Interpreting Exponential Functions

In 2020, the number of confirmed COVID-19 cases in the United States doubled roughly every three days during the early weeks of the outbreak. Epidemiologists described this as exponential growth — not as a compliment, but as a warning. The math behind that warning is the same math you use when a bank pays you interest, when a population of bacteria grows in a lab, or when a car loses value every year. The formula is $f(x) = a \cdot b^x$, and every number in it means something specific.

Start with $a$. When $x = 0$, you get $f(0) = a \cdot b^0 = a \cdot 1 = a$. So $a$ is always the starting value — the amount you have before any growth or decay happens. In the COVID example, $a$ would be the number of cases on day zero. In a savings account, $a$ is the amount you deposit on the first day. It is sometimes called the initial value or the y-intercept, because the graph of the function crosses the y-axis at $(0, a)$.

Now look at $b$. This is the base, and it controls what happens each time $x$ increases by 1. Every step forward in $x$ multiplies the output by another factor of $b$. If $b > 1$, the function grows. If $0 < b < 1$, the function decays — it shrinks toward zero. The value of $b$ is sometimes called the growth factor or the decay factor.

Here is how to read $b$ as a percent. If a population grows by 6% each year, you keep 100% of what you had and add 6%, so $b = 1.06$. If a car loses 12% of its value each year, you keep 88%, so $b = 0.88$. The percent change is always hiding just underneath the surface of $b$: subtract 1 from $b$ and you have the rate of change as a decimal.

The red curve is $f(x) = 2 \cdot 1.5^x$. It starts at 2 and grows by 50% each step. The blue curve is $g(x) = 100 \cdot 0.85^x$. It starts at 100 and shrinks by 15% each step. Both curves share the same basic shape — one rises, one falls — but the starting point and the steepness come directly from $a$ and $b$.

Here is a worked example pulling a real-world meaning out of both numbers.

A town's population is modeled by $P(t) = 4500 \cdot 1.03^t$, where $t$ is the number of years since 2010. Identify the initial population, the growth rate, and the population in 2015.

Now a decay example. This time the value of $b$ is less than 1.

A laptop purchased for $800 loses value each year. Its value is modeled by $V(t) = 800 \cdot 0.75^t$, where $t$ is the number of years after purchase. Identify the starting value, the decay rate, and the value after 3 years.

Compound interest follows exactly this structure. When a bank pays you interest that compounds annually, your balance after $t$ years is

where $P$ is the principal (the amount you deposit), $r$ is the annual interest rate as a decimal, and $t$ is the number of years. This is exactly $f(x) = a \cdot b^x$ with $a = P$ and $b = 1 + r$. A savings account with $1000 at 4% annual interest becomes $A = 1000 \cdot 1.04^t$. After 10 years: KaTeX can only parse string typed expression. The bank did not add 4% of $1000 ten times — it multiplied by 1.04 ten times, which is a bigger number.