Intermediate Algebra

11.1 Exponential Functions

Recall the graphs of two functions which are inverses of each other mirror each other about the line y = x.

Inverse functions also have the following property:

Consider two functions f and g which are inverses of each other. Insert a number into one of the two functions and call this number A.

Exactly one number is returned; call this number B.

Now take the number B and insert it into the other function, the output is the number A.

What one function does, the inverse function undoes and returns the original value.

Let's look at an example of this property. Consider f(x) and g(x) which are inverse functions.

Input the number 3 into the first function.

The number 9 is returned.

Now insert the number 9 into the second function.

The original value of 3 is returned.

The number 3 into f(x) produced 9, and the number 9 into g(x) returned 3.
Since f(x) and g(x) are inverses, their graphs are symmetric about the line y = x.

This is how inverse functions operate, and Exponential and log functions are inverse functions.

Let's start with exponential functions.

What is an exponential function?

Exponential Function

Let b be a real number other than 0 or 1.
The exponential function with base b is y = b^x.

This means that any number, other than 0 or 1, to the power of x is an exponential function.

The reason for the exception of 1 and 0 is as follows:
y = 1^x is the same as y = 1 which is a horizontal line at 1.
y = 0^x is the same as y = 0 which is a horizontal line at 0.

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Recall the reflection of a point about the x and y-axis.

The point (2,3) reflected about the x-axis is (2, -3).

The point (2,3) reflected about the y-axis is (-2,3).

The point (2,3) reflected about the x and y-axis is (-2,-3).

Property

The point (x,y) reflected about the x-axis is the point (x, -y).
The point (x,y) reflected about the y-axis is the point (-x, y).
The point (x,y) reflected about the x and y-axis is the point (-x, -y).

Property

A function y = f(x) has all points reflected across the x-axis when y = - f(x).

In other words, when there is a negative in FRONT of a function, all points mirror about the x-axis.
All points (x,y) change to (x,-y).

Property

A function y = f(x) has all points reflected across the y-axis when y = f(-x).

In other words, when there is a negative in front of the "x inside a function", all points mirror about the y-axis.
All points (x,y) change to (-x,y).

Example 1:

Graph y = 2^x

Solution:

First, we will graph the function by plotting points. After we have plotted points for a couple of exponential functions, a short cut will be demonstrated. It is important to grasp the long way of graphing functions, which is plotting points, so that the short cut can be understood.

Recall the following:

If y = 2^x, and x = -1, then .

If y = 2^x, and x = -2, then .

y = 2^x


This shape is the general shape of all exponential functions!

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Example 2:

Graph y = 3^x

Solution:

y = 3^x


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Both of the graphs in Example 1 and Example 2 have the same general shape. Notice the left hand side of the graph gets closer and closer to the x-axis, but it never quite touches. This is called an asymptote. An asymptote is a barrier that a graph cannot touch or pass. The graph gets very very close to the asymptote, but it does not touch it!

All exponential functions asymptote toward some line.
The functions in Example 1 and Example 2 asymptote toward the line y = 0.

To use the shortcut method, we must first make a couple of observations.

First, when we insert x = 0 into y = 2^x and y = 3^x, the value of y = 1 is returned for both functions.

This is not unique to these two functions. Any base, other than zero, raised to the power of zero is always equal to one.
If b is a real number and b > 1, then b⁰ = 1.

This means that all exponential functions of the form y = b^x pass through the point (0,1).

Second, observe when x = 1 is inserted into the function, the base is returned.
If y = 2^x, and x = 1, then y = 2. The point (1,2) is on y = 2^x.
If y = 3^x, and x = 1, then y = 3. The point (1,3) is on y = 3^x.
If y = 4^x, and x = 1, then y = 4. The point (1,4) is on y = 4^x.

In general all exponential functions pass through the point (1, base) or (1,b).

Property

All functions of the form y = b^x pass through the points (0,1) and (1,b).

With this information let's graph y = 4^x, using the shortcut method.
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Example 3:

Graph y = 4^x

Solution:

The points (0,1) and (1,4) are points on the graph.    The rest of the graph is drawn in by knowing the shape of the exponential function.

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You can now graph the functions y = 5^x, y = 6^x, and y = 7^x without plotting points.

The model points for y = 5^x, are (0,1) and (1,5).
The model points for y = 6^x, are (0,1) and (1,6).
The model points for y = 7^x, are (0,1) and (1,7).

Parham’s Model Point Method
The Exponential Function

Let a,b,c be positive real numbers.
The exponential function with base b is: y = b^{x - a} + c.
The model function is y = b^x, and the model points are (0,1),(1, b).

