# 8.6: Exponential Functions

**At Grade**Created by: CK-12

## Learning Objectives

- Graph an exponential function.
- Compare graphs of exponential functions.
- Analyze the properties of exponential functions.

## Introduction

A colony of bacteria has a population of three thousand at noon on Monday. During the next week, the colony’s population doubles every day. What is the population of the bacteria colony just before midnight on Saturday?

At first glance, this seems like a problem you could solve using a geometric sequence. And you could, if the bacteria population doubled all at once every day; since it doubled every day for five days, the final population would be 3000 times \begin{align*}2^5\end{align*}

But bacteria don’t reproduce all at once; their population grows slowly over the course of an entire day. So how do we figure out the population after five *and a half* days?

## Exponential Functions

Exponential functions are a lot like geometrical sequences. The main difference between them is that a geometric sequence is **discrete** while an exponential function is **continuous**.

**Discrete** means that the sequence has values only at distinct points (the 1st term, 2nd term, etc.)

**Continuous** means that the function has values for all possible values of \begin{align*}x\end{align*}

The problem with the bacteria is an example of a continuous function. Here’s an example of a discrete function:

*An ant walks past several stacks of Lego blocks. There is one block in the first stack, 3 blocks in the \begin{align*}2^{nd}\end{align*} stack and 9 blocks in the \begin{align*}3^{rd}\end{align*} stack. In fact, in each successive stack there are triple the number of blocks than in the previous stack.*

In this example, each stack has a distinct number of blocks and the next stack is made by adding a certain number of whole pieces all at once. More importantly, however, there are no values of the sequence **between** the stacks. You can’t ask how high the stack is between the \begin{align*}2^{nd}\end{align*} and \begin{align*}3^{rd}\end{align*} stack, as no stack exists at that position!

As a result of this difference, we use a geometric series to describe quantities that have values at discrete points, and we use exponential functions to describe quantities that have values that change continuously.

When we graph an exponential function, we draw the graph with a solid curve to show that the function has values at any time during the day. On the other hand, when we graph a geometric sequence, we draw discrete points to signify that the sequence only has value at those points but not in between.

Here are graphs for the two examples above:

The formula for an exponential function is similar to the formula for finding the terms in a geometric sequence. An exponential function takes the form

\begin{align*}y = A \cdot b^x\end{align*}

where \begin{align*}A\end{align*} is the starting amount and \begin{align*}b\end{align*} is the amount by which the total is multiplied every time. For example, the bacteria problem above would have the equation \begin{align*}y = 3000 \cdot 2^x\end{align*}.

## Compare Graphs of Exponential Functions

Let’s graph a few exponential functions and see what happens as we change the constants in the formula. The basic shape of the exponential function should stay the same—but it may become steeper or shallower depending on the constants we are using.

First, let’s see what happens when we change the value of \begin{align*}A\end{align*}.

**Example 1**

*Compare the graphs of* \begin{align*}y = 2^x\end{align*} *and* \begin{align*}y = 3 \cdot 2^x\end{align*}.

**Solution**

Let’s make a table of values for both functions.

\begin{align*}x\end{align*} | \begin{align*}y = 2^x\end{align*} | \begin{align*}y = 3 \cdot 2^x\end{align*} |
---|---|---|

-3 | \begin{align*}\frac{1}{8}\end{align*} | \begin{align*}y = 3 \cdot 2^{-3} = 3 \cdot \frac{1}{2^3} = \frac{3}{8}\end{align*} |

-2 | \begin{align*}\frac{1}{4}\end{align*} | \begin{align*}y = 3 \cdot 2^{-2} = 3 \cdot \frac{1}{2^2} = \frac{3}{4}\end{align*} |

-1 | \begin{align*}\frac{1}{2}\end{align*} | \begin{align*}y = 3 \cdot 2^{-1} = 3 \cdot \frac{1}{2^1} = \frac{3}{2}\end{align*} |

0 | 1 | \begin{align*}y = 3 \cdot 2^0 = 3\end{align*} |

1 | 2 | \begin{align*}y = 3 \cdot 2^1 = 6\end{align*} |

2 | 4 | \begin{align*}y = 3 \cdot 2^2 = 3 \cdot 4 = 12\end{align*} |

3 | 8 | \begin{align*}y = 3 \cdot 2^3 = 3 \cdot 8 = 24\end{align*} |

Now let's use this table to graph the functions.

