Figure 1. Electron micrograph of E.Coli bacteria (credit: “Mattosaurus,” Wikimedia Commons)

Focus in on a square centimeter of your skin. Look closer. Closer still. If you could look closely enough, you would see hundreds of thousands of microscopic organisms. They are bacteria, and they are not only on your skin, but in your mouth, nose, and even your intestines. In fact, the bacterial cells in your body at any given moment outnumber your own cells. But that is no reason to feel bad about yourself. While some bacteria can cause illness, many are healthy and even essential to the body.

Bacteria commonly reproduce through a process called binary fission, during which one bacterial cell splits into two. When conditions are right, bacteria can reproduce very quickly. Unlike humans and other complex organisms, the time required to form a new generation of bacteria is often a matter of minutes or hours, as opposed to days or years.^[1]

For simplicity’s sake, suppose we begin with a culture of one bacterial cell that can divide every hour. The table below shows the number of bacterial cells at the end of each subsequent hour. We see that the single bacterial cell leads to over one thousand bacterial cells in just ten hours! And if we were to extrapolate the table to twenty-four hours, we would have over 16 million!

Exponential Functions

In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential functions. Both types of functions have numerous real-world applications when it comes to modeling and interpreting data.

India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.^[2] If this rate continues, the population of India will exceed China’s population by the year 2031. When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth.

Evaluate Exponential Functions

Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:

• Let b = –9 and [latex]x=\frac{1}{2}[/latex]. Then [latex]f\left(x\right)=f\left(\frac{1}{2}\right)={\left(-9\right)}^{\frac{1}{2}}=\sqrt{-9}[/latex], which is not a real number.

Why do we limit the base to positive values other than 1? Because base 1 results in the constant function. Observe what happens if the base is 1:

• Let b = 1. Then [latex]f\left(x\right)={1}^{x}=1[/latex] for any value of x.

To evaluate an exponential function with the form [latex]f\left(x\right)={b}^{x}[/latex], we simply substitute x with the given value, and calculate the resulting power. For example:

Let [latex]f\left(x\right)={2}^{x}[/latex]. What is [latex]f\left(3\right)[/latex]?

To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:

Let [latex]f\left(x\right)=30{\left(2\right)}^{x}[/latex]. What is [latex]f\left(3\right)[/latex]?

Note that if the order of operations were not followed, the result would be incorrect:

[latex]f\left(3\right)=30{\left(2\right)}^{3}\ne {60}^{3}=216,000[/latex]

Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.

In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function [latex]A\left(x\right)=100+50x[/latex]. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]B\left(x\right)=100{\left(1+0.5\right)}^{x}[/latex].

A few years of growth for these companies are illustrated below.

The graphs comparing the number of stores for each company over a five-year period are shown in below. We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.

Figure 2. The graph shows the numbers of stores Companies A and B opened over a five-year period.

Notice that the domain for both functions is [latex]\left[0,\infty \right)[/latex], and the range for both functions is [latex]\left[100,\infty \right)[/latex]. After year 1, Company B always has more stores than Company A.

Now we will turn our attention to the function representing the number of stores for Company B, [latex]B\left(x\right)=100{\left(1+0.5\right)}^{x}[/latex]. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\left(x\right)=100{\left(1.5\right)}^{x}[/latex], where 100 is the initial value, 1.5 is called the base, and x is called the exponent.

 Find the Equation of an Exponential Function

In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a and b, and evaluate the function.

How To: Given two data points, write an exponential model.

1. If one of the data points has the form [latex]\left(0,a\right)[/latex], then a is the initial value. Using a, substitute the second point into the equation [latex]f\left(x\right)=a{\left(b\right)}^{x}[/latex], and solve for b.
2. If neither of the data points have the form [latex]\left(0,a\right)[/latex], substitute both points into two equations with the form [latex]f\left(x\right)=a{\left(b\right)}^{x}[/latex]. Solve the resulting system of two equations in two unknowns to find a and b.
3. Using the a and b found in the steps above, write the exponential function in the form [latex]f\left(x\right)=a{\left(b\right)}^{x}[/latex].

