4.2: Graphs of Exponential Functions
Graphing 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 \(f(x)=b^x\) whose base is greater than one. We’ll use the function \(f(x)=2^x\). Observe how the output values in Table \(\PageIndex{1}\) change as the input increases by \(1\).
| \(x\) | \(−3\) | \(−2\) | \(−1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|---|---|
| \(f(x)=2^x\) | \(\dfrac{1}{8}\) | \(\dfrac{1}{4}\) | \(\dfrac{1}{2}\) | \(1\) | \(2\) | \(4\) | \(8\) |
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 \(f(x)=ab^x\), \(b\) 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 \(\PageIndex{1}\) shows the exponential growth function \(f(x)=2^x\).
The domain of \(f(x)=2^x\) is all real numbers, the range is \((0,\infty)\), and the horizontal asymptote is \(y=0\).
To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form \(f(x)=b^x\) whose base is between zero and one. We’ll use the function \(g(x)={\left(\dfrac{1}{2}\right)}^x\). Observe how the output values in Table \(\PageIndex{2}\) change as the input increases by \(1\).
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|---|---|
| \(g(x)={\left(\dfrac{1}{2}\right)}^x\) | \(8\) | \(4\) | \(2\) | \(1\) | \(\dfrac{1}{2}\) | \(\dfrac{1}{4}\) | \(\dfrac{1}{8}\) |
Again, because the input is increasing by \(1\), each output value is the product of the previous output and the base, or constant ratio \(\dfrac{1}{2}\).
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.
Figure \(\PageIndex{2}\) shows the exponential decay function, \(g(x)={\left(\dfrac{1}{2}\right)}^x\).
The domain of \(g(x)={(\dfrac{1}{2})}^x\) is all real numbers, the range is \((0,\infty)\),and the horizontal asymptote is \(y=0\).
An exponential function with the form \(f(x)=b^x\), \(b>0\), \(b≠1\),has these characteristics:
- one-to-one function
- horizontal asymptote: \(y=0\)
- domain: \((–\infty, \infty)\)
- range: \((0,\infty)\)
- x- intercept: none
- y- intercept: \((0,1)\)
- increasing if \(b>1\)
- decreasing if \(b<1\)
Figure \(\PageIndex{3}\) compares the graphs of exponential growth and decay functions.
- Create a table of points.
- Plot at least \(3\) point from the table, including the y -intercept \((0,1)\).
- Draw a smooth curve through the points.
- State the domain, \((−\infty,\infty)\),the range, \((0,\infty)\),and the horizontal asymptote, \(y=0\).
Sketch a graph of \(f(x)=0.25^x\). State the domain, range, and asymptote.
Solution
Before graphing, identify the behavior and create a table of points for the graph.
| \(x\) | \(−3\) | \(−2\) | \(−1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|---|---|
| \(f(x)={0.25}^x\) | \(64\) | \(16\) | \(4\) | \(1\) | \(0.25\) | \(0.0625\0 | \(0.015625\) |
- 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 as in Table \(\PageIndex{3}\).
- Plot the y -intercept, \((0,1)\),along with two other points. We can use \((−1,4)\) and \((1,0.25)\).
Draw a smooth curve connecting the points as in Figure \(\PageIndex{4}\).
The domain is \((−\infty,\infty)\); the range is \((0,\infty)\); the horizontal asymptote is \(y=0\).
Sketch the graph of \(f(x)=4^x\). State the domain, range, and asymptote.
- Answer
-
The domain is \((−\infty,\infty)\); the range is \((0,\infty)\); the horizontal asymptote is \(y=0\).
Graphing Transformations of Exponential Functions
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 \(f(x)=b^x\) 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 \(f(x)=b^x\), giving us a vertical shiftd d units in the same direction as the sign. For example, if we begin by graphing a parent function, \(f(x)=2^x\), we can then graph two vertical shifts alongside it, using \(d=3\): the upward shift, \(g(x)=2^x+3\) and the downward shift, \(h(x)=2^x−3\). Both vertical shifts are shown in Figure \(\PageIndex{5}\).
Observe the results of shifting \(f(x)=2^x\) vertically:
- The domain, \((−\infty,\infty)\) remains unchanged.
-
When the function is shifted up \(3\) units to \(g(x)=2^x+3\):
- The y- intercept shifts up \(3\) units to \((0,4)\).
- The asymptote shifts up \(3\) units to \(y=3\).
- The range becomes \((3,\infty)\).
-
When the function is shifted down \(3\) units to \(h(x)=2^x−3\):
- The y- intercept shifts down \(3\) units to \((0,−2)\).
- The asymptote also shifts down \(3\) units to \(y=−3\).
- The range becomes \((−3,\infty)\).
Graphing a Horizontal Shift
The next transformation occurs when we add a constant \(c\) to the input of the parent function \(f(x)=b^x\), giving us a horizontal shift \(c\) units in the opposite direction of the sign. For example, if we begin by graphing the parent function \(f(x)=2^x\), we can then graph two horizontal shifts alongside it, using \(c=3\): the shift left, \(g(x)=2^{x+3}\), and the shift right, \(h(x)=2^{x−3}\). \(h(x)=2^{x−3}\). Both horizontal shifts are shown in Figure \(\PageIndex{6}\).
Observe the results of shifting \(f(x)=2^x\) horizontally:
- The domain, \((−\infty,\infty)\),remains unchanged.
- The asymptote, \(y=0\),remains unchanged.
