5.7.3 Describing Graphs

Compare your answer:

$v=p(1-\frac{1}{4}{)}^t$

$v=p(\frac{3}{4}{)}^t$

Where:

$v=$value of the car after $t$ years.

$p=$ initial price of the car.

$t=$ time in years.

$14=$ fraction by which the car loses its value every year.

Compare your answer:

$v=p(3{)}^{\frac{t}{8}}$

Where:

$v$ is the current value of the stock after t years.

$p$ is the initial value of the stock.

$t$is the number of years since the initial value.

In this equation, we raise 3 to the power of $(\frac{t}{8})$ to calculate the value after $t$ years. This represents the stock tripling in value approximately every 8 years.

Compare your answer:

$v=s(2{)}^{\frac{t}{4}}$

$v=$ value of the stock after $t$ years.

$s=$ initial price of the stock.

$t=$ time in years.

$\frac{t}{4}=$ the number of doubling event for every four years.

Compare your answer:

$v=p(1-\frac{2}{5}{)}^t$

Where:

$v$ is the current value of the laptop after t years.

$p$ is the initial purchase value of the laptop.

$t$ is the number of years since the purchase.

In this equation, the laptop loses $\frac{2}{5}$ (or $\frac{3}{5}$ of its remaining value) every year. We raise $(\frac{3}{5})$ to the power of $t$ to calculate the value after $t$ years.