Example 3: Writing an Exponential Model When the Initial Value Is Known

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

Show Solution

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation [latex]N\left(t\right)=80{b}^{t}[/latex] to find b:

[latex]\begin{align}N\left(t\right)& =80{b}^{t} \\ 180& =80{b}^{6} && \text{Substitute using point }\left(6, 180\right). \\ \frac{9}{4} & ={b}^{6} && \text{Divide and write in lowest terms}. \\ b & ={\left(\frac{9}{4}\right)}^{\frac{1}{6}} && \text{Isolate }b\text{ using properties of exponents}. \\ b & \approx 1.1447 && \text{Round to 4 decimal places}. \end{align}[/latex]

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is [latex]N\left(t\right)=80{\left(1.1447\right)}^{t}[/latex]. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem, [latex]\left(0,\text{ 8}0\right)[/latex] and [latex]\left(\text{6},\text{ 18}0\right)[/latex]. We can also see that the domain for the function is [latex]\left[0,\infty \right)[/latex], and the range for the function is [latex]\left[80,\infty \right)[/latex].

Figure 3. Graph showing the population of deer over time, [latex]N\left(t\right)=80{\left(1.1447\right)}^{t}[/latex], t years after 2006

Try It

A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013 the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N of wolves over time t.

Show Solution

[latex]\left(0,129\right)[/latex] and [latex]\left(2,236\right);N\left(t\right)=129{\left(\text{1}\text{.3526}\right)}^{t}[/latex]

Example 4: Writing an Exponential Model When the Initial Value is Not Known

Find an exponential function that passes through the points [latex]\left(-2,6\right)[/latex] and [latex]\left(2,1\right)[/latex].

Show Solution

Because we don’t have the initial value, we substitute both points into an equation of the form [latex]f\left(x\right)=a{b}^{x}[/latex], and then solve the system for a and b.

• Substituting [latex]\left(-2,6\right)[/latex] gives [latex]6=a{b}^{-2}[/latex]
• Substituting [latex]\left(2,1\right)[/latex] gives [latex]1=a{b}^{2}[/latex]

Use the first equation to solve for a in terms of b:

[latex]\begin{align}&6=ab^{-2}\\&\frac{6}{b^{-2}}=a && \text{Divide.} \\&a=6b^2 && \text{Use properties of exponents to rewrite the denominator.}\end{align}[/latex]

Substitute a in the second equation, and solve for b:

[latex]\begin{align}&1=ab^2 \\ &1=6b^2b^2=6b^4 && \text{Substitute } a=6b^2. \\ &b=\left(\frac{1}{6}\right)^\frac{1}{4} && \text{Use properties of exponents to isolate }b. \\ & b\approx0.6389 && \text{Round to 4 decimal places}. \end{align}[/latex]

Use the value of b in the first equation to solve for the value of a:

[latex]a=6b^2\approx6(0.6389)^2\approx2.4492[/latex]

Thus, the equation is [latex]f\left(x\right)=2.4492{\left(0.6389\right)}^{x}[/latex].

We can graph our model to check our work. Notice that the graph below passes through the initial points given in the problem, [latex]\left(-2,\text{ 6}\right)[/latex] and [latex]\left(2,\text{ 1}\right)[/latex]. The graph is an example of an exponential decay function.

Figure 4. The graph of [latex]f\left(x\right)=2.4492{\left(0.6389\right)}^{x}[/latex] models exponential decay.

Try It

Given the two points [latex]\left(1,3\right)[/latex] and [latex]\left(2,4.5\right)[/latex], find the equation of the exponential function that passes through these two points.

Show Solution

[latex]f\left(x\right)=2{\left(1.5\right)}^{x}[/latex]

Try It

Q & A

Do two points always determine a unique exponential function?

Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x, which in many real world cases involves time.

How To: Given the graph of an exponential function, write its equation.

1. First, identify two points on the graph. Choose the y-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.
2. If one of the data points is the y-intercept [latex]\left(0,a\right)[/latex] , then a is the initial value. Using a, substitute the second point into the equation [latex]f\left(x\right)=a{\left(b\right)}^{x}[/latex], and solve for b.
3. If neither of the data points have the form [latex]\left(0,a\right)[/latex], substitute both points into two equations with the form [latex]f\left(x\right)=a{\left(b\right)}^{x}[/latex]. Solve the resulting system of two equations in two unknowns to find a and b.
4. Write the exponential function, [latex]f\left(x\right)=a{\left(b\right)}^{x}[/latex].

Example 5: Writing an Exponential Function Given Its Graph

Find an equation for the exponential function graphed in Figure 5.