-
The
y-
intercept shifts such that:
- When the function is shifted left \(3\) units to \(g(x)=2^{x+3}\),the y -intercept becomes \((0,8)\). This is because \(2^{x+3}=(8)2^x\),so the initial value of the function is \(8\).
- When the function is shifted right \(3\) units to \(h(x)=2^{x−3}\),the y -intercept becomes \((0,\dfrac{1}{8})\). Again, see that \(2^{x−3}=(\dfrac{1}{8})2^x\),so the initial value of the function is \(\dfrac{1}{8}\).
For any constants \(c\) and \(d\),the function \(f(x)=b^{x+c}+d\) shifts the parent function \(f(x)=b^x\)
- 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 \((0,b^c+d)\).
- The horizontal asymptote becomes \(y=d\).
- The range becomes \((d,\infty)\).
- The domain, \((−\infty,\infty)\),remains unchanged.
- Draw the horizontal asymptote \(y=d\).
- Identify the shift as \((−c,d)\). Shift the graph of \(f(x)=b^x\) left \(c\) units if \(c\) is positive, and right \(c\) units if \(c\) is negative.
- Shift the graph of \(f(x)=b^x\) up \(d\) units if \(d\) is positive, and down \(d\) units if \(d\) is negative.
- State the domain, \((−\infty,\infty)\),the range, \((d,\infty)\),and the horizontal asymptote \(y=d\).
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 \(f(x)=b^x\) by a constant \(|a|>0\). For example, if we begin by graphing the parent function \(f(x)=2^x\),we can then graph the stretch, using \(a=3\),to get \(g(x)=3{(2)}^x\) as shown on the left in Figure \(\PageIndex{8}\), and the compression, using \(a=\dfrac{1}{3}\),to get \(h(x)=\dfrac{1}{3}{(2)}^x\) as shown on the right in Figure \(\PageIndex{8}\).
For any factor \(a>0\),the function \(f(x)=a{(b)}^x\)
- is stretched vertically by a factor of \(a\) if \(|a|>1\).
- is compressed vertically by a factor of \(a\) if \(|a|<1\).
- has a y -intercept of \((0,a)\).
- has a horizontal asymptote at \(y=0\), a range of \((0,\infty)\), and a domain of \((−\infty,\infty)\),which are unchanged from the parent function.
Sketch a graph of \(f(x)=4{(\dfrac{1}{2})}^x\). State the domain, range, and asymptote.
Solution
Before graphing, identify the behavior and key points on the graph.
- Since \(b=\dfrac{1}{2}\) 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 \(f(x)={(\dfrac{1}{2})}^x\) will be stretched by a factor of \(4\).
-
Create a table of points as shown in Table \(\PageIndex{4}\).
Table \(\PageIndex{4}\) \(x\) \(−3\) \(−2\) \(−1\) \(0\) \(1\) \(2\) \(3\) \(f(x)=4{(\dfrac{1}{2})}^x\) \(32\) \(16\) \(8\) \(4\) \(2\) \(1\) \(0.5\) - Plot the y- intercept, \((0,4)\),along with two other points. We can use \((−1,8)\) and \((1,2)\).
Draw a smooth curve connecting the points, as shown in Figure \(\PageIndex{9}\).
The domain is \((−\infty,\infty)\); the range is \((0,\infty)\); the horizontal asymptote is \(y=0\).
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 \(f(x)=b^x\) 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 \(f(x)=2^x\), we can then graph the two reflections alongside it. The reflection about the x -axis, \(g(x)=−2^x\),is shown on the left side of Figure \(\PageIndex{10}\), and the reflection about the y-axis \(h(x)=2^{−x}\), is shown on the right side of Figure \(\PageIndex{10}\).
The function \(f(x)=−b^x\)
- reflects the parent function \(f(x)=b^x\) about the x -axis.
- has a y -intercept of \((0,−1)\).
- has a range of \((−\infty,0)\)
- has a horizontal asymptote at \(y=0\) and domain of \((−\infty,\infty)\),which are unchanged from the parent function.
The function \(f(x)=b^{−x}\)
- reflects the parent function \(f(x)=b^x\) about the y -axis.
- has a y -intercept of \((0,1)\), a horizontal asymptote at \(y=0\), a range of \((0,\infty)\), and a domain of \((−\infty,\infty)\), which are unchanged from the parent function.
Summarizing Translations of the Exponential Function
Now that we have worked with each type of translation for the exponential function, we can summarize them in Table \(\PageIndex{6}\) to arrive at the general equation for translating exponential functions.
A translation of an exponential function has the form
\(f(x)=ab^{x+c}+d\)
Where the parent function, \(y=b^x\), \(b>1\),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.
Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range.
\(f(x)=e^x\) is vertically stretched by a factor of \(2\) , reflected across the y -axis, and then shifted up \(4\) units.
Solution
We want to find an equation of the general form f(x)=abx+c+d. f(x)=abx+c+d. We use the description provided to finda, a, b,b, c,c, and d. d.
- We are given the parent function \(f(x)=e^x\), 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: \(e^{−x}\).
- The graph is shifted vertically 4 units, so \(d=4\).
Substituting in the general form we get,
\(f(x)=ab^{x+c}+d\)
\(=2e^{−x+0}+4\)
\(=2e^{−x}+4\)
The domain is \((−\infty,\infty)\); the range is \((4,\infty)\); the horizontal asymptote is \(y=4\).