**
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.
**

**
Horizontal and vertical shifts: The next examples discuss the
difference between the graph of f(x) and the graph of f(x + A)
+ B.
**

**
Example 9:** 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 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.How would you physically shift (move) the graph of so that it would be superimposed on the graph of ? After you move the graph, where would the point (0, 1) be located?

6.Describe what you can tell about the relationship between the graphs from just their equations.

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

2.You can see that neither of the graphs cross 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 5.00247875218 because .

3.The point , rounded to (2, 7.4) for graphing purposes, is located on the graph of . The point , rounded to (2, 5) for graphing purposes, is located on the graph of .

4.Both graphs have the same shape. It appears from the graph, that the graph of is a result of shifting the graph of to the right and upward.

5.After we move the graph of to the right 6 units and up 5 units, it is superimposed on the graph of . The point (0, 1) on the graph of would first be shifted to the right 6 units and up 5 units to (0 + 6, 1 + 5) or (6, 6).

6.Since the exponents differ in each equation by a constant, there will be a horizontal shift. By setting x - 6 to 0, you can tell that the shift is to the right 6 units. You can also tell that the equations differ by a constant. This means there will also be vertical shift of 5 units up

**Example 10:** 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 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 physically move (shift) the graph of so that it would be superimposed on the graph of . Where would the point (0, 1) on the graph of wind up on after the move?

1.The graph of is located in quadrants I and II. The graph of is also located in quadrants I, III, and IV.

2.The graph of does not cross the x-axis because there is no value of x that would cause to equal zero.

The graph of crosses the x-axis at -9.09861228867 because that is the solution when we set :

3.The graph of crosses the y-axis at 1, and the graph of crosses the y-axis at -2.99966453737 because .

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

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

5.Both graphs have the same shape. The graph of appears to the right and above the graph of .

Whenever the exponents differ by a constant, there is a horizontal shift in the graphs. Whenever the constant terms differ in the equations, there is a vertical shift in the graphs.

From the equation, you can see that the horizontal shift is to the right 8 units, and the vertical shift is down 3 units. It does no make any difference what you do first.

Therefore, shift the graph of to the right 8 units and down 3 units. The point (0, 1) would be moved to the right 8 units and down 3 units to (0 + 8, 1 - 3) or (8, - 2).

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

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Contact us

Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA

users online during the last hour