Figure 5

Show Solution

We can choose the y-intercept of the graph, [latex]\left(0,3\right)[/latex], as our first point. This gives us the initial value, [latex]a=3[/latex]. Next, choose a point on the curve some distance away from [latex]\left(0,3\right)[/latex] that has integer coordinates. One such point is [latex]\left(2,12\right)[/latex].

[latex]\begin{align}y&=a{b}^{x} && \text{Write the general form of an exponential equation}. \\ y&=3{b}^{x} && \text{Substitute the initial value 3 for }a. \\ 12&=3{b}^{2} && \text{Substitute in 12 for }y\text{ and 2 for }x. \\ 4&={b}^{2} && \text{Divide by 3}. \\ b&=\pm 2 && \text{Take the square root}.\end{align}[/latex]

Because we restrict ourselves to positive values of b, we will use b = 2. Substitute a and b into the standard form to yield the equation [latex]f\left(x\right)=3{\left(2\right)}^{x}[/latex].

Try It

Find an equation for the exponential function graphed in Figure 6.

Figure 6

Show Solution

[latex]f\left(x\right)=\sqrt{2}{\left(\sqrt{2}\right)}^{x}[/latex]. Answers may vary due to round-off error. The answer should be very close to [latex]1.4142{\left(1.4142\right)}^{x}[/latex].

Try It

How To: Given two points on the curve of an exponential function, use a graphing calculator to find the equation.

1. Press [STAT].
2. Clear any existing entries in columns L1 or L2.
3. In L1, enter the x-coordinates given.
4. In L2, enter the corresponding y-coordinates.
5. Press [STAT] again. Cursor right to CALC, scroll down to ExpReg (Exponential Regression), and press [ENTER].
6. The screen displays the values of a and b in the exponential equation [latex]y=a\cdot {b}^{x}[/latex].

Example 6: Using a Graphing Calculator to Find an Exponential Function

Use a graphing calculator to find the exponential equation that includes the points [latex]\left(2,24.8\right)[/latex] and [latex]\left(5,198.4\right)[/latex].

Show Solution

Follow the guidelines above. First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4.

Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The values a = 6.2 and b = 2 will be displayed. The exponential equation is [latex]y=6.2\cdot {2}^{x}[/latex].

Try It

Use a graphing calculator to find the exponential function that includes the points (3, 75.98) and (6, 481.07).

Show Solution

[latex]y\approx 12\cdot {1.85}^{x}[/latex]

 Use Compound Interest Formulas

Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account.

The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.

We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t, principal P, APR r, and number of compounding periods in a year n:

For example, observe the table below, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.

Evaluate exponential functions with base e

Investigating Continuous Growth

So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used as the base for exponential functions. Exponential models that use e as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.

A General Note: The Continuous Growth/Decay Formula

For all real numbers t, and all positive numbers a and r, continuous growth or decay is represented by the formula

[latex]A\left(t\right)=a{e}^{rt}[/latex]

where

• a is the initial value,
• r is the continuous growth rate per unit time,
• and t is the elapsed time.

If r > 0, then the formula represents continuous growth. If r < 0, then the formula represents continuous decay.

For business applications, the continuous growth formula is called the continuous compounding formula and takes the form

[latex]A\left(t\right)=P{e}^{rt}[/latex]

where

• P is the principal or the initial invested,
• r is the growth or interest rate per unit time,
• and t is the period or term of the investment.

How To: Given the initial value, rate of growth or decay, and time t, solve a continuous growth or decay function.

1. Use the information in the problem to determine a, the initial value of the function.
2. Use the information in the problem to determine the growth rate r.
   1. If the problem refers to continuous growth, then r > 0.
   2. If the problem refers to continuous decay, then r < 0.
3. Use the information in the problem to determine the time t.
4. Substitute the given information into the continuous growth formula and solve for A(t).

Example 10: Calculating Continuous Growth

A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year?

Show Solution

Since the account is growing in value, this is a continuous compounding problem with growth rate r = 0.10. The initial investment was $1,000, so P = 1000. We use the continuous compounding formula to find the value after t = 1 year:

[latex]\begin{align}A\left(t\right) & =P{e}^{rt}&& \text{Use the continuous compounding formula}.\\ & =1000{\left(e\right)}^{0.1} && \text{Substitute known values for }P, r,\text{ and }t. \\ & \approx 1105.17&& \text{Use a calculator to approximate}. \end{align}[/latex]

The account is worth $1,105.17 after one year.

Try It

A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?

Show Solution

$3,659,823.44

Example 11: Calculating Continuous Decay

Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?

