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College Algebra
Course: College Algebra > Unit 7
Lesson 4: Graphing exponential growth and decayGraphing exponential growth & decay
Graphing the exponential functions y=27⋅(⅓)ˣ and y=-30⋅2ˣ.
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Would the asymptote for an exponential function ever be anything but zero?(76 votes)
Yes. for example y=(2^x)+3
The horizontal asymptote is now y=3(78 votes)
Is there previous lesson about asymptotes, or was it introduced it in this video the first time ?(41 votes)
he didn't explain the asymptote at all?(22 votes)- He probably assumed that everyone watching this knew what that was from a previous lesson, but I couldn't find anything myself about intro. to asymptotes in Algebra I.
An asymptote is basically a point where the line approaches but never meets, no matter how close it gets. For example, this is sort of unrelated to a graph situation but, if you take the number 1 and keep dividing it by 1/2 you will never get to zero.(27 votes)
How does this relate to the real world(7 votes)
JFails,
Exponential growth comes up a lot in business (investments), in social science (population growth), and in virology (how quickly a virus spreads).
Exponential decay comes up in audio engineering (sound levels decrease exponentially over longer distance), sports tournaments (if a tournament starts with 32 teams, how many rounds are there until the championship), archaeology(radiocarbon dating), and climate studies (atmospheric pressure decreases exponentially with increased altitude.)(19 votes)
In other higher level math courses, lines can occasionally cross a horizontal asymptote. Is this possible when graphing exponential functions?(8 votes)
Is it possible to move the asymptote so that it's not zero? What changes to the equation do we have to make? And also, in the video with negative 30, the graph is decaying. How might we have a graph that is growing but the value of the y-intercept is still negative?(5 votes)
All functions can be shifted horizontally and/or vertically.
In general 𝑓(𝑥 − ℎ) + 𝑘 shifts 𝑓(𝑥) ℎ units to the right and 𝑘 units up.
Also, 𝑓(−𝑥) reflects 𝑓(𝑥) over the 𝑥-axis
and −𝑓(𝑥) reflects 𝑓(𝑥) over the 𝑦-axis.
– – –
As you mention, ℎ(𝑥) = 27⋅(1∕3)^𝑥 has a horizontal asymptote at 𝑦 = 0.
If we want to move that to, say, 𝑦 = −4,
then instead of graphing 𝑦 = ℎ(𝑥)
we should graph 𝑦 = ℎ(𝑥) − 4 = 27⋅(1∕3)^𝑥 − 4.
– – –
To make 𝑔(𝑥) = −30⋅2^𝑥 growing instead of decaying,
we can reflect it over the 𝑥-axis
by graphing 𝑦 = −𝑔(𝑥) = 30⋅2^𝑥
This of course changes the 𝑦-intercept to (0, 30), so if we still want it to have a negative 𝑦-intercept we could move it down maybe 40 units by graphing
𝑦 = −𝑔(𝑥) − 40 = 30⋅2^𝑥 − 40(7 votes)
- how would you graph a function with a negative base like y= -5^x(4 votes)
I'm assuming you mean y=-(5)^x (the positioning of the negative sign is very important). In that case, you'll plot the graph of y=5^x and then reflect it over the y axis.
In the case where you want to take the exponent of a negative value, it gets a lot more tricky, because sometimes, the function is not defined for negative values. for x=0.5, y=(-5)^x is not defined on the real line (you're taking the square root of a negative number).(8 votes)
- Is the asymptote always horizontal and on the zero line? Also how do you figure where the asymptote lies. Lastly, what is the asymptote and is this asked or used on the SAT. Whoever answers my questions will get an upvote from me! Please help me out here.(3 votes)
No, there are vertical and other asymptotes as well. For exponential functions, the basic parent function is y=2^x which has a asymptote at x=0, but if it is shifted up or down by adding a constant (y = 2^x + k), the asymptote also shifts to x=k. I do not know what all is on the SAT, but if you have a rational function whose parent function is y = 1/x, you have a horizontal asymptote at x=0 and a vertical asymptote at y=0. These both can be shifted by adding (h,k) so that y = 1/(x-h) + k.(10 votes)
How do I find where the asymptote goes?
Thanks! :)(3 votes)- Think about if the graph moves up or down by adding something after the exponent. So if you have y=1/3(2)^x, there is nothing after, so asymptote is a y=0. If you have y=1/3(2)^x + 5, the asymptote would be at y=5 or y=1/3(2)^x -3, the asymptote would be at y=-3.(6 votes)
- You say that the line at some point will get really darn close to 0, but will it ever reach 0?(3 votes)
It never will equal zero; instead, it will get infinitely close to zero.
Another example of a function with an asymptote at y=0 is 1/x. If x=10, you get 0.1. If x=1000, you get 0.001. If x=1,000,000,000, you get 0.000000001, which is pretty small but still not zero. No matter how large x is, you will never have exactly 0.(6 votes)
Video transcript
- [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.