The exponential function reflected about the x axis is: y = -b^{x - a} + c, and the model points are (0, -1),(1, -b).
The exponential function reflected about the y axis is: y = b^{-x + a} + c, and the model points are (0,1),(-1, b).
The exponential function reflected about the origin is: y = -b^{-x + a} + c, and the model points are (0,-1),(-1, -b).

The equation of the horizontal asymptote is y = c.
The shift is (a, c).

The numbers a and c describes the shift of each model point on the graph away from the origin in the x and y directions.

The shift in the x direction is found by setting the exponent on the base equal to zero.

If y = b^{x - a} + c, the shift in the x direction is found by setting x - a = 0.

x - a = 0

Solve for x.

x = a

The shift in the x direction is a units.

Thus, every point on the graph is shifted to the left a units if a is positive, and to the right a units if a is negative.

The shift in the y direction is found by determining the value of c.

Every point on the graph is shifted up c units if c is positive, and down c units if c is negative.

Therefore the shift is (a,c).

Notice the following functions and their asymptotes.

y = 2^x
The asymptote is the line y = 0.

y = 3^{x + 2} + 1
The asymptote is the line y = 1.

y = 2^{x - 1} - 3
The asymptote is the line y = - 3.

In general, the asymptote is the line y = c for y = b^{x + a} + c.

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Example 4:

Graph y = 2^{x - 1} - 3

Solution:
The model function is y = 2^x.
The model points are (0,1) and (1,2).

The graph of y = 2^{x - 1} - 3 is shaped just like the graph of y = 2^x, except it is shifted away from the model points.

We need the shift, the new model points and the equation of the asymptote for
y = 2^{x - 1} - 3.

The Shift:

To find the shift in the x direction, set the exponent x - 1 equal to zero and solve.

x - 1 = 0.

x = 1

Thus the shift in the x direction is 1 unit.

The value of c is -3, so the shift in the y direction is -3 units.

Thus the shift is (1, -3).

The New Points:

The model points (0,1) and (1,2) are each moved one unit to the right and three units down.
The two new model points, which are the model points shifted away from the origin, for the graph of

y = 2^{x - 1} - 3 are obtained by adding the shift to the model points.

Thus the new model points are (1, -2)(2, -1).

The Equation of the Asymptote:

The value of c is -3 in y = b^{x + a} + c, so the equation of the horizontal asymptote is y = - 3.

Now that we know the location of the new model points, the equation of the asymptote and the general shape of the exponential function, the graph can be drawn for
y = 2^{x - 1} - 3.

Plot the points (1,-2) (2,-1), and draw in the horizontal asymptote y = - 3.

Connect the two points and then head toward the horizontal asymptote getting closer and closer, but don't quite touch the line. The other side of the graph climbs exponentially so it rises quickly to the right.

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Example 5:

Graph the exponential function y = 3^{x + 2} + 1.

Solution:

The model function is y = 3^x, and the model points are (0,1) and (1,3).

The shift in the x direction is obtained by setting the exponent x + 2 = 0.

x + 2 = 0

x = - 2

Thus the shift in the x direction is - 2 units.

The shift in the y direction is 1 unit.

The shift is (-2, 1).

To obtain the new model points, add the shift to the model points.

The new model points are (-2, 2)(-1, 4).

The asymptote is the line y = 1.

Plot the points (-2,2) (-1,4), and draw in the horizontal asymptote.

Connect the two points and then head toward the horizontal asymptote getting closer and closer, but don't quite touch the line. The other side of the graph climbs exponentially so it rises quickly to the right.

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Example 6:

Graph y = 4^{x - 3} + 2

Solution:
The model function is y = 4^x, and the model points are (0, 1)(1, 4).

The shift is (3,2), and the asymptote is the line y = 2.

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The above examples are exponential functions with a positive base. Let's see what happens to an exponential function with a negative base. Again we will first plot points, make some observations as to the affect of the negative on the base, and then move to the shortcut method.

Example 7:

Graph y = - 2^x



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The negative base of y = -2^x, reflects every point of the graph of y = 2^x, across the x-axis.

The graph of y = - f(x) contains all the points of y = f(x), reflected about the x-axis.
The graph of y = -2^x contains all the points of y = 2^x, reflected about the x-axis.

Recall the point (x,y) reflected about the x-axis is the point (x, -y).
The graph y = 2^x passes through the points (0,1) and (1,2).

The graph y = - 2^x passes through the points (0,-1) and (1,- 2).

In general, the exponential function with negative base of the form y = - b^x, passes through the points (0,-1) and (1, -b).

A graph of y = 3^x has the model points of (0,1) and (1, 3).