Show Solution

Since the substance is decaying, the rate, 17.3%, is negative. So, r = –0.173. The initial amount of radon-222 was 100 mg, so a = 100. We use the continuous decay formula to find the value after t = 3 days:

[latex]\begin{align}A\left(t\right)& =a{e}^{rt}&& \text{Use the continuous growth formula}.\\ & =100{e}^{-0.173\left(3\right)} && \text{Substitute known values for }a, r,\text{ and }t.\\ & \approx 59.5115&& \text{Use a calculator to approximate}.\end{align}[/latex]

So 59.5115 mg of radon-222 will remain.

Try It

Using the data in Example 9, how much radon-222 will remain after one year?

Show Solution

3.77E-26 (This is calculator notation for the number written as [latex]3.77\times {10}^{-26}[/latex] in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)

Try It

Working with an equation that describes a real-world situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events.

Graph exponential functions

Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form [latex]f\left(x\right)={b}^{x}[/latex] whose base is greater than one. We’ll use the function [latex]f\left(x\right)={2}^{x}[/latex]. Observe how the output values in the table below change as the input increases by 1.

x	–3	–2	–1	0	1	2	3
[latex]f\left(x\right)={2}^{x}[/latex]	[latex]\frac{1}{8}[/latex]	[latex]\frac{1}{4}[/latex]	[latex]\frac{1}{2}[/latex]	1	2	4	8

Each output value is the product of the previous output and the base, 2. We call the base 2 the constant ratio. In fact, for any exponential function with the form [latex]f\left(x\right)=a{b}^{x}[/latex], b is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a.

Notice from the table that

• the output values are positive for all values of x;
• as x increases, the output values increase without bound; and
• as x decreases, the output values grow smaller, approaching zero.

Figure 1 shows the exponential growth function [latex]f\left(x\right)={2}^{x}[/latex].

Figure 1. Notice that the graph gets close to the x-axis, but never touches it.

The domain of [latex]f\left(x\right)={2}^{x}[/latex] is all real numbers, the range is [latex]\left(0,\infty \right)[/latex], and the horizontal asymptote is [latex]y=0[/latex].

To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form [latex]f\left(x\right)={b}^{x}[/latex] whose base is between zero and one. We’ll use the function [latex]g\left(x\right)={\left(\frac{1}{2}\right)}^{x}[/latex]. Observe how the output values in the table below change as the input increases by 1.

x	–3	–2	–1	0	1	2	3
[latex]g\left(x\right)=\left(\frac{1}{2}\right)^{x}[/latex]	8	4	2	1	[latex]\frac{1}{2}[/latex]	[latex]\frac{1}{4}[/latex]	[latex]\frac{1}{8}[/latex]

Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio [latex]\frac{1}{2}[/latex].

Notice from the table that

• the output values are positive for all values of x;
• as x increases, the output values grow smaller, approaching zero; and
• as x decreases, the output values grow without bound.

The graph shows the exponential decay function, [latex]g\left(x\right)={\left(\frac{1}{2}\right)}^{x}[/latex].

Figure 2. The domain of [latex]g\left(x\right)={\left(\frac{1}{2}\right)}^{x}[/latex] is all real numbers, the range is [latex]\left(0,\infty \right)[/latex], and the horizontal asymptote is [latex]y=0[/latex].

A General Note: Characteristics of the Graph of the Parent Function f(x) = b^x

An exponential function with the form [latex]f\left(x\right)={b}^{x}[/latex], [latex]b>0[/latex], [latex]b\ne 1[/latex], has these characteristics:

• one-to-one function
• horizontal asymptote: [latex]y=0[/latex]
• domain: [latex]\left(-\infty , \infty \right)[/latex]
• range: [latex]\left(0,\infty \right)[/latex]
• x-intercept: none
• y-intercept: [latex]\left(0,1\right)[/latex]
• increasing if [latex]b>1[/latex]
• decreasing if [latex]b<1[/latex]

Compare the graphs of exponential growth and decay functions.

How To: Given an exponential function of the form [latex]f\left(x\right)={b}^{x}[/latex], graph the function.

1. Create a table of points.
2. Plot at least 3 point from the table, including the y-intercept [latex]\left(0,1\right)[/latex].
3. Draw a smooth curve through the points.
4. State the domain, [latex]\left(-\infty ,\infty \right)[/latex], the range, [latex]\left(0,\infty \right)[/latex], and the horizontal asymptote, [latex]y=0[/latex].