We can see that the function \begin{align*}y = 3 \cdot 2^x\end{align*} is bigger than the function \begin{align*}y = 2^x\end{align*}. In both functions, the value of \begin{align*}y\end{align*} doubles every time \begin{align*}x\end{align*} increases by one. However, \begin{align*}y = 3 \cdot 2^x\end{align*} “starts” with a value of 3, while \begin{align*}y = 2^x\end{align*} “starts” with a value of 1, so it makes sense that \begin{align*}y = 3 \cdot 2^x\end{align*} would be bigger as its values of \begin{align*}y\end{align*} keep getting doubled.

Similarly, if the starting value of \begin{align*}A\end{align*} is smaller, the values of the entire function will be smaller.

**Example 2**

*Compare the graphs of \begin{align*}y =2^x\end{align*} and \begin{align*}y = \frac{1}{3} \cdot 2^x\end{align*}.*

**Solution**

Let’s make a table of values for both functions.

\begin{align*}x\end{align*} | \begin{align*}y = 2^x\end{align*} | \begin{align*}y = \frac{1}{3} \cdot 2^x\end{align*} |
---|---|---|

-3 | \begin{align*}\frac{1}{8}\end{align*} | \begin{align*}y = \frac{1}{3} \cdot 2^{-3} = \frac{1}{3} \cdot \frac{1}{2^3} = \frac{1}{24}\end{align*} |

-2 | \begin{align*}\frac{1}{4}\end{align*} | \begin{align*}y = \frac{1}{3} \cdot 2^{-2} = \frac{1}{3} \cdot \frac{1}{2^2} = \frac{1}{12}\end{align*} |

-1 | \begin{align*}\frac{1}{2}\end{align*} | \begin{align*}y = \frac{1}{3} \cdot 2^{-1} = \frac{1}{3} \cdot \frac{1}{2^1} = \frac{1}{6}\end{align*} |

0 | 1 | \begin{align*}y = \frac{1}{3} \cdot 2^0 = \frac{1}{3}\end{align*} |

1 | 2 | \begin{align*}y = \frac{1}{3} \cdot 2^1 = \frac{2}{3}\end{align*} |

2 | 4 | \begin{align*}y = \frac{1}{3} \cdot 2^2 = \frac{1}{3} \cdot 4 = \frac{4}{3}\end{align*} |

3 | 8 | \begin{align*}y = \frac{1}{3} \cdot 2^3 = \frac{1}{3} \cdot 8 = \frac{8}{3}\end{align*} |

Now let's use this table to graph the functions.

As we expected, the exponential function \begin{align*}y = \frac{1}{3} \cdot 2^x\end{align*} is smaller than the exponential function \begin{align*}y = 2^x\end{align*}.

So what happens if the starting value of \begin{align*}A\end{align*} is negative? Let’s find out.

**Example 3**

*Graph the exponential function* \begin{align*}y = -5 \cdot 2^x\end{align*}.

**Solution**

Let’s make a table of values:

\begin{align*}x\end{align*} | \begin{align*}y = -5 \cdot 2^x\end{align*} |
---|---|

-2 | \begin{align*}- \frac{5}{4}\end{align*} |

-1 | \begin{align*}- \frac{5}{2}\end{align*} |

0 | -5 |

1 | -10 |

2 | -20 |

3 | -40 |

Now let's graph the function:

This result shouldn’t be unexpected. Since the starting value is negative and keeps doubling over time, it makes sense that the value of \begin{align*}y\end{align*} gets farther from zero, but in a negative direction. The graph is basically just like the graph of \begin{align*}y = 5 \cdot 2^x\end{align*}, only mirror-reversed about the \begin{align*}x-\end{align*}axis.