A graph of y = - 3^x has the model points of (0,-1) and (1, -3).
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Example 8:

Graph y = - 2^{x - 3} + 4

Solution:

The base is negative so the model function is y = - 2^x, and thus the points (0,1) and (1,2) are reflected about the x-axis and become (0,-1) and (1, -2).

The shift is (3,4).
To find the new model points, add the shift to the model points.

The asymptote is the horizontal line y = 4.

Since the base is negative, the graph has the shape of the exponential function reflected across the x-axis.

Plot the points (3,3) (4,2), and draw in the horizontal asymptote.

Connect the two points and then head toward the horizontal asymptote getting closer and closer, but don't quite touch the line. The other side of the graph decreases quickly to the right.

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What about when the base is a fraction or there is a negative in the exponent of the x?

Example 9:

Graph

Solution:

Notice the function of is the same as y = 2^{- x}, and is in the form y = f(-x).

Recall: A function y = f(-x) contains all points reflected across the y-axis for the graph of y = f(x).
This means the graph of y = 2^x reflected across the y-axis produces the graph of y = 2^{- x}.

Plot points to graph the function.


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Example 10:

Graph y = 3^{- x + 1} - 1

Solution:

Since the base is 3, the original model points are (0,1),(1,3).
Reflect these points about the y-axis, which produce the model points (0,1),(-1,3).

The shift in the x direction is obtained by setting - x + 1 = 0.

The shift is (1,-1).

The asymptote is y = - 1.

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Example 11:

Graph y = 2^{- x - 3} - 4

Solution:

Since the base is 2, the original model points are (0,1),(1,2).

These points are reflected about the y-axis which produce the model points (0,1),(-1,2).

The shift in the x direction is obtained by setting - x - 3 = 0.

- x - 3 = 0
- x = 3
x = - 3

The shift is (-3,-4).

The asymptote is y = - 4.

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Example 12:

Graph y = - 3^{- x + 1} + 2

Solution:

Since the base is 3, the original model points are (0,1),(1,3).    These points are reflected about the x and the y-axis which produce the model points (0,-1),(-1,-3).

The shift in the x direction is obtained by setting - x + 1 = 0.

- x + 1 = 0
- x = - 1
x = 1

The shift is (1, 2).

The asymptote is y = 2.

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In all the above examples, the bases of the exponential functions have been rational numbers. The base can be an irrational number.

Recall p is an irrational number which is approximately 3.14. There are many more irrational numbers. A commonly used irrational numbers is e. The number e is approximately 2.7.

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Example 13:

Graph y = e^x

Solution:

The model points are (0,1) and (1,e). Since e is approximately 2.7, the model points are (0,1) and (1,e).

Since the form of the equation is a number to a power of x, it is an exponential function.

The function y = e^x can be written as y = e^{x + 0} + 0.

The shift is (0,0).

The line y = 0 is the horizontal asymptote.

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Example 14:

Graph y = e^{x - 2} + 3

Solution:

The model points are (0,1) and (1,e).
Since e is approximately 2.7, the model points are (0,1) and (1, 2.7).

The shift is (2,3).

The asymptote is y = 3.

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SUMMARY OF EXPONENTIAL FUNCTIONS:

Positive Base:

If y = +b^+x and b > 1, then the model points are (0,1) and (1,b).

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Negative Base:

If y = -b^+x and b > 1, then the model points are (0,-1) and (1,-b).

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Negative Exponent:

If    and b > 1, then the model points are (0,1) and (-1,b).

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Negative Base and Negative Exponent:

If y = -b^-x and b > 1, then the model points are (0,-1) and (-1,-b).

Shift:

The shift of y = b^{x - a} + c is (a,c).

The horizontal asymptote of the exponential function is the line y = c.

Student #1:
Graph y = 2^{x + 3} - 2
Use graph paper to graph the above equation, and then click Answer to view the solution.

Answer

Student #2:
Graph y = - 3^{x + 1} + 2
Use graph paper to graph the above equation, and then click Answer to view the solution.

Answer

Student #3:
Graph y = 3^{-x + 4} + 1
Use graph paper to graph the above equation, and then click Answer to view the solution.

Answer

Student #4:
Graph y = - 2^{-x - 1} + 3
Use graph paper to graph the above equation, and then click Answer to view the solution.

Answer

Student #5:
Graph y = e^{x - 3} - 1
Use graph paper to graph the above equation, and then click Answer to view the solution.

Answer

Homework Problems

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Student #1:

Solution:

y = 2^{x + 3} - 2

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Student #2:

Solution:

y = -3^{x + 1} + 2

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Student #3:

Solution:

y = 3^{-x + 4} + 1

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Student #4:

Solution:

y = - 2^{-x - 1} + 3

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Student #5:

Solution:

y = e^{x - 3} - 1

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