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Current time:0:00Total duration:4:06

CC Math: HSF.IF.C.7, HSF.IF.C.7e

- [Voiceoer] This is from the graph basic exponential functions on Khan Academy. They asked us graph the
following exponential function. And they give us the function, h of x is equal to 27 times 1/3 to the x. So our initial value is 27
and 1/3 is our common ratio. It's written in a
standard exponential form. And it gives us some graphing tool where we can define these two points and we can also define a horizontal asymptote to construct our function. And these three things
are enough to define to graph an exponential if we know that it is an exponential function. So let's think about it a little bit. So the easiest thing
that I could think of is, well, let's think about its initial value. Its initial value is going
to be x equal to zero. X equal zeros, 1/3 to the
zero power is just one and so you're just left with
27 times one or just 27. That's what we call this number here when you've written in this form. You call this the initial value. So when x is equal to zero, h of x is equal to 27. Now we're graphing y=h(x). So now let's graph another point. So let's think about it a little bit. When x is equal to one, when x is equal to one, what is h of x? It's gonna be 1/3 to the first power which is just 1/3. And so 1/3 times 27 is gonna be nine. So when x is one, h of one is nine. And we can verify. And now let's just think about, let's think about the asymptote. So what's gonna happen here when x becomes really, really, really, really,
really, really, really big? If I take 1/3 to like a
really large exponent to, say to the 10th power
or to the 100th power or to the 1000th power, this
thing right over here is going to start approaching
zero as x becomes much, much, much, much larger. And so something that is
approaching zero times 27, well, that's going to
approach zero as well. So we're gonna have a
horizontal asymptote at zero. And you can verify that this works for more than just a two
points we talk about. When x is equal to, this is
telling us that the graph, y equals h of x goes to
the point two comma three. So h of two should be equal to three. And you can verify that
that is indeed the case. If x is two, 1/3 squared is nine. Oh sorry, 1/3 squared is 1/9 times 27 is three. And we see that right over here. When x is two, h of two is three. So I feel pretty good about that. Let's do another one of this. So graph the following
exponential function. So same logic. When x is zero, the g
of zero is just gonna boiled out to that initial value. And so you scroll down. Initial value is negative 30. And so let's think about
when x is equal to one. When x is equal to one, two to
the first power is just two. And so two times negative
30 is negative 60. So when x is equal to one, the value of the graph is negative 60. Now let's think about this asymptote over that, it should sit. So let's think about what
happens when x becomes really, really, really, really,
really, really negative. So when x is really negative, so two to the negative one power is 1/2. Two to the negative two is 1/4. Two to the negative three is 1/8. As it gets larger and larger negative or higher magnitude negative values or in other words, as x becomes more and more and more negative, two to that power is
going to approach zero. And so negative 30 times
something approaching zero is going to approach zero. So this asymptote is in the right place, a horizontal asymptote as x
approaches negative infinity. As we move further and
further to the left, the value of a function
is going to approach zero. Now we can see it kind of
approaches zero from below. We can see they approaches below because we already looked
at the initial value and we used that common ratio
to find one other point. Hopefully, you found that interesting.