Now, let’s compare exponential functions whose bases \begin{align*}(b)\end{align*} are different.

**Example 4**

*Graph the following exponential functions on the same graph*: \begin{align*}y = 2^x, y = 3^x, y = 5^x, y = 10^x\end{align*}.

**Solution**

First we’ll make a table of values for all four functions.

\begin{align*}x\end{align*} | \begin{align*}y = 2^x\end{align*} | \begin{align*}y = 3^x\end{align*} | \begin{align*}y = 5^x\end{align*} | \begin{align*}y = 10^x\end{align*} |
---|---|---|---|---|

-2 | \begin{align*}\frac{1}{4}\end{align*} | \begin{align*}\frac{1}{9}\end{align*} | \begin{align*}\frac{1}{25}\end{align*} | \begin{align*}\frac{1}{100}\end{align*} |

-1 | \begin{align*}\frac{1}{2}\end{align*} | \begin{align*}\frac{1}{3}\end{align*} | \begin{align*}\frac{1}{5}\end{align*} | \begin{align*}\frac{1}{10}\end{align*} |

0 | 1 | 1 | 1 | 1 |

1 | 2 | 3 | 5 | 10 |

2 | 4 | 9 | 25 | 100 |

3 | 8 | 27 | 125 | 1000 |

Now let's graph the functions.

Notice that for \begin{align*}x = 0\end{align*}, all four functions equal 1. They all “start out” at the same point, but the ones with higher values for \begin{align*}b\end{align*} grow faster when \begin{align*}x\end{align*} is positive—and also shrink faster when \begin{align*}x\end{align*} is negative.

Finally, let’s explore what happens for values of \begin{align*}b\end{align*} that are less than 1.

**Example 5**

*Graph the exponential function* \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^x\end{align*}.

**Solution**

Let’s start by making a table of values. (Remember that a fraction to a negative power is equivalent to its reciprocal to the same positive power.)

\begin{align*}x\end{align*} | \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^x\end{align*} |
---|---|

-3 | \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{-3} = 5 \cdot 2^3 = 40\end{align*} |

-2 | \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{-2} = 5 \cdot 2^2 = 20\end{align*} |

-1 | \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{-1} = 5 \cdot 2^1 = 10\end{align*} |

0 | \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{0} = 5 \cdot 1 = 5\end{align*} |

1 | \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{1} = \frac{5}{2}\end{align*} |

2 | \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{2} = \frac{5}{4}\end{align*} |

Now let's graph the function.

This graph looks very different than the graphs from the previous example! What’s going on here?

When we raise a number greater than 1 to the power of \begin{align*}x\end{align*}, it gets bigger as \begin{align*}x\end{align*} gets bigger. But when we raise a number smaller than 1 to the power of \begin{align*}x\end{align*}, it gets *smaller* as \begin{align*}x\end{align*} gets bigger—as you can see from the table of values above. This makes sense because multiplying any number by a quantity less than 1 always makes it smaller.

So, when the base \begin{align*}b\end{align*} of an exponential function is between 0 and 1, the graph is like an ordinary exponential graph, only decreasing instead of increasing. Graphs like this represent **exponential decay** instead of **exponential growth**. Exponential decay functions are used to describe quantities that decrease over a period of time.

When \begin{align*}b\end{align*} can be written as a fraction, we can use the Property of Negative Exponents to write the function in a different form. For instance, \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{x}\end{align*} is equivalent to \begin{align*}5 \cdot 2^{-x}\end{align*}. These two forms are both commonly used, so it’s important to know that they are equivalent.

**Example 6**

*Graph the exponential function* \begin{align*}y = 8 \cdot 3^{-x}\end{align*}.