Example 1: Sketching the Graph of an Exponential Function of the Form f(x) = b^x

Sketch a graph of [latex]f\left(x\right)={0.25}^{x}[/latex]. State the domain, range, and asymptote.

Show Solution

Before graphing, identify the behavior and create a table of points for the graph.

• Since b = 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y = 0.
• Create a table of points.
  

  x	–3	–2	–1	0	1	2	3
  [latex]f\left(x\right)={0.25}^{x}[/latex]	64	16	4	1	0.25	0.0625	0.015625

• Plot the y-intercept, [latex]\left(0,1\right)[/latex], along with two other points. We can use [latex]\left(-1,4\right)[/latex] and [latex]\left(1,0.25\right)[/latex].

Draw a smooth curve connecting the points.

Figure 4. The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(0,\infty \right)[/latex]; the horizontal asymptote is [latex]y=0[/latex].

Try It

Sketch the graph of [latex]f\left(x\right)={4}^{x}[/latex]. State the domain, range, and asymptote.

Show Solution

The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(0,\infty \right)[/latex]; the horizontal asymptote is [latex]y=0[/latex].

Graph exponential functions using transformations

Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function [latex]f\left(x\right)={b}^{x}[/latex] without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied.

Graphing a Vertical Shift

The first transformation occurs when we add a constant d to the parent function [latex]f\left(x\right)={b}^{x}[/latex], giving us a vertical shift d units in the same direction as the sign. For example, if we begin by graphing a parent function, [latex]f\left(x\right)={2}^{x}[/latex], we can then graph two vertical shifts alongside it, using [latex]d=3[/latex]: the upward shift, [latex]g\left(x\right)={2}^{x}+3[/latex] and the downward shift, [latex]h\left(x\right)={2}^{x}-3[/latex]. Both vertical shifts are shown in Figure 5.

Figure 5

Observe the results of shifting [latex]f\left(x\right)={2}^{x}[/latex] vertically:

• The domain, [latex]\left(-\infty ,\infty \right)[/latex] remains unchanged.
• When the function is shifted up 3 units to [latex]g\left(x\right)={2}^{x}+3[/latex]:
  • The y-intercept shifts up 3 units to [latex]\left(0,4\right)[/latex].
  • The asymptote shifts up 3 units to [latex]y=3[/latex].
  • The range becomes [latex]\left(3,\infty \right)[/latex].
• When the function is shifted down 3 units to [latex]h\left(x\right)={2}^{x}-3[/latex]:
  • The y-intercept shifts down 3 units to [latex]\left(0,-2\right)[/latex].
  • The asymptote also shifts down 3 units to [latex]y=-3[/latex].
  • The range becomes [latex]\left(-3,\infty \right)[/latex].

Graphing a Horizontal Shift

The next transformation occurs when we add a constant c to the input of the parent function [latex]f\left(x\right)={b}^{x}[/latex], giving us a horizontal shift c units in the opposite direction of the sign. For example, if we begin by graphing the parent function [latex]f\left(x\right)={2}^{x}[/latex], we can then graph two horizontal shifts alongside it, using [latex]c=3[/latex]: the shift left, [latex]g\left(x\right)={2}^{x+3}[/latex], and the shift right, [latex]h\left(x\right)={2}^{x - 3}[/latex]. Both horizontal shifts are shown in Figure 6.

Figure 6

Observe the results of shifting [latex]f\left(x\right)={2}^{x}[/latex] horizontally:

• The domain, [latex]\left(-\infty ,\infty \right)[/latex], remains unchanged.
• The asymptote, [latex]y=0[/latex], remains unchanged.
• The y-intercept shifts such that:
  • When the function is shifted left 3 units to [latex]g\left(x\right)={2}^{x+3}[/latex], the y-intercept becomes [latex]\left(0,8\right)[/latex]. This is because [latex]{2}^{x+3}=\left(8\right){2}^{x}[/latex], so the initial value of the function is 8.
  • When the function is shifted right 3 units to [latex]h\left(x\right)={2}^{x - 3}[/latex], the y-intercept becomes [latex]\left(0,\frac{1}{8}\right)[/latex]. Again, see that [latex]{2}^{x - 3}=\left(\frac{1}{8}\right){2}^{x}[/latex], so the initial value of the function is [latex]\frac{1}{8}[/latex].

