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Graphing Exponential Functions

We now define exponential functions. We assume that $a^{x}$ $a^{x}$ has meaning for any real number x and any positive real number a and that the laws of exponents still hold, though we will not prove them here.

We require the base to be positive in order to avoid the imaginary numbers that would occur by taking even roots of negative numbers—an example is ${(- 1)}^{1 / 2}$ ${(- 1)}^{1 / 2}$ , the square root of −1, which is not a real number. The restriction $a \neq 1$ $a \neq 1$ is made to exclude the constant function $f (x) = 1^{x} = 1$ $f (x) = 1^{x} = 1$ , which does not have an inverse that is a function because it is not one-to-one.

The following are examples of exponential functions:

f (x) = 2^{x}, f (x) = {(\frac{1}{2})}^{x}, f (x) = {(3.57)}^{x} .

$f (x) = 2^{x}, f (x) = {(\frac{1}{2})}^{x}, f (x) = {(3.57)}^{x} .$

Note that, in contrast to functions like $f (x) = x^{5}$ $f (x) = x^{5}$ and $f (x) = x^{1 / 2}$ $f (x) = x^{1 / 2}$ in which the variable is the base of an exponential expression, the variable in an exponential function is in the exponent.

Let’s now consider graphs of exponential functions.

Example 1

Graph the exponential function $y = f (x) = 2^{x}$ $y = f (x) = 2^{x}$ .

Solution

We compute some function values and list the results in a table.

`x`	`y`	(`x`, `y`)
`x`	$y = f (x) = 2^{x}$ $y = f (x) = 2^{x}$	(`x`, `y`)
0	1	(0, 1)
1	2	(1, 2)
2	4	(2, 4)
3	8	(3, 8)
−1	$\frac{1}{2}$ $\frac{1}{2}$	$(- 1, \frac{1}{2})$ $(- 1, \frac{1}{2})$
−2	$\frac{1}{4}$ $\frac{1}{4}$	$(- 2, \frac{1}{4})$ $(- 2, \frac{1}{4})$
−3	$\frac{1}{8}$ $\frac{1}{8}$	$(- 3, \frac{1}{8})$ $(- 3, \frac{1}{8})$

\begin{array}{l} \begin{array}{l} f (0) = 2^{0} = 1; \\ f (1) = 2^{1} = 2; \\ f (2) = 2^{2} = 4; \\ f (3) = 2^{3} = 8; \end{array} & \begin{array}{l} f (- 1) = 2^{- 1} = \frac{1}{2^{1}} = \frac{1}{2}; \\ f (- 2) = 2^{- 2} = \frac{1}{2^{2}} = \frac{1}{4}; \\ f (- 3) = 2^{- 3} = \frac{1}{2^{3}} = \frac{1}{8} . \end{array} \end{array}

$\begin{array}{l} \begin{array}{l} f (0) = 2^{0} = 1; \\ f (1) = 2^{1} = 2; \\ f (2) = 2^{2} = 4; \\ f (3) = 2^{3} = 8; \end{array} & \begin{array}{l} f (- 1) = 2^{- 1} = \frac{1}{2^{1}} = \frac{1}{2}; \\ f (- 2) = 2^{- 2} = \frac{1}{2^{2}} = \frac{1}{4}; \\ f (- 3) = 2^{- 3} = \frac{1}{2^{3}} = \frac{1}{8} . \end{array} \end{array}$

Next, we plot these points and connect them with a smooth curve. Be sure to plot enough points to determine how steeply the curve rises.

Note that as x increases, the function values increase without bound. As x decreases, the function values decrease, getting close to 0. That is, as $x \to - \infty, y \to 0$ $x \to - \infty, y \to 0$ . Thus the x-axis, or the line $y = 0$ $y = 0$ , is a horizontal asymptote. As the x-inputs decrease, the curve gets closer and closer to this line, but does not cross it.

Now Try Exercise 11.

Example 2

Graph the exponential function $y = f (x) = {(\frac{1}{2})}^{x}$ $y = f (x) = {(\frac{1}{2})}^{x}$ .

