f(t)=200, 0≤ t ≤ 0.25 f(t)=200(1.02)^t, 0.25< t ≤ 5.25
Practice makes perfect
We are told about what Charmaine does with money she earned from selling her bicycle. Let's label the number of years since she sold her bicycle t. We will also label the amount of money she has as y. The function will be divided into two parts.
Up to three months after she sold the bicycle.
Three months after she sold the bicycle.
Up to Three Months
Three months equals a quarter of a year, or t=0.25 From the exercise we know that for the first three months she kept the money in her drawer. Since the money did not earn interest or change in amount, we get the following function from t=0 to t=0.25.
f(t)=200, 0≤ t ≤ 0.25
After Three Months
After three months had passed, Charmaine put the money in a five-year Certificate of Deposit. It then earned an annual interest of 2 %. This means that after 3 months, the money can be described by an exponential function with an initial value of 200 and a multiplier of 1.02.
f(t)=200(1.02)^t, 0.25< t ≤5.25
Combining the Functions
If we combine these functions we get a piecewise function that describes the growth of the money.
f(t)=200, 0≤ t ≤ 0.25 f(t)=200(1.02)^t, 0.25