A General Note: Shifts of the Parent Function [latex]f\left(x\right)={b}^{x}[/latex]

For any constants c and d, the function [latex]f\left(x\right)={b}^{x+c}+d[/latex] shifts the parent function [latex]f\left(x\right)={b}^{x}[/latex]

• vertically d units, in the same direction of the sign of d.
• horizontally c units, in the opposite direction of the sign of c.
• The y-intercept becomes [latex]\left(0,{b}^{c}+d\right)[/latex].
• The horizontal asymptote becomes y = d.
• The range becomes [latex]\left(d,\infty \right)[/latex].
• The domain, [latex]\left(-\infty ,\infty \right)[/latex], remains unchanged.

How To: Given an exponential function with the form [latex]f\left(x\right)={b}^{x+c}+d[/latex], graph the translation.

1. Draw the horizontal asymptote y = d.
2. Identify the shift as [latex]\left(-c,d\right)[/latex]. Shift the graph of [latex]f\left(x\right)={b}^{x}[/latex] left c units if c is positive, and right [latex]c[/latex] units if c is negative.
3. Shift the graph of [latex]f\left(x\right)={b}^{x}[/latex] up d units if d is positive, and down d units if d is negative.
4. State the domain, [latex]\left(-\infty ,\infty \right)[/latex], the range, [latex]\left(d,\infty \right)[/latex], and the horizontal asymptote [latex]y=d[/latex].

Example 2: Graphing a Shift of an Exponential Function

Graph [latex]f\left(x\right)={2}^{x+1}-3[/latex]. State the domain, range, and asymptote.

Show Solution

We have an exponential equation of the form [latex]f\left(x\right)={b}^{x+c}+d[/latex], with [latex]b=2[/latex], [latex]c=1[/latex], and [latex]d=-3[/latex].

Draw the horizontal asymptote [latex]y=d[/latex], so draw [latex]y=-3[/latex].

Identify the shift as [latex]\left(-c,d\right)[/latex], so the shift is [latex]\left(-1,-3\right)[/latex].

Shift the graph of [latex]f\left(x\right)={b}^{x}[/latex] left 1 units and down 3 units.

Figure 7. The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(-3,\infty \right)[/latex]; the horizontal asymptote is [latex]y=-3[/latex].

Try It

Graph [latex]f\left(x\right)={2}^{x - 1}+3[/latex]. State domain, range, and asymptote.

Show Solution

The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(3,\infty \right)[/latex]; the horizontal asymptote is y = 3.

Try It

How To: Given an equation of the form [latex]f\left(x\right)={b}^{x+c}+d[/latex] for [latex]x[/latex], use a graphing calculator to approximate the solution.

• Press [Y=]. Enter the given exponential equation in the line headed “Y₁=.”
• Enter the given value for [latex]f\left(x\right)[/latex] in the line headed “Y₂=.”
• Press [WINDOW]. Adjust the y-axis so that it includes the value entered for “Y₂=.”
• Press [GRAPH] to observe the graph of the exponential function along with the line for the specified value of [latex]f\left(x\right)[/latex].
• To find the value of x, we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x for the indicated value of the function.

Example 3: Approximating the Solution of an Exponential Equation

Solve [latex]42=1.2{\left(5\right)}^{x}+2.8[/latex] graphically. Round to the nearest thousandth.

Show Solution

Press [Y=] and enter [latex]1.2{\left(5\right)}^{x}+2.8[/latex] next to Y₁=. Then enter 42 next to Y2=. For a window, use the values –3 to 3 for x and –5 to 55 for y. Press [GRAPH]. The graphs should intersect somewhere near x = 2.

For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess?) To the nearest thousandth, [latex]x\approx 2.166[/latex].

Try It

Solve [latex]4=7.85{\left(1.15\right)}^{x}-2.27[/latex] graphically. Round to the nearest thousandth.

Show Solution

[latex]x\approx -1.608[/latex]

Graphing a Stretch or Compression

While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression occurs when we multiply the parent function [latex]f\left(x\right)={b}^{x}[/latex] by a constant [latex]|a|>0[/latex]. For example, if we begin by graphing the parent function [latex]f\left(x\right)={2}^{x}[/latex], we can then graph the stretch, using [latex]a=3[/latex], to get [latex]g\left(x\right)=3{\left(2\right)}^{x}[/latex] as shown on the left in Figure 8, and the compression, using [latex]a=\frac{1}{3}[/latex], to get [latex]h\left(x\right)=\frac{1}{3}{\left(2\right)}^{x}[/latex] as shown on the right in Figure 8.

