As we discussed in the previous section, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. 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.
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 f ( x ) = b x f ( x ) = b x whose base is greater than one. We’ll use the function f ( x ) = 2 x . f ( x ) = 2 x . Observe how the output values in Table 1 change as the input increases by 1. 1.
x x | − 3 − 3 | − 2 − 2 | − 1 − 1 | 0 0 | 1 1 | 2 2 | 3 3 |
f ( x ) = 2 x f ( x ) = 2 x | 1 8 1 8 | 1 4 1 4 | 1 2 1 2 | 1 1 | 2 2 | 4 4 | 8 8 |
Each output value is the product of the previous output and the base, 2. 2. We call the base 2 2 the constant ratio. In fact, for any exponential function with the form f ( x ) = a b x , f ( x ) = a b x , b 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 . a .
Notice from the table that
Figure 1 shows the exponential growth function f ( x ) = 2 x . f ( x ) = 2 x .
Figure 1 Notice that the graph gets close to the x-axis, but never touches it.The domain of f ( x ) = 2 x f ( x ) = 2 x is all real numbers, the range is ( 0 , ∞ ) , ( 0 , ∞ ) , and the horizontal asymptote is y = 0. y = 0.
To get a sense of the behavior of exponential decay , we can create a table of values for a function of the form f ( x ) = b x f ( x ) = b x whose base is between zero and one. We’ll use the function g ( x ) = ( 1 2 ) x . g ( x ) = ( 1 2 ) x . Observe how the output values in Table 2 change as the input increases by 1. 1.
x x | -3 -3 | -2 -2 | -1 -1 | 0 0 | 1 1 | 2 2 | 3 3 |
g ( x ) = ( 1 2 ) x g ( x ) = ( 1 2 ) x | 8 8 | 4 4 | 2 2 | 1 1 | 1 2 1 2 | 1 4 1 4 | 1 8 1 8 |
Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio 1 2 . 1 2 .
Notice from the table that
Figure 2 shows the exponential decay function, g ( x ) = ( 1 2 ) x . g ( x ) = ( 1 2 ) x .
The domain of g ( x ) = ( 1 2 ) x g ( x ) = ( 1 2 ) x is all real numbers, the range is ( 0 , ∞ ) , ( 0 , ∞ ) , and the horizontal asymptote is y = 0. y = 0.
An exponential function with the form f ( x ) = b x , f ( x ) = b x , b > 0 , b > 0 , b ≠ 1 , b ≠ 1 , has these characteristics:
Figure 3 compares the graphs of exponential growth and decay functions.
Given an exponential function of the form f ( x ) = b x , f ( x ) = b x , graph the function.
Sketch a graph of f ( x ) = 0.25 x . f ( x ) = 0.25 x . State the domain, range, and asymptote.
Before graphing, identify the behavior and create a table of points for the graph.
x x | − 3 − 3 | − 2 − 2 | − 1 − 1 | 0 0 | 1 1 | 2 2 | 3 3 |
f ( x ) = 0.25 x f ( x ) = 0.25 x | 64 64 | 16 16 | 4 4 | 1 1 | 0.25 0.25 | 0.0625 0.0625 | 0.015625 0.015625 |
Draw a smooth curve connecting the points as in Figure 4.
The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 0 , ∞ ) ; ( 0 , ∞ ) ; the horizontal asymptote is y = 0. y = 0.
Sketch the graph of f ( x ) = 4 x . f ( x ) = 4 x . State the domain, range, and asymptote.
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 f ( x ) = b x f ( x ) = b x 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.
The first transformation occurs when we add a constant d d to the parent function f ( x ) = b x , f ( x ) = b x , giving us a vertical shift d d units in the same direction as the sign. For example, if we begin by graphing a parent function, f ( x ) = 2 x , f ( x ) = 2 x , we can then graph two vertical shifts alongside it, using d = 3 : d = 3 : the upward shift, g ( x ) = 2 x + 3 g ( x ) = 2 x + 3 and the downward shift, h ( x ) = 2 x − 3. h ( x ) = 2 x − 3. Both vertical shifts are shown in Figure 5.
Observe the results of shifting f ( x ) = 2 x f ( x ) = 2 x vertically:
The next transformation occurs when we add a constant c c to the input of the parent function f ( x ) = b x , f ( x ) = b x , giving us a horizontal shift c c units in the opposite direction of the sign. For example, if we begin by graphing the parent function f ( x ) = 2 x , f ( x ) = 2 x , we can then graph two horizontal shifts alongside it, using c = 3 : c = 3 : the shift left, g ( x ) = 2 x + 3 , g ( x ) = 2 x + 3 , and the shift right, h ( x ) = 2 x − 3 . h ( x ) = 2 x − 3 . Both horizontal shifts are shown in Figure 6.
