Math Core

A linear function grows by adding the same amount at each step. Many things in the real world grow differently: a rumor spreads, a savings account earns interest, a population doubles. These grow by multiplying by the same factor at each step, and the functions that describe them are called exponential functions.

Adding versus multiplying

Compare two patterns, each starting at $3$ :

$x$	$0$	$1$	$2$	$3$	$4$	$5$
add $2$ each step	$3$	$5$	$7$	$9$	$11$	$13$
multiply by $2$ each step	$3$	$6$	$12$	$24$	$48$	$96$

The first row is linear: $y = 2x + 3$ . The second row starts slower than you might expect, then races ahead. After $x$ steps you've multiplied $3$ by $2$ a total of $x$ times, so $y = 3 \cdot 2^x$ . The variable is in the exponent, which is where the name comes from.

Definition

Exponential function

An exponential function has the form

f(x) = a \cdot b^x

where $a \ne 0$ , $b > 0$ and $b \ne 1$ . The number $a$ is the initial value (the $y$ -intercept, since $f(0) = a \cdot b^0 = a$ ). The number $b$ is the base, or growth factor: each time $x$ increases by $1$ , $f(x)$ is multiplied by $b$ .

The base must be positive and not $1$ . With $b = 1$ the function would be the constant $a$ , and a negative base would make values flip signs and be undefined for inputs like $x = \tfrac{1}{2}$ .

Evaluating an exponential function

Substitute the input, then follow the order of operations: exponent first, then multiply by $a$ .

Worked example: Plugging in

Let $f(x) = 5 \cdot 3^x$ . Find $f(2)$ , $f(0)$ and $f(-1)$ .

\begin{aligned} f(2) &= 5 \cdot 3^2 = 5 \cdot 9 = 45 \\ f(0) &= 5 \cdot 3^0 = 5 \cdot 1 = 5 \\ f(-1) &= 5 \cdot 3^{-1} = 5 \cdot \tfrac{1}{3} = \tfrac{5}{3} \end{aligned}

Common mistake

Don't multiply $a$ and $b$ before applying the exponent. In $f(2) = 5 \cdot 3^2$ , the exponent belongs only to $3$ . Computing $15^2 = 225$ is wrong; the correct value is $5 \cdot 9 = 45$ .

Growth and decay

The base decides the shape of the graph (assuming $a > 0$ ).

If $b > 1$ , the function shows exponential growth: it rises from left to right, faster and faster.
If $0 < b < 1$ , the function shows exponential decay: it falls from left to right, leveling off toward $0$ .

Here are $y = 2^x$ (growth) and $y = \left(\tfrac{1}{2}\right)^x$ (decay). Both pass through $(0, 1)$ because $a = 1$ .

y = 2^x rises to the right (growth). y = (1/2)^x falls to the right (decay). Both cross the y-axis at (0, 1).Open in grapher →

Notice that the two graphs are mirror images across the $y$ -axis. That's because $\left(\tfrac{1}{2}\right)^x = 2^{-x}$ .

Features of the graph

Look at $y = 2^x$ as $x$ gets very negative: $2^{-5} = \tfrac{1}{32}$ , $2^{-10} = \tfrac{1}{1024}$ . The values get closer and closer to $0$ but never reach it, since a power of a positive number is never $0$ or negative.

Graph of f(x) = a · bˣ with a > 0

$y$ -intercept: $(0, a)$ .
Domain: all real numbers. You can use any input as an exponent.
Range: $y > 0$ . The graph stays above the $x$ -axis.
Asymptote: the line $y = 0$ . The graph gets closer and closer to it on one side without touching it.
Direction: rising (growth) if $b > 1$ ; falling (decay) if $0 < b < 1$ .

The value of $a$ stretches the graph. For $y = 3 \cdot 2^x$ every output is $3$ times the output of $y = 2^x$ , so the graph crosses the $y$ -axis at $3$ instead of $1$ .

y = 3 · 2^x (steeper curve) is y = 2^x stretched upward by a factor of 3. Its y-intercept is 3.Open in grapher →

Recognizing exponential patterns in tables

In a table with $x$ -values that go up by $1$ :

If the outputs have a common difference, the function is linear.
If the outputs have a common ratio (each output divided by the one before gives the same number), the function is exponential. That ratio is $b$ .

Worked example: Linear or exponential?

Decide whether each table is linear or exponential, and write its equation.

$x$	$0$	$1$	$2$	$3$
Table A	$80$	$40$	$20$	$10$
Table B	$80$	$70$	$60$	$50$

Table A: $\dfrac{40}{80} = \dfrac{20}{40} = \dfrac{10}{20} = \dfrac{1}{2}$ . The common ratio is $\tfrac{1}{2}$ and the value at $x = 0$ is $80$ , so $y = 80\left(\tfrac{1}{2}\right)^x$ . This is exponential decay.

Table B: each output is $10$ less than the one before. That's a common difference of $-10$ , so it's linear: $y = -10x + 80$ .

Writing an equation from two points

If you know the output at $x = 0$ and at $x = 1$ , you have everything you need: $a = f(0)$ and $b = \dfrac{f(1)}{f(0)}$ .

Worked example: From a graph

An exponential graph passes through $(0, 4)$ and $(1, 12)$ . Write its equation and find $f(3)$ .

The $y$ -intercept gives $a = 4$ . Going from $x = 0$ to $x = 1$ multiplies the output by $\dfrac{12}{4} = 3$ , so $b = 3$ .

f(x) = 4 \cdot 3^x, \qquad f(3) = 4 \cdot 3^3 = 4 \cdot 27 = 108.

Tip

An exponential growth function eventually overtakes every linear function, no matter how steep the line. For example, $2^x$ is smaller than $100x$ at first, but by $x = 10$ it's $1024 > 1000$ , and it only pulls further ahead.