Figure 8. (a) [latex]g\left(x\right)=3{\left(2\right)}^{x}[/latex] stretches the graph of [latex]f\left(x\right)={2}^{x}[/latex] vertically by a factor of 3. (b) [latex]h\left(x\right)=\frac{1}{3}{\left(2\right)}^{x}[/latex] compresses the graph of [latex]f\left(x\right)={2}^{x}[/latex] vertically by a factor of [latex]\frac{1}{3}[/latex].

A General Note: Stretches and Compressions of the Parent Function f(x) = b^x

For any factor a > 0, the function [latex]f\left(x\right)=a{\left(b\right)}^{x}[/latex]

• is stretched vertically by a factor of a if [latex]|a|>1[/latex].
• is compressed vertically by a factor of a if [latex]|a|<1[/latex].
• has a y-intercept of [latex]\left(0,a\right)[/latex].
• has a horizontal asymptote at [latex]y=0[/latex], a range of [latex]\left(0,\infty \right)[/latex], and a domain of [latex]\left(-\infty ,\infty \right)[/latex], which are unchanged from the parent function.

Example 4: Graphing the Stretch of an Exponential Function

Sketch a graph of [latex]f\left(x\right)=4{\left(\frac{1}{2}\right)}^{x}[/latex]. State the domain, range, and asymptote.

Show Solution

Before graphing, identify the behavior and key points on the graph.

• Since [latex]b=\frac{1}{2}[/latex] is between zero and one, the left tail of the graph will increase without bound as x decreases, and the right tail will approach the x-axis as x increases.
• Since a = 4, the graph of [latex]f\left(x\right)={\left(\frac{1}{2}\right)}^{x}[/latex] will be stretched by a factor of 4.
• Create a table of points.
  

  x	–3	–2	–1	0	1	2	3
  [latex]f\left(x\right)=4\left(\frac{1}{2}\right)^{x}[/latex]	32	16	8	4	2	1	0.5

• Plot the y-intercept, [latex]\left(0,4\right)[/latex], along with two other points. We can use [latex]\left(-1,8\right)[/latex] and [latex]\left(1,2\right)[/latex].

Draw a smooth curve connecting the points.

Figure 9. The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(0,\infty \right)[/latex]; the horizontal asymptote is y = 0.

Try It

Sketch the graph of [latex]f\left(x\right)=\frac{1}{2}{\left(4\right)}^{x}[/latex]. State the domain, range, and asymptote.

Show Solution

The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(0,\infty \right)[/latex]; the horizontal asymptote is [latex]y=0[/latex]. 

Try It

Graphing Reflections

In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x-axis or the y-axis. When we multiply the parent function [latex]f\left(x\right)={b}^{x}[/latex] by –1, we get a reflection about the x-axis. When we multiply the input by –1, we get a reflection about the y-axis. For example, if we begin by graphing the parent function [latex]f\left(x\right)={2}^{x}[/latex], we can then graph the two reflections alongside it. The reflection about the x-axis, [latex]g\left(x\right)={-2}^{x}[/latex], is shown on the left side, and the reflection about the y-axis [latex]h\left(x\right)={2}^{-x}[/latex], is shown on the right side.

A General Note: Reflections of the Parent Function f(x) = b^x

The function [latex]f\left(x\right)=-{b}^{x}[/latex]

• reflects the parent function [latex]f\left(x\right)={b}^{x}[/latex] about the x-axis.
• has a y-intercept of [latex]\left(0,-1\right)[/latex].
• has a range of [latex]\left(-\infty ,0\right)[/latex]
• has a horizontal asymptote at [latex]y=0[/latex] and domain of [latex]\left(-\infty ,\infty \right)[/latex], which are unchanged from the parent function.

The function [latex]f\left(x\right)={b}^{-x}[/latex]

• reflects the parent function [latex]f\left(x\right)={b}^{x}[/latex] about the y-axis.
• has a y-intercept of [latex]\left(0,1\right)[/latex], a horizontal asymptote at [latex]y=0[/latex], a range of [latex]\left(0,\infty \right)[/latex], and a domain of [latex]\left(-\infty ,\infty \right)[/latex], which are unchanged from the parent function.

Example 5: Writing and Graphing the Reflection of an Exponential Function

Find and graph the equation for a function, [latex]g\left(x\right)[/latex], that reflects [latex]f\left(x\right)={\left(\frac{1}{4}\right)}^{x}[/latex] about the x-axis. State its domain, range, and asymptote.