Observe the results of shifting f ( x ) = 2 x f ( x ) = 2 x horizontally:
For any constants c c and d , d , the function f ( x ) = b x + c + d f ( x ) = b x + c + d shifts the parent function f ( x ) = b x f ( x ) = b x
Given an exponential function with the form f ( x ) = b x + c + d , f ( x ) = b x + c + d , graph the translation.
Graph f ( x ) = 2 x + 1 − 3. f ( x ) = 2 x + 1 − 3. State the domain, range, and asymptote.
We have an exponential equation of the form f ( x ) = b x + c + d , f ( x ) = b x + c + d , with b = 2 , b = 2 , c = 1 , c = 1 , and d = − 3. d = − 3.
Draw the horizontal asymptote y = d y = d , so draw y = −3. y = −3.
Identify the shift as ( − c , d ) , ( − c , d ) , so the shift is ( − 1 , −3 ) . ( − 1 , −3 ) .
Shift the graph of f ( x ) = b x f ( x ) = b x left 1 units and down 3 units.
The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( − 3 , ∞ ) ; ( − 3 , ∞ ) ; the horizontal asymptote is y = −3. y = −3.
Graph f ( x ) = 2 x − 1 + 3. f ( x ) = 2 x − 1 + 3. State domain, range, and asymptote.
Given an equation of the form f ( x ) = b x + c + d f ( x ) = b x + c + d for x , x , use a graphing calculator to approximate the solution.
Solve 42 = 1.2 ( 5 ) x + 2.8 42 = 1.2 ( 5 ) x + 2.8 graphically. Round to the nearest thousandth.
Press [Y=] and enter 1.2 ( 5 ) x + 2.8 1.2 ( 5 ) x + 2.8 next to Y1=. Then enter 42 next to Y2=. For a window, use the values –3 to 3 for x x and –5 to 55 for y . y . Press [GRAPH]. The graphs should intersect somewhere near x = 2. 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, x ≈ 2.166. x ≈ 2.166.
Solve 4 = 7.85 ( 1.15 ) x − 2.27 4 = 7.85 ( 1.15 ) x − 2.27 graphically. Round to the nearest thousandth.
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 f ( x ) = b x f ( x ) = b x by a constant | a | > 0. | a | > 0. For example, if we begin by graphing the parent function f ( x ) = 2 x , f ( x ) = 2 x , we can then graph the stretch, using a = 3 , a = 3 , to get g ( x ) = 3 ( 2 ) x g ( x ) = 3 ( 2 ) x as shown on the left in Figure 8, and the compression, using a = 1 3 , a = 1 3 , to get h ( x ) = 1 3 ( 2 ) x h ( x ) = 1 3 ( 2 ) x as shown on the right in Figure 8.
Figure 8 (a) g ( x ) = 3 ( 2 ) x g ( x ) = 3 ( 2 ) x stretches the graph of f ( x ) = 2 x f ( x ) = 2 x vertically by a factor of 3. 3. (b) h ( x ) = 1 3 ( 2 ) x h ( x ) = 1 3 ( 2 ) x compresses the graph of f ( x ) = 2 x f ( x ) = 2 x vertically by a factor of 1 3 . 1 3 .
For any factor a > 0 , a > 0 , the function f ( x ) = a ( b ) x f ( x ) = a ( b ) x
Sketch a graph of f ( x ) = 4 ( 1 2 ) x . f ( x ) = 4 ( 1 2 ) x . State the domain, range, and asymptote.
Before graphing, identify the behavior and key points on the graph.
x x | − 3 − 3 | − 2 − 2 | − 1 − 1 | 0 0 | 1 1 | 2 2 | 3 3 |
f ( x ) = 4 ( 1 2 ) x f ( x ) = 4 ( 1 2 ) x | 32 32 | 16 16 | 8 8 | 4 4 | 2 2 | 1 1 | 0.5 0.5 |
Draw a smooth curve connecting the points, as shown in Figure 9.
The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 0 , ∞ ) ; ( 0 , ∞ ) ; the horizontal asymptote is y = 0. y = 0.
Sketch the graph of f ( x ) = 1 2 ( 4 ) x . f ( x ) = 1 2 ( 4 ) x . State the domain, range, and asymptote.
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 f ( x ) = b x f ( x ) = b x by −1 , −1 , we get a reflection about the x-axis. When we multiply the input by −1 , −1 , we get a reflection about the y-axis. For example, if we begin by graphing the parent function f ( x ) = 2 x , f ( x ) = 2 x , we can then graph the two reflections alongside it. The reflection about the x-axis, g ( x ) = −2 x , g ( x ) = −2 x , is shown on the left side of Figure 10, and the reflection about the y-axis h ( x ) = 2 − x , h ( x ) = 2 − x , is shown on the right side of Figure 10.
