 to the power of 𝑡? Is it option B) 𝑓 of 𝑡 is equal to 0.15 multiplied by 28000 to the power of 𝑡? Option C) 𝑓 of 𝑡 is equal to 28000 multiplied by 0.85 to the power of 𝑡? Or option D) 𝑓 of 𝑡 is equal to 0.85 multiplied by 28000 to the power of 𝑡?
We’re told in the question that the car depreciates by 15 percent each year. This means that its value decreases by 15 percent per year. If the initial value is 100 percent, then after one year the value will be 85 percent of this initial value, as 100 minus 15 is equal to 85. The word percent means out of 100. Therefore, 85 percent written as a fraction is equal to 85 over, or out of, 100.
The line in a fraction means divide. And dividing by 100 moves all the digits two places to the right. Therefore, 85 percent is also equivalent to the decimal 0.85. The value of the car after year one will be 28000 multiplied by 0.85. At the end of year two, the car will be valued at 85 percent of the value at the end of your one. Therefore, we need to multiply the end of year one value by 0.85. This gives us 28000 multiplied by 0.85 multiplied by 0.85. We can rewrite this as 28000 multiplied by 0.85 squared.
This pattern will continue for year three, where the new value of the car will be 28000 multiplied by 0.85 cubed, or to the power of three. The original, or initial, value of 28000 will not change, but the power of the multiplier, in this case 0.85, will increase each year. This means that after 𝑡 years the value of the car will be 28000 multiplied by 0.85 to the power of 𝑡. The correct answer is option C.
This is often referred to as the compound interest formula, where the new value can be calculated by multiplying the original value by a multiplier to the power of the number of years. In this case, our original value was 28000 dollars, the multiplier was 0.85 as the car was depreciating by 15 percent, and the number of years was 𝑡.
