Exponential Functions: Graphs and Examples

An exponential function has the variable in the exponent. Its form is $f(x) = a \cdot b^x$ , where $a$ is the starting value and $b$ is the number you multiply by each time $x$ goes up by 1. For example, $f(x) = 3 \cdot 2^x$ doubles at every step.

Before this lesson: You should know function notation, the exponent rules, and zero and negative exponents.

Extra practice: The Algebra 1 Unit 6 Practice Pack has printable problems, answer keys and a cheat sheet for this unit.

What Makes a Function Exponential

Fold a sheet of paper in half. It is now 2 layers thick. Fold again and you have 4 layers, then 8, then 16. Each fold multiplies the layers by 2.

After $x$ folds, there are $2^x$ layers. The number of folds is in the exponent. That is what makes the function exponential.

A linear function adds the same amount each step. An exponential function multiplies by the same amount each step. That number is the base, $b$ .

The Exponential Function Form

f(x) = a \cdot b^x

Part	Meaning	In f(x) = 3 · 2ˣ
$a$	The starting value, or $y$ -intercept. It is the output when $x = 0$ . It can’t be 0.	$a = 3$
$b$	The base. Multiply by it each time $x$ goes up by 1. It must be positive and not 1.	$b = 2$
$x$	The input. It sits in the exponent.	$x$

If $b$ is greater than 1, the function grows. This is exponential growth. If $b$ is between 0 and 1, the function shrinks. This is exponential decay.

How to Tell If a Table Is Exponential

Check that the $x$ -values go up by the same amount, such as 1.
Divide each $y$ -value by the one before it. This is the ratio.
If every ratio is the same, the table is exponential. That ratio is $b$ .
Find $a$ , the $y$ -value when $x = 0$ . Then write $f(x) = a \cdot b^x$ .

Worked Examples

Example 1: Spot the Exponential Functions

Which are exponential functions? (a) $y = 4 \cdot 3^x$ , (b) $y = x^3$ , (c) $y = 10(0.5)^x$ , (d) $y = 5x + 2$ and (e) $y = 6 \cdot 1^x$ .

y = 4 \cdot 3^x

Step 1. (a) The variable is in the exponent. Here $a = 4$ and $b = 3$ . Yes, it is exponential.

y = x^3

Step 2. (b) The variable is in the base, not the exponent. No.

y = 10(0.5)^x

Step 3. (c) The variable is in the exponent. Here $a = 10$ and $b = 0.5$ . Since $0 \lt 0.5 \lt 1$ , it is decay. Yes.

y = 5x + 2

Step 4. (d) There is no exponent on x. This is linear. It adds 5 each step. No.

1^x = 1

Step 5. (e) 1 to any power is 1.

y = 6

Step 6. The function is a flat line, not exponential. No.

Answer: (a) and (c)

Example 2: Evaluate an Exponential Function

Let $f(x) = 4 \cdot 3^x$ . Find $f(2)$ , $f(0)$ and $f(-1)$ .

f(2) = 4 \cdot 3^2

Step 1. Substitute $x = 2$ (see function notation).

4 \cdot 9 = 36

Step 2. Do the exponent first, then multiply (see order of operations).

f(0) = 4 \cdot 3^0

Step 3. Substitute $x = 0$ .

4 \cdot 1 = 4

Step 4. Any nonzero number to the zero power is 1. This gives the starting value, $a$ .

f(-1) = 4 \cdot 3^{-1}

Step 5. Substitute $x = -1$ .

4 \cdot \frac{1}{3}

Step 6. Rewrite the negative exponent as a fraction (see zero and negative exponents).

4 \cdot \frac{1}{3} = \frac{4}{3}

Step 7. Multiply. The output is still positive.

Answer: $f(2) = 36$ , $f(0) = 4$ , $f(-1) = \dfrac{4}{3}$

Example 3: Write a Function from a Table

Decide if each table is exponential. If it is, write its function.

$x$	0	1	2	3
Table A	5	10	20	40
Table B	81	27	9	3
Table C	2	5	8	11

\frac{10}{5} = 2, \quad \frac{20}{10} = 2, \quad \frac{40}{20} = 2

Step 1. Table A: divide each y-value by the one before it.

f(x) = 5 \cdot 2^x

Step 2. The ratios match, so $b = 2$ . At $x = 0$ , $y = 5$ , so $a = 5$ .

\frac{27}{81} = \tfrac{1}{3}, \quad \frac{9}{27} = \tfrac{1}{3}, \quad \frac{3}{9} = \tfrac{1}{3}

Step 3. Table B: divide each y-value by the one before it.

f(x) = 81\left(\tfrac{1}{3}\right)^x

Step 4. The ratios match, so $b = \tfrac{1}{3}$ . At $x = 0$ , $y = 81$ . This is decay.