Solution

Before we plot points and draw the curve, note that

y = f (x) = {(\frac{1}{2})}^{x} = {(2^{- 1})}^{x} = 2^{- x} .

$y = f (x) = {(\frac{1}{2})}^{x} = {(2^{- 1})}^{x} = 2^{- x} .$

Points of $g (x) = 2^{x}$ $g (x) = 2^{x}$	Points of $f (x) = {(\frac{1}{2})}^{x} = 2^{- x}$ $f (x) = {(\frac{1}{2})}^{x} = 2^{- x}$
(0, 1)	(0, 1)
(1, 2)	(−1, 2)
(2, 4)	(−2, 4)
(3, 8)	(−3, 8)
$(- 1, \frac{1}{2})$ $(- 1, \frac{1}{2})$	$(1, \frac{1}{2})$ $(1, \frac{1}{2})$
$(- 2, \frac{1}{4})$ $(- 2, \frac{1}{4})$	$(2, \frac{1}{4})$ $(2, \frac{1}{4})$
$(- 3, \frac{1}{8})$ $(- 3, \frac{1}{8})$	$(3, \frac{1}{8})$ $(3, \frac{1}{8})$

This tells us that this graph is a reflection of the graph of $y = 2^{x}$ $y = 2^{x}$ across the y-axis. For example, if (3, 8) is a point of the graph of $g (x) = 2^{x}$ $g (x) = 2^{x}$ , then (−3, 8) is a point of the graph of $f (x) = 2^{- x}$ $f (x) = 2^{- x}$ . Selected points are listed in the table at left.

Next, we plot these points and connect them with a smooth curve.

Note that as x increases, the function values decrease, getting close to 0. The x-axis, $y = 0$ $y = 0$ , is the horizontal asymptote. As x decreases, the function values increase without bound.

Now Try Exercise 15.

To graph other types of exponential functions, keep in mind the ideas of translation, stretching, and reflection. All these concepts allow us to visualize the graph before drawing it.

Example 3

Graph each of the following. Before doing so, describe how each graph can be obtained from the graph of $f (x) = 2^{x}$ $f (x) = 2^{x}$ .

$f (x) = 2^{x - 2}$ $f (x) = 2^{x - 2}$
$f (x) = 2^{x} - 4$ $f (x) = 2^{x} - 4$
$f (x) = 5 - {0.5}^{x}$ $f (x) = 5 - {0.5}^{x}$

Solution

The graph of $f (x) = 2^{x - 2}$ $f (x) = 2^{x - 2}$ is the graph of $y = 2^{x}$ $y = 2^{x}$ shifted right 2 units.

`x`	$f (x)$ $f (x)$
−1	$\frac{1}{8}$ $\frac{1}{8}$
0	$\frac{1}{4}$ $\frac{1}{4}$
1	$\frac{1}{2}$ $\frac{1}{2}$
2	1
3	2
4	4
5	8

The graph of $f (x) = 2^{x} - 4$ $f (x) = 2^{x} - 4$ is the graph of $y = 2^{x}$ $y = 2^{x}$ shifted down 4 units.

`x`	$f (x)$ $f (x)$
−2	$- 3 \frac{3}{4}$ $- 3 \frac{3}{4}$
−1	$- 3 \frac{1}{2}$ $- 3 \frac{1}{2}$
0	−3
1	−2
2	0
3	4

The graph of $f (x) = 5 - {0.5}^{x} = 5 - {(\frac{1}{2})}^{x} = 5 - 2^{- x}$ $f (x) = 5 - {0.5}^{x} = 5 - {(\frac{1}{2})}^{x} = 5 - 2^{- x}$ is a reflection of the graph of $y = 2^{x}$ $y = 2^{x}$ across the y-axis, followed by a reflection across the x-axis and then a shift up 5 units.

`x`	$f (x)$ $f (x)$
−3	−3
−2	1
−1	3
0	4
1	$4 \frac{1}{2}$ $4 \frac{1}{2}$
2	$4 \frac{3}{4}$ $4 \frac{3}{4}$

Now Try Exercises 27 and 33.

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Table of Contents for Graphing Exponential Functions

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Table of Contents for
Graphing Exponential Functions