A 12% increase means you keep 100% of the value and add 12% more. How do you write that as a single decimal multiplier applied each year?

If you multiply by the same factor every year, what kind of function is this (linear or exponential), and where should the time $t$ appear in the equation?

Try $t=1$ and $t=2$ in the answer choices. Which equation actually makes the value 12% larger than before each year, starting from $540$?

In Desmos, type the four equations as functions of $t$, for example:

• A(t) = 540(1.12)^t
• B(t) = 540(0.88)^t
• C(t) = 540(1.12)^(t/12)
• D(t) = 540(1+0.12t) This will graph all four models on the same axes.

Use the keypad or click on each graph at $t=0$ and see its $y$-value. The correct model must give $V(0)=540$, since the coin is worth $540$ today.

Now look at $t=1$ and $t=2$ on each graph (or by typing A(1), A(2), etc.). The correct equation will show the value multiplying by the same factor each year so that the value at $t=1$ is 12% more than at $t=0$, and the value at $t=2$ is 12% more than at $t=1$.

The coin’s value increases by 12% each year. A fixed percentage increase applied repeatedly over equal time intervals is modeled by an exponential function, not a linear one. So we are looking for an expression of the form

$\text{(initial value)} \times (\text{growth factor})^t$.

A 12% increase means each year the new value is the old value plus 12% of the old value.

• 12% as a decimal is $0.12$.
• New value each year $= \text{old value} \times (1 + 0.12) = \text{old value} \times 1.12$.

So the growth factor per year is $1.12$.

Start with the initial value of $540$ dollars.

• After 1 year: $540 \times 1.12$.
• After 2 years: multiply by $1.12$ again: $540 \times 1.12 \times 1.12 = 540(1.12)^2$.
• After 3 years: multiply by $1.12$ again: $540(1.12)^2 \times 1.12 = 540(1.12)^3$.

You can see the exponent matches the number of years.

Following the same pattern, after $t$ years, the value is the initial $540$ multiplied by $1.12$ once for each year, which is $540(1.12)^t$.

So the correct equation is $V(t)=540(1.12)^t$, which corresponds to choice A.