Figure 10 (a) g ( x ) = − 2 x g ( x ) = − 2 x reflects the graph of f ( x ) = 2 x f ( x ) = 2 x about the x-axis. (b) g ( x ) = 2 − x g ( x ) = 2 − x reflects the graph of f ( x ) = 2 x f ( x ) = 2 x about the y-axis.
The function f ( x ) = − b x f ( x ) = − b x
The function f ( x ) = b − x f ( x ) = b − x
Find and graph the equation for a function, g ( x ) , g ( x ) , that reflects f ( x ) = ( 1 4 ) x f ( x ) = ( 1 4 ) x about the x-axis. State its domain, range, and asymptote.
Since we want to reflect the parent function f ( x ) = ( 1 4 ) x f ( x ) = ( 1 4 ) x about the x-axis, we multiply f ( x ) f ( x ) by − 1 − 1 to get, g ( x ) = − ( 1 4 ) x . g ( x ) = − ( 1 4 ) x . Next we create a table of points as in Table 5.
x x | − 3 − 3 | − 2 − 2 | − 1 − 1 | 0 0 | 1 1 | 2 2 | 3 3 |
g ( x ) = − ( 1 4 ) x g ( x ) = − ( 1 4 ) x | − 64 − 64 | − 16 − 16 | − 4 − 4 | − 1 − 1 | − 0.25 − 0.25 | − 0.0625 − 0.0625 | − 0.0156 − 0.0156 |
Plot the y-intercept, ( 0 , −1 ) , ( 0 , −1 ) , along with two other points. We can use ( −1 , −4 ) ( −1 , −4 ) and ( 1 , −0.25 ) . ( 1 , −0.25 ) .
Draw a smooth curve connecting the points:
The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( − ∞ , 0 ) ; ( − ∞ , 0 ) ; the horizontal asymptote is y = 0. y = 0.
Find and graph the equation for a function, g ( x ) , g ( x ) , that reflects f ( x ) = 1.25 x f ( x ) = 1.25 x about the y-axis. State its domain, range, and asymptote.
Now that we have worked with each type of translation for the exponential function, we can summarize them in Table 6 to arrive at the general equation for translating exponential functions.
A translation of an exponential function has the form
f ( x ) = a b x + c + d f ( x ) = a b x + c + dWhere the parent function, y = b x , y = b x , b > 1 , b > 1 , is
Note the order of the shifts, transformations, and reflections follow the order of operations.
Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range.
We want to find an equation of the general form f ( x ) = a b x + c + d . f ( x ) = a b x + c + d . We use the description provided to find a , a , b , b , c , c , and d . d .
Substituting in the general form we get,
f ( x ) = a b x + c + d = 2 e − x + 0 + 4 = 2 e − x + 4 f ( x ) = a b x + c + d = 2 e − x + 0 + 4 = 2 e − x + 4
The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 4 , ∞ ) ; ( 4 , ∞ ) ; the horizontal asymptote is y = 4. y = 4.
Write the equation for function described below. Give the horizontal asymptote, the domain, and the range.
Access this online resource for additional instruction and practice with graphing exponential functions.
What role does the horizontal asymptote of an exponential function play in telling us about the end behavior of the graph?
What is the advantage of knowing how to recognize transformations of the graph of a parent function algebraically?
The graph of f ( x ) = 3 x f ( x ) = 3 x is reflected about the y-axis and stretched vertically by a factor of 4. 4. What is the equation of the new function, g ( x ) ? g ( x ) ? State its y-intercept, domain, and range.
The graph of f ( x ) = ( 1 2 ) − x f ( x ) = ( 1 2 ) − x is reflected about the y-axis and compressed vertically by a factor of 1 5 . 1 5 . What is the equation of the new function, g ( x ) ? g ( x ) ? State its y-intercept, domain, and range.
The graph of f ( x ) = 10 x f ( x ) = 10 x is reflected about the x-axis and shifted upward 7 7 units. What is the equation of the new function, g ( x ) ? g ( x ) ? State its y-intercept, domain, and range.
The graph of f ( x ) = ( 1.68 ) x f ( x ) = ( 1.68 ) x is shifted right 3 3 units, stretched vertically by a factor of 2 , 2 , reflected about the x-axis, and then shifted downward 3 3 units. What is the equation of the new function, g ( x ) ? g ( x ) ? State its y-intercept (to the nearest thousandth), domain, and range.
The graph of f x = - 1 2 ( 1 4 ) x - 2 + 4 f x = - 1 2 ( 1 4 ) x - 2 + 4 is shifted downward 4 4 units, and then shifted left 2 2 units, stretched vertically by a factor of 4 , 4 , and reflected about the x-axis. What is the equation of the new function, g ( x ) ? g ( x ) ? State its y-intercept, domain, and range.
For the following exercises, graph the function and its reflection about the y-axis on the same axes, and give the y-intercept.
f ( x ) = 3 ( 1 2 ) x f ( x ) = 3 ( 1 2 ) x