Show Solution

Since we want to reflect the parent function [latex]f\left(x\right)={\left(\frac{1}{4}\right)}^{x}[/latex] about the x-axis, we multiply [latex]f\left(x\right)[/latex] by –1 to get, [latex]g\left(x\right)=-{\left(\frac{1}{4}\right)}^{x}[/latex]. Next we create a table of points.

[latex]x[/latex]	–3	–2	–1	0	1	2	3
[latex]g\left(x\right)=-\left(\frac{1}{4}\right)^{x}[/latex]	–64	–16	–4	–1	–0.25	–0.0625	–0.0156

Plot the y-intercept, [latex]\left(0,-1\right)[/latex], along with two other points. We can use [latex]\left(-1,-4\right)[/latex] and [latex]\left(1,-0.25\right)[/latex].

Draw a smooth curve connecting the points:

Figure 11. The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(-\infty ,0\right)[/latex]; the horizontal asymptote is [latex]y=0[/latex].

Try It

Find and graph the equation for a function, [latex]g\left(x\right)[/latex], that reflects [latex]f\left(x\right)={1.25}^{x}[/latex] about the y-axis. State its domain, range, and asymptote.

Show Solution

The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(0,\infty \right)[/latex]; the horizontal asymptote is [latex]y=0[/latex].

Try It

Summarizing Translations of the Exponential Function

Now that we have worked with each type of translation for the exponential function, we can summarize them to arrive at the general equation for translating exponential functions.

Translations of the Parent Function [latex]f\left(x\right)={b}^{x}[/latex]
Translation	Form
Shift Horizontally c units to the left Vertically d units up	[latex]f\left(x\right)={b}^{x+c}+d[/latex]
Stretch and Compress Stretch if |a|>1 Compression if 0<|a|<1	[latex]f\left(x\right)=a{b}^{x}[/latex]
Reflect about the x-axis	[latex]f\left(x\right)=-{b}^{x}[/latex]
Reflect about the y-axis	[latex]f\left(x\right)={b}^{-x}={\left(\frac{1}{b}\right)}^{x}[/latex]
General equation for all translations	[latex]f\left(x\right)=a{b}^{x+c}+d[/latex]

A General Note: Translations of Exponential Functions

A translation of an exponential function has the form

[latex] f\left(x\right)=a{b}^{x+c}+d[/latex]

Where the parent function, [latex]y={b}^{x}[/latex], [latex]b>1[/latex], is

• shifted horizontally c units to the left.
• stretched vertically by a factor of |a| if |a| > 0.
• compressed vertically by a factor of |a| if 0 < |a| < 1.
• shifted vertically d units.
• reflected about the x-axis when a < 0.

Note the order of the shifts, transformations, and reflections follow the order of operations.

Example 6: Writing a Function from a Description

Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range.

[latex]f\left(x\right)={e}^{x}[/latex] is vertically stretched by a factor of 2, reflected across the y-axis, and then shifted up 4 units.

Show Solution

We want to find an equation of the general form [latex] f\left(x\right)=a{b}^{x+c}+d[/latex]. We use the description provided to find a, b, c, and d.

• We are given the parent function [latex]f\left(x\right)={e}^{x}[/latex], so b = e.
• The function is stretched by a factor of 2, so a = 2.
• The function is reflected about the y-axis. We replace x with –x to get: [latex]{e}^{-x}[/latex].
• The graph is shifted vertically 4 units, so d = 4.

Substituting in the general form we get,

[latex]\begin{align} f\left(x\right)& =a{b}^{x+c}+d \\ & =2{e}^{-x+0}+4\\ & =2{e}^{-x}+4 \end{align}[/latex]

The domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(4,\infty \right)[/latex]; the horizontal asymptote is [latex]y=4[/latex].

Try It

Write the equation for function described below. Give the horizontal asymptote, the domain, and the range.

[latex]f\left(x\right)={e}^{x}[/latex] is compressed vertically by a factor of [latex]\frac{1}{3}[/latex], reflected across the x-axis and then shifted down 2 units.

Show Solution

[latex]f\left(x\right)=-\frac{1}{3}{e}^{x}-2[/latex]; the domain is [latex]\left(-\infty ,\infty \right)[/latex]; the range is [latex]\left(-\infty ,2\right)[/latex]; the horizontal asymptote is [latex]y=2[/latex].

 

Key Equations