**Solution**

Here is our table of values and the graph of the function.

\begin{align*}x\end{align*} | \begin{align*}y = 8 \cdot 3^{-x}\end{align*} |
---|---|

-3 | \begin{align*}y = 8 \cdot 3^{-(-3)} = 8 \cdot 3^3 = 216\end{align*} |

-2 | \begin{align*}y = 8 \cdot 3^{-(-2)} = 8 \cdot 3^2 = 72\end{align*} |

-1 | \begin{align*}y = 8 \cdot 3^{-(-1)} = 8 \cdot 3^1 = 24\end{align*} |

0 | \begin{align*}y = 8 \cdot 3^{0} = 8\end{align*} |

1 | \begin{align*}y = 8 \cdot 3^{-1} = \frac{8}{3}\end{align*} |

2 | \begin{align*}y = 8 \cdot 3^{-2} = \frac{8}{9}\end{align*} |

**Example 7**

*Graph the functions \begin{align*}y = 4^x\end{align*} and \begin{align*}y = 4^{-x}\end{align*} on the same coordinate axes.*

**Solution**

Here is the table of values for the two functions. Looking at the values in the table, we can see that the two functions are “backwards” of each other, in the sense that the values for the two functions are reciprocals.

\begin{align*}x\end{align*} | \begin{align*}y = 4^x\end{align*} | \begin{align*}y = 4^{-x}\end{align*} |
---|---|---|

-3 | \begin{align*}y = 4^{-3} = \frac{1}{64}\end{align*} | \begin{align*}y = 4^{-(-3)} = 64\end{align*} |

-2 | \begin{align*}y = 4^{-2} = \frac{1}{16}\end{align*} | \begin{align*}y = 4^{-(-2)} = 16\end{align*} |

-1 | \begin{align*}y = 4^{-1} = \frac{1}{4}\end{align*} | \begin{align*}y = 4^{-(-1)} = 4\end{align*} |

0 | \begin{align*}y = 4^0 = 1\end{align*} | \begin{align*}y = 4^0 = 1\end{align*} |

1 | \begin{align*}y = 4^1 = 4\end{align*} | \begin{align*}y = 4^{-1} = \frac{1}{4}\end{align*} |

2 | \begin{align*}y = 4^2 = 16\end{align*} | \begin{align*}y = 4^{-2} = \frac{1}{16}\end{align*} |

3 | \begin{align*}y = 4^3 = \frac{1}{64}\end{align*} | \begin{align*}y = 4^{-3} = \frac{1}{64}\end{align*} |

Here is the graph of the two functions. Notice that the two functions are mirror images of each other if the mirror is placed vertically on the \begin{align*}y-\end{align*}axis.

In the next lesson, you’ll see how exponential growth and decay functions can be used to represent situations in the real world.

## Review Questions

Graph the following exponential functions by making a table of values.

- \begin{align*}y = 3^x\end{align*}
- \begin{align*}y = 5 \cdot 3^x\end{align*}
- \begin{align*}y = 40 \cdot 4^x\end{align*}
- \begin{align*}y = 3 \cdot 10^x\end{align*}

Graph the following exponential functions.

- \begin{align*}y = \left ( \frac{1}{5} \right )^x\end{align*}
- \begin{align*}y = 4 \cdot \left ( \frac{2}{3} \right )^x\end{align*}
- \begin{align*}y = 3^{-x}\end{align*}
- \begin{align*}y = \frac{3}{4} \cdot 6^{-x}\end{align*}
- Which two of the eight graphs above are mirror images of each other?
- What function would produce a graph that is the mirror image of the one in problem 4?
- How else might you write the exponential function in problem 5?
- How else might you write the function in problem 6?

Solve the following problems.

- A chain letter is sent out to 10 people telling everyone to make 10 copies of the letter and send each one to a new person.
- Assume that everyone who receives the letter sends it to ten new people and that each cycle takes a week. How many people receive the letter on the sixth week?
- What if everyone only sends the letter to 9 new people? How many people will then get letters on the sixth week?

- Nadia received $200 for her \begin{align*}10^{th}\end{align*} birthday. If she saves it in a bank account with 7.5% interest compounded yearly, how much money will she have in the bank by her \begin{align*}21^{st}\end{align*} birthday?

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