Lesson Explainer: Graphs of Exponential Functions Mathematics

In this explainer, we will learn how to sketch and identify the graphical transformations of exponential functions.

Exponential functions are extremely important in mathematics and have many real-world applications. Examples in science include modeling population growth and radioactive decay. In finance, exponential functions are used, for example, to model investments with specific compound interest requirements.

Definition: Exponential Functions

A function of the form 𝑓(π‘₯)=𝑏, where π‘βˆˆβ„>0 and 𝑏≠1, is an exponential function.

𝑏 is called the base of the exponential function, and the domain, that is, the set of possible π‘₯-values, is the real numbers, ℝ.

As an example, the function 𝑓(π‘₯)=3, shown in the graph below, is an exponential function with base 3.

Exponential functions can take many different forms. For example, 𝑦=βˆ’2 and 𝑔(𝑑)=7ο€Ό14 are considered exponential functions of the form 𝑐⋅𝑏, where π‘βˆˆβ„. If the independent variable is time, 𝑑, as in 𝑔(𝑑) above, then 𝑐 is called the initial value, that is, the value of the function when 𝑑=0. We will see more examples of exponential functions of the form 𝑦=𝑐⋅𝑏 a little later.

However, the thing we notice about all three of these examples, which is in fact their defining feature, is that the variable is the exponent. This is in contrast to algebraic functions, for example, 𝑓(π‘₯)=3π‘₯, where the exponent (here, 2) is a constant and the base (π‘₯) is a variable. In an exponential function, the reverse is true: the base is constant and the exponent is the variable.

For an exponential function of the form 𝑓(π‘₯)=𝑏, there are two general directions the function can go as the independent variable, π‘₯, increases. These depend on the value of the base, 𝑏: if 0<𝑏<1, the graph decreases as π‘₯ increases, and the function represents exponential decay. On the other hand, if 𝑏>1, then the graph increases as π‘₯ increases, and the function represents exponential growth.

Definition: Exponential Growth and Exponential Decay

For an exponential function of the form 𝑓(π‘₯)=𝑏, with π‘βˆˆβ„>0 and 𝑏≠1,

  • if 0<𝑏<1, then 𝑓(π‘₯) represents exponential decay;
  • if 𝑏>1, then 𝑓(π‘₯) represents exponential growth.

A further feature of the graph of an exponential function of the form 𝑓(π‘₯)=𝑏 is that it will always pass through the point (0,1). This is because when π‘₯=0, we have 𝑓(0)=𝑏=1.

Example 1: Finding the Point of 𝑦-Intersection of an Exponential Function

Determine the point at which the graph of the function 𝑓(π‘₯)=6 crosses the 𝑦-axis.

Answer

The graph of a function in the π‘₯𝑦-plane crosses the 𝑦-axis when π‘₯ is equal to zero. Substituting π‘₯=0 into our function then will give us the value of the 𝑦-intercept.

In our case, we have the function, 𝑓(π‘₯)=6, and substituting π‘₯=0 gives us 𝑓(0)=6. Since we know that any nonzero real number raised to the power zero is equal to one, we must have 𝑓(0)=1. Hence, the 𝑦-intercept for this function is at the point (0,1).

Now, let’s revisit exponential functions of the form 𝑐𝑓(π‘₯)=𝑐⋅𝑏. These represent a vertical stretch of the function 𝑓(π‘₯)=𝑏 by the scale factor 𝑐. The 𝑦-intercept is also scaled to the initial value, 𝑐, because when π‘₯=0, 𝑓(0)=𝑐⋅𝑏=π‘οŠ¦. Hence, the graph now passes through the point (0,𝑐). Below, we have graphs of our original function, 𝑓(π‘₯)=𝑏, and two further functions, 𝑓(π‘₯)=𝑐⋅𝑓(π‘₯), for values of 𝑐>0 and π‘Ž>1βˆΆπ‘=π‘Ž, and 𝑐=1π‘Ž.

If 𝑐<0 in our function 𝑓(π‘₯)=𝑐⋅𝑏, then the graph of 𝑓(π‘₯)=𝑏 is reflected in the horizontal axis and stretched vertically by a factor of 𝑐 units. The 𝑦-intercept, or initial value, is again scaled to 𝑦=𝑐, as shown below with 𝑐=βˆ’π‘Ž, 𝑐=βˆ’1π‘Ž, and π‘Ž>0.