\frac{5}{2}, \quad \frac{8}{5}

Step 5. Table C: divide each y-value by the one before it.

Step 6. The ratios don’t match, so it is not exponential. It adds 3 each time, so it is linear.

Answer: A is $f(x) = 5 \cdot 2^x$ ; B is $f(x) = 81\left(\tfrac{1}{3}\right)^x$ ; C is not exponential.

Example 4: Graph Growth and Decay

Graph $f(x) = 2^x$ and $g(x) = 8\left(\tfrac{1}{2}\right)^x$ . Give the domain and range.

f(-1) = 2^{-1}

Step 1. Make a table for each function. For example, substitute −1 into f.

2^{-1} = \tfrac{1}{2}

Step 2. A negative exponent means one over the power.

g(-1) = 8 \cdot 2

Step 3. Substitute −1 into g. One-half to the −1 power is 2.

8 \cdot 2 = 16

Step 4. Multiply.

$x$	−1	0	1	2	3
$f(x) = 2^x$	1/2	1	2	4	8
$g(x) = 8\left(\tfrac{1}{2}\right)^x$	16	8	4	2	1

Step 5. Plot the points and join them with a smooth curve. $f$ rises from left to right. $g$ falls.

f(x) = 2ˣ grows and g(x) = 8(1/2)ˣ decays. Both stay above the x-axis.

Step 6. Read the y-intercepts: $(0, 1)$ and $(0, 8)$ .

Step 7. Each curve gets closer and closer to the x-axis but never touches it. The line $y = 0$ is called a horizontal asymptote.

y \gt 0

Step 8. Any x-value works, so the domain is all real numbers. The outputs are always positive, so this is the range. (See domain and range.)

Answer: For both functions, the domain is all real numbers and the range is $y \gt 0$ .

Example 5: Word Problem

A new video has 300 views. The number of views triples every day for the first week. Write a function for the views $V$ after $d$ days. Then find the views on day 4. On which day does it first pass 20,000 views?

V(d) = 300 \cdot 3^d

Step 1. The start is 300, so $a = 300$ . Tripling means $b = 3$ .

V(4) = 300 \cdot 3^4

Step 2. Substitute $d = 4$ .

300 \cdot 81

Step 3. Find $3^4 = 81$ .

300 \cdot 81 = 24{,}300

Step 4. Multiply.

V(3) = 300 \cdot 3^3 = 300 \cdot 27 = 8{,}100

Step 5. Check the day before.

Step 6. 8,100 is less than 20,000, so day 4 is the first day over.

Answer: $V(d) = 300 \cdot 3^d$ . Day 4 has 24,300 views, and day 4 is the first day over 20,000.

Common Mistakes

Mistake	Example of the error	Fix
Multiplying a and b first	$3 \cdot 2^x = 6^x$	Do the exponent first: $3 \cdot 2^2 = 3 \cdot 4 = 12$ , not $6^2 = 36$
Mixing up x² and 2ˣ	Calling $y = x^2$ exponential	Exponential means the variable is in the exponent
Checking differences instead of ratios	2, 6, 18 has differences 4 and 12, so “not a pattern”	Divide: $\frac{6}{2} = 3$ and $\frac{18}{6} = 3$ . It is exponential with $b = 3$
Thinking the graph crosses the x-axis	“ $2^x = 0$ when $x$ is very negative”	$2^x$ gets close to 0 but is always positive
Using the first table value as a	Table starts at $x = 1$ , $y = 6$ , so $a = 6$	$a$ is the value at $x = 0$ . Work back one step by dividing by $b$

Growth vs. Decay at a Glance

Feature	Growth ( $b \gt 1$ )	Decay ( $0 \lt b \lt 1$ )
Example	$y = 2^x$	$y = \left(\tfrac{1}{2}\right)^x$
Graph from left to right	Rises, faster and faster	Falls, slower and slower
Each step	Multiply by more than 1	Multiply by a number between 0 and 1
$y$ -intercept	$(0, a)$	$(0, a)$
Horizontal asymptote	$y = 0$	$y = 0$
Domain and range (when a > 0)	All real numbers; $y \gt 0$	All real numbers; $y \gt 0$

Real growth and decay are often given as a percent, like “grows 5% per year.” That is the next lesson, Exponential Growth and Decay. Later in this unit, Linear vs. Exponential Functions compares a constant ratio with a constant average rate of change. The domain and range ideas come from domain and range.

