# GRAPHS OF EXPONENTIAL FUNCTIONS

GRAPHS OF EXPONENTIAL FUNCTIONS

By Nancy Marcus

In this section we will illustrate, interpret, and discuss the graphs of exponential functions. We will also illustrate how you can use graphs to HELP you solve exponential problems and check your answers.

Reflection across the y-axis and horizontal shift: The next examples discuss the difference between the graph of f(x) and the graph of f(C - x)

Example 7: Graph the function and the function on the same rectangular coordinate system. and answer the following questions about each graph:

1.In what quadrants is the graph of the function located? In what quadrants is the graph of the function located?

2.What is the x-intercept and the y-intercept of the graph of the function ? What is the x-intercept and the y-intercept of the graph of the function ?

3.Find the point (2, f(2)) on the graph of and find (2, g(2)) on the graph of . What do these two points have in common?

4.Describe the relationship between the two graphs.

5.How would you shift (move) the graph of so that it would be superimposed on the graph of ? Where would the point (0, 1) be located after the move?

1.You can see that the both graphs are located in quadrants I and II. Therefore, both function values will always be positive.

2.You can see that neither of the graphs crosses the x-axis; therefore, neither of the graphs has an x-intercept. Notice that the graph of f(x) crosses the y-axis at 1 because .

The graph of g(x) crosses the y-axis at 1,096.63315843 because .

3.The point , rounded to (2, 7.4) for graphing purposes, is located on the graph of . The point , rounded to (2, 148.4) for graphing purposes, is located on the graph of .

4.Both graphs have the same shape. There is a reflection of across the y-axis because the graphs are pointing in different directions. There is also a horizontal shift to the right.

5.From the equation, you might be tempted to say that the graph of is nothing more than the graph of shifted to the right 7 units. Good thing that you resisted that temptation because you would be wrong.

The minus in front of the x indicates that the graph of will be reflected over the y-axis. The 7 means that there will a horizontal shift of 7 units; but which way?

You can answer the questions in two ways: One, rewrite by factoring out a - 1 from the exponent: . Now you can see that the shift is to the right 7 units. Two, set the exponent equal to 0: 7-x=0 when x = 7. since the 7 is positive, you know that the shift is to the right.

The point (0, 1) on the graph of is reflected over the y-axis. In the reflection, there is no change in the point since it is located on the y-axis. The graph is then shifted right to (0 + 7, 1) or (7, 1).

The point (2, 7.4) is first reflected across the y-axis to (-2, 7.4) and then shifted to the right 7 to (5, 7.4) You can check that this later point is located on the graph of by direct substitution. If you had first shifted the f(x) graph to the right 7 units and then reflected it across the y-axis, the point (2, 7,.4) on f(x) would have been reflected to (9, 7.4) and then shifted to (-9, 7.4). This later point is not located on the graph of g(x).

Rule to remember: When you are reflecting about the y-axis, perform the reflection before you do any horizontal shifting.

Example 8: Graph the function and the function on the same rectangular coordinate system. Answer the following questions about each graph:

1.In what quadrants is the graph of the function located? In what quadrants is the graph of the function located?

2.What is the x-intercept and the y-intercept on the graph of the function ? What is the x-intercept and the y-intercept on the graph of the function ?

3.Find the point (2, f(2)) on the graph of and find (2, g(2)) on the graph of . What do these two points have in common?

4.Describe the relationship between the two graphs.

5.Describe how you would move the graph of so that it is superimposed on the graph of . Where would the point (0, 1) on the graph of wind up on after the move?

6.Describe how you can determine the movement of the f(x) to g(x) from the two equations.

1.The graphs of and are also located in quadrants I and II. This means that both function values will always be positive.

2.Neither graph crosses the x-axis; therefore, there are no x-intercepts. The graph of crosses the y-axis at 1 because

The graph of crosses the y-axis at 0.0000453999297625 because .

3.The point , rounded to (2, 7.4) for graphing purposes, is located on the graph of .

The point , rounded to (2, 0) for graphing purposes, is located on the graph of .

4.Both graphs have the same shape. The graph of appears to open in a direction opposite to the direction of the graph of

, indicating a reflection across the y-axis.

There also appears to be some kind of horizontal shift. The minus sign in front of the x in the exponents indicates that there is a reflection across the y-axis. Rewrite the equation

by factoring out a - 1 from the exponent to read .

From the equation, you can see that the horizontal shift will be to the left 7 units and there will be a reflection over the y-axis. It makes a difference what you do first. For example, we know that the point (2, 7.4) is located on the graph of f(x). If we were to shift the point to the left 10 units, the point would wind up at (-8, 7.4). If were then to reflect the point, the ultimate location would be (8, 7.4). However, if we were to first reflect the point (2, 7.4) across the y-axis, it would wind up at (-2, 7.4). If we were to then shift it to the left 10 units, the ultimate location of the point would be (-12, 7.4). These points are not the same, so the order makes a difference. What move is correct. Let's substitute both points in the equation

. First let's try (8, 7.4). Since , the point (8, 7.4) is incorrect and the sequence of moves to that position is wrong. Now let's try (-12, 7.4). Since , the point (-12, 7.4) is correct and the sequence of moves to that position is correct.

To summarize, when you are reflecting a graph across the y-axis and shifting horizontally, do the reflection first.

Therefore, reflect the graph of over the y-axis and then shift (move) the reflected graph left 7 units. The point (0, 1) doesn't change in the reflection because it is located on the y-axis. After the horizontal shift, the point would be moved left 7 units to (0 - 7, 1) or (- 7, 1).

Suppose the point (a, b) was located on the graph of f(x) to the right of the y-axis. Where would that point be located after the move? After the reflection, the point would be located at (- a, b). After the shift to the left 10 units, the point would be located at ( -a - 10, b).

If you would like to review another example, click on Example.

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Author: Nancy Marcus

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