Note that although the function 𝑓(π‘₯)=𝑐⋅𝑏 is decreasing for 𝑐<0, if the base 𝑏>1, we still refer to this as exponential growth, since the exponential component of the function 𝑏 is increasing. Another way to think about this is that the magnitude of the function is increasing.

Example 2: Identifying Graphs of Exponential Functions

Which of the graphs is that of 𝑓(π‘₯)=2β‹…3?

Answer

One way of approaching this question is by choosing some values of π‘₯ and calculating their corresponding 𝑦-values for the exponential function 𝑓(π‘₯)=2β‹…3. We can then determine which of the given graphs matches our input and output values.

Choosing first π‘₯=0, and substituting this into our function 𝑓(π‘₯), we have 𝑓(0)=2β‹…3. Since any nonzero real number raised to the power zero is equal to 1, this equates to 𝑓(0)=2β‹…1=2. Therefore, the first point on the graph of our function has the coordinates (0,2). However, we see from our graph that all four of options A, B, C, and D pass through the point (0,2).

Hence, we will need to investigate at least one more point. Let’s consider next the value of 𝑓(π‘₯) when π‘₯=1. In this case, we have 𝑓(1)=2β‹…3=2β‹…3=6. Our second point, therefore, has the coordinates (1,6).

Marking this on our graph, we see that this coincides with graph B. The only graph that passes through both of the calculated points (0,2) and (1,6) is option B. Hence, the function 𝑓(π‘₯)=2β‹…3 is represented by graph B.

Note that, in this case, the exponential 3 has been stretched vertically by a factor of 2 (as it is multiplied by 2). Also, since the base of the function 𝑓(π‘₯) is 3, which is greater than 1, we know that 𝑓(π‘₯) models exponential growth.

Let’s look at another example of identifying graphs of exponential functions of the form 𝑓(π‘₯)=𝑐⋅𝑏.

Example 3: Identifying Graphs of Exponential Equations

Which of the following graphs represents the function 𝑦=βˆ’4(2)?

Answer

We are given the equation 𝑦=βˆ’4(2), and we want to determine which of the given graphs represents this equation. To do this, let’s choose some values of π‘₯ and calculate their 𝑦-coordinates. We will then be able to see which of the graphs passes through these points and therefore matches the equation.

Choosing first π‘₯=0, and substituting this into the equation, we find 𝑦=βˆ’4ο€Ή2=βˆ’4β‹…1=βˆ’4. Hence, our graph must pass through the point (0,βˆ’4). Two of the given graphs, options A and D, pass through this point.

Hence, we can eliminate options B and C, since they do not pass through the calculated point. Inspecting our two remaining graphs, A and D, one way to distinguish between them is to consider an π‘₯-value that gives different 𝑦-values in each of options A and D. For example, we see that when π‘₯=1 in graph A, 𝑦 is approximately equal to βˆ’8. Hence, graph A passes through the point (1,βˆ’8). However, when π‘₯=1 in graph D, the curve passes through the point (1,βˆ’2).

Substituting π‘₯=1 into our equation, we have 𝑦=βˆ’4ο€Ή2=βˆ’4β‹…2=βˆ’8. Hence, the point (1,βˆ’8) must lie on our graph. Only the graph in option A passes through this point. Hence, the equation 𝑦=βˆ’4(2) is represented by graph A.

The graph of an exponential function 𝑓(π‘₯)=𝑏 will always have a horizontal asymptote at 𝑦=0. That is, in the case of exponential functions, the graph will approach the horizontal axis but never touch it. Similarly, if the function is shifted up or down by adding or subtracting a constant, 𝑐>0, so that 𝑓(π‘₯)=𝑏±𝑐, the asymptote will be at 𝑦=±𝑐.

In our next example, we see how an exponential function with a nonzero asymptote may be identified from a graph.

Example 4: Identifying Exponential Functions from Graphs

Which of the following could be the equation of the curve?

  1. 𝑦=βˆ’4(1+3)
  2. 𝑦=4(1+3)
  3. 𝑦=4(1βˆ’3)οŠ±ο—
  4. 𝑦=4(1+3)οŠ±ο—
  5. 𝑦=4(1βˆ’3).

