SOLUTION: Could you please take the time to help me solve this problem: Sketch the following exponential functions: f(x) = 2x, g(x) = 3x, and h(x) = ex, where e is the natural number. Identi

Algebra ->  Exponents -> SOLUTION: Could you please take the time to help me solve this problem: Sketch the following exponential functions: f(x) = 2x, g(x) = 3x, and h(x) = ex, where e is the natural number. Identi      Log On


   



Question 107258: Could you please take the time to help me solve this problem: Sketch the following exponential functions: f(x) = 2x, g(x) = 3x, and h(x) = ex, where e is the natural number. Identify all intercept points as well as the behaviors of the functions when x is approaching plus and infinity.Thank you so much for helping me.
Thank you for helping me.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
You are given the three functions:
.
f%28x%29=+2%5Ex
.
g%28x%29+=+3%5Ex
.
and
.
h%28x%29+=+e%5Ex
.
You are to make the graphs of y+=+2%5Ex, y+=+3%5Ex, and y+=+e%5Ex.
.
Some points to notice. Since e is approximately 2.718281828 for thinking about its graph you
could say it is the graph of y+=+2.7%5Ex. Look at how similar these three graphs will
be ... one is the graph of y+=+2%5Ex ... another is y+=+2.7%5Ex ... and the last
one is y+=+3%5Ex
.
So if you understand the procedure for one of the graphs, you can use that same procedure
on the other two graphs. Let's look at the graph of y+=+2%5Ex. We can get an idea
of its graph by picking values for x and calculating y. Suppose we let x = +1. What is
the corresponding value of y? Go to the equation and substitute +1 for x to get:
.
y+=+2%5Ex+=+2%5E1+=+2
.
So you can plot the point (+1,+2) as a point on the graph.
.
Next let x = +2 and get the corresponding value of y:
.
y+=+2%5Ex+=+2%5E2+=+4
.
The point (2,4) is on the graph
.
Next let x = +3 and get the corresponding value of y:
.
y+=+2%5Ex+=+2%5E3+=+8
.
The point (3,8) is on the graph
.
One more ...
.
Next let x = +4 and get the corresponding value of y:
.
y+=+2%5Ex+=+2%5E4+=+16
.
The point (4,16) is on the graph
.
If you plot those points, you should have a pretty good idea of what is happening when
x is positive and is getting bigger.
.
What about when x = 0? Any number raised to the exponent zero results in 1 as the answer.
Therefore, for all three of the graphs you are to make, when x is set equal to zero the
corresponding value of y is +1. Therefore (0, +1) is a point on all three of your graphs.
.
Finally, what happens when x is a negative value? Remember that a negative exponent can
be converted to a positive exponent by putting the quantity in the denominator of a fraction
whose numerator is 1. Let's see what happens to y+=+2%5Ex when x is a negative
number.
.
Begin by setting x = -1. The equation becomes:
.
y+=+2%5Ex+=+2%5E%28-1%29+=+1%2F2
.
So the point (-1, 1/2) is on the graph.
.
Next set x = -2. The equation becomes:
.
y+=+2%5Ex+=+2%5E%28-2%29+=+1%2F%282%5E2%29+=+1%2F4
.
So the point (-2, 1/4) is on the graph.
.
Then set x = -3. The equation becomes:
.
y+=+2%5Ex+=+2%5E%28-3%29+=+1%2F%282%5E3%29+=+1%2F8
.
So the point (-3, 1/8) is on the graph.
.
Next set x = -4. The equation becomes:
.
y+=+2%5Ex+=+2%5E%28-4%29+=+1%2F%282%5E4%29+=+1%2F16
.
So the point (-4, 1/16) is on the graph.
.
By this time you can tell that as the value of x moves to the left in the negative
domain, the value of y gets smaller and smaller.
.
You can apply this same methodology to the other two functions. [Note that a scientific
calculator can be used to find values of e%5Ex but for your graphs you can just raise
2.72 to powers of x that you choose and that will be close enough to see what is going on
with the graph.]
.
When you get done with the three graphs they should look like this:
.
graph%28800%2C800%2C-5%2C5%2C-1%2C20%2C2%5Ex%2C3%5Ex%2C%282.7%29%5Ex%29
.
Where the "red" graph is the graph of y+=+2%5Ex and the "green" graph is the graph
of y+=+3%5Ex and the "purple" graph is the graph of y+=+e%5Ex.
.
Notice that the "purple" graph is a path between the other two graphs.
.
Hope this is enough info that you can work your way through the entire problem.
.