Practice Problems

Is $y = 7(1.5)^x$ an exponential function? If so, give $a$ and $b$ , and say if it is growth or decay.
Let $f(x) = 2 \cdot 5^x$ . Find $f(3)$ .
Let $f(x) = 6\left(\tfrac{1}{2}\right)^x$ . Find $f(-2)$ .
Write a function for this table: $x$ = 0, 1, 2, 3 and $y$ = 3, 12, 48, 192.
Is this table exponential? $x$ = 1, 2, 3, 4 and $y$ = 1, 4, 9, 16.
Give the $y$ -intercept of $f(x) = 9\left(\tfrac{2}{3}\right)^x$ . Is it growth or decay?
What is the range of $f(x) = 5 \cdot 2^x$ ?
An exponential function passes through $(0, 4)$ and $(1, 12)$ . Write its equation.
Let $h(x) = 160\left(\tfrac{1}{2}\right)^x$ . Find $h(4)$ .
A lab starts with 25 cells. The number of cells doubles every hour. Write a function for the cells after h hours. How many cells are there after 5 hours?

Show answers

The variable is in the exponent, so yes

a = 7, \quad b = 1.5

Read the parts.

Since $1.5 \gt 1$ , it is growth
Yes. It is exponential growth with a = 7 and b = 1.5.

f(3) = 2 \cdot 5^3

Substitute x = 3.

2 \cdot 125

Do the exponent first.

2 \cdot 125 = 250

Multiply.

f(-2) = 6\left(\tfrac{1}{2}\right)^{-2}

Substitute x = −2.

6 \cdot 2^2

A negative exponent flips the fraction.

6 \cdot 4 = 24

Find $2^2 = 4$ and multiply.

\frac{12}{3} = 4, \quad \frac{48}{12} = 4, \quad \frac{192}{48} = 4

Find the ratios.

f(x) = 3 \cdot 4^x

The ratios match, so b = 4. At x = 0, y = 3, so a = 3.

\frac{4}{1} = 4, \quad \frac{9}{4}, \quad \frac{16}{9}

Find the ratios.

The ratios don’t match
No. This table is $y = x^2$ .

f(0) = 9\left(\tfrac{2}{3}\right)^0 = 9 \cdot 1 = 9

Substitute x = 0.

Since $0 \lt \tfrac{2}{3} \lt 1$ , it is decay
The y-intercept is (0, 9), and it is decay.

The starting value 5 is positive, and every power of 2 is positive
So every output is positive, and the curve gets close to 0 but never reaches it
The range is $y \gt 0$ .

a = 4

The point (0, 4) gives the starting value.

b = \frac{12}{4} = 3

Divide the outputs to find b.

So $f(x) = 4 \cdot 3^x$ .

h(4) = 160\left(\tfrac{1}{2}\right)^4

Substitute x = 4.

\left(\tfrac{1}{2}\right)^4 = \tfrac{1}{16}

Find the power.

160 \cdot \tfrac{1}{16} = 10

Multiply.

10.

C(h) = 25 \cdot 2^h

The start is 25 and the cells double, so a = 25 and b = 2.

C(5) = 25 \cdot 2^5

Substitute h = 5.

25 \cdot 32 = 800

Find $2^5 = 32$ and multiply.

There are 800 cells after 5 hours.

Want more practice? The Algebra 1 Unit 6 Practice Pack has 49 pages of problems on this and the other lessons in the unit, with step-by-step answers and a cheat sheet.

FAQ

What makes a function exponential?

The variable is in the exponent, as in $f(x) = a \cdot b^x$ . In a table, the outputs are multiplied by the same number each time $x$ goes up by 1.

What is the horizontal asymptote of an exponential function?

For $f(x) = a \cdot b^x$ , it is the line $y = 0$ . The graph gets closer and closer to it but never touches it.

Why can’t the base be negative or 1?

With $b = 1$ , every output is the same, so the graph is a flat line. With a negative base, the signs jump back and forth, as in $(-2)^1 = -2$ and $(-2)^2 = 4$ , so there is no smooth curve.

How do you tell growth from decay?

Look at the base. If $b \gt 1$ , it is growth. If $b$ is between 0 and 1, it is decay.

Summary

An exponential function has the form $f(x) = a \cdot b^x$ , with the variable in the exponent.
$a$ is the starting value and the $y$ -intercept. $b$ is the number you multiply by each step.
Growth when $b \gt 1$ ; decay when $0 \lt b \lt 1$ .
In a table, the ratios between outputs are all the same.
The graph never touches its horizontal asymptote, $y = 0$ .

Previous lesson: Zero and Negative Exponents

Next lesson: Exponential Growth and Decay

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