Answer

We notice first that the given graph appears to go through the point (0,0). This means that, in the equation of the curve, when π‘₯=0, 𝑦=0. We can test whether this is the case or not for each of the given options. Substituting π‘₯=0 into the right-hand side of each option and evaluating gives us ABCDEβˆΆβˆ’4ο€Ή1+3=βˆ’4(1+1)β‰ 0∢4ο€Ή1+3=4(1+1)β‰ 0∢4ο€Ή1βˆ’3=4(1βˆ’1)=0βœ“βˆΆ4ο€Ή1+3=4(1+1)β‰ 0∢4ο€Ή1βˆ’3=4(1βˆ’1)=0βœ“οŠ¦οŠ¦οŠ±οŠ¦οŠ±οŠ¦οŠ¦Γ—Γ—Γ—

Options C and E are the only two equations whose graphs go through the point (0,0). Hence, we can discount options A, B, and D.

Our next step is to consider the general behavior of the graph. We see that, for positive π‘₯-values, the 𝑦-values very quickly become extremely large and negative. Let’s try substituting a positive π‘₯-value, for example π‘₯=1, into each of our remaining options, C and E: CEβˆΆπ‘¦=4ο€Ή1βˆ’3=83>0,βˆΆπ‘¦=4ο€Ή1βˆ’3=βˆ’8<0.

We see that equation C gives a positive value when π‘₯=1, which does not match with the given graph. However, for equation E, when π‘₯=1, 𝑦=βˆ’8, which does agree with the behavior of the given graph. Hence, equation E is the equation of the curve.

Examining the equation of the curve, 𝑦=4(1βˆ’3), distributing the parentheses and rearranging, we have 𝑦=βˆ’4β‹…3+4. This means that the exponential, 3, is vertically stretched by a factor of 4 (the 4 multiplying 3), reflected in the horizontal axis (because of the negative), and then shifted up by 4 units (because of the addition of 4). The horizontal asymptote is then at 𝑦=4.

In our final example, we consider the graph of a function representing exponential decay.

Example 5: Identifying Graphs of Exponential Equations

Which of the following graphs represents the function 𝑦=ο€Ό14οˆο—?

Answer

We know that our function 𝑦=ο€Ό14οˆο— is an exponential function, since it is of the form 𝑦=𝑓(π‘₯)=𝑏. We know also that, by definition, the graph of such a function passes through the point (0,1) and has a horizontal asymptote at 𝑦=0.

This means that we can eliminate graph A, since it satisfies neither of these conditions: it does not pass through the point (0,1), and it does not get closer and closer to, but never touch, the line 𝑦=0 In fact, graph A is that of a quadratic function.

Considering graphs B and C, although graph B appears to get closer and closer to the line 𝑦=0, neither graph passes through the point (0,1).

In fact, neither of these two graphs appears to exist for values of π‘₯<0, which is not the case for our exponential function. We can therefore eliminate graphs B and C.

This leaves us with graphs D and E, both of which pass through the point (0,1) and have an asymptote at 𝑦=0.

To determine which of these represents our function, 𝑦=ο€Ό14οˆο—, we must consider a further property of exponential functions of the form 𝑦=𝑓(π‘₯)=𝑏. That is,

  • if 𝑏>1,𝑓(π‘₯) models exponential growth;
  • if 0<𝑏<1,𝑓(π‘₯) models exponential decay.

In our case, 𝑏=14. Hence, our function models exponential decay. This means that as the independent variable, π‘₯, increases, the function decreases in value. We see that, in graph D, the opposite occurs: as π‘₯ increases, the function also increases. Hence, we can eliminate graph D.

Since the function represented in graph E decreases as π‘₯ increases, this represents exponential decay. Therefore, graph E represents the equation 𝑦=ο€Ό14οˆο—.

Let’s finish by reminding ourselves of some of the key points about the graphs of exponential functions.

Key Points

  • A function of the form 𝑓(π‘₯)=𝑏, where π‘βˆˆβ„>0 and 𝑏≠1, is an exponential function.
  • 𝑏>1 models exponential growth;
    0<𝑏<1 models exponential decay.
  • The graph of 𝑓(π‘₯)=𝑏 passes through the point (0,1) and has a horizontal asymptote at 𝑦=0.
  • 𝑓(π‘₯)=𝑐⋅𝑏 is an exponential function passing through the point (0,𝑐), that is, the 𝑦-intercept or 𝑦=𝑐.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.