WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about rational functions and vertical asymptotes.
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The next few lessons are going to be about rational functions and asymptotes; and they are pronounced aa-sim-tohts.
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They are spelled kind of funny--asymptotes--but we want to know how to pronounce them; they are pronounced aa-sim-tohts, asymptotes.
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In exploring these ideas, we will see some very interesting behavior; and we will learn why it occurs.
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But before you start these lessons, it is important that you have a reasonable understanding of polynomials.
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Polynomials are going to be central to this, as we are about to see in the definition of a rational function.
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You will need to know their basics; you will need to know how to factor them to find zeroes and roots.
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And you will also need to know polynomial division.
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If you haven't watched the previous section of lessons on polynomials, you might find it helpful to watch
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the three lessons Introduction to Polynomials, Roots/Zeroes of Polynomials, and Polynomial Division;
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or just the ones that you need specific help with, because those ideas are going to definitely come up as we explore this stuff.
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Also, it should be mentioned that we would be working only with the real numbers; so we are back to working just with the reals.
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And while we briefly worked with the complex numbers in the last couple of lessons, it is back to the reals.
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We are not going to see anything other than real numbers anymore.
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If we would have solutions in the complex, too bad; we are not going to really care about them right now.
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We are back to focusing just on the reals; they are still pretty interesting--there is lots of stuff to explore in the reals.
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You can leave complex numbers to a future, later course in math.
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All right, first let's define a rational function: a rational function is the quotient of two polynomials.
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So, f(x) is our rational function; it is equal to n(x) divided by d(x), where n(x) and d(x) are both polynomials, and d(x) is not equal to 0.
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And by "not equal to 0," I mean that d(x) isn't 0, like straight "this is zero, all the time, forever."
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So, d(x) = 0, the constant function, just 0, all the time, forever--that is not allowed.
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But d(x) can have roots; we can say that d(x) can have roots.
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For example, d(x) could be x² - 1, where d would be 0 at positive 1 and -1; that is OK.
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d(x) can have roots, but it is not going to be the constant function of 0 all the time.
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That is what is not allowed, because then our function would be broken, completely, everywhere.
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But we are allowed to have slight breaks occasionally, when we end up having roots appear in the denominator.
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So, it is just a normal polynomial, but not just 0; that would be bad.
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All right, some examples: g(x) = (2x + 1)/(x² - 4) or 1/x³ or (x⁴ - 3x²)/(x + 2).
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In all of these cases, we have a polynomial divided by a polynomial--as simple as that.
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You might wonder why they are called rational functions--what is so rational about them?
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Remember: many lessons back, when we have talked about the idea of sets, we called fractions made up of integers,
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things like 3/5 or -47/2, the rational numbers, because they seem to be made in a fairly rational, sensible way.
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Thus, we are using a similar name, because the rational functions are built similarly.
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Rational numbers are built out of division, and rational functions are built out of division.
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So, we are using a similar name; cool.
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Since a rational function f(x) = n(x)/d(x) is inherently built out of the operation of division, we have to watch out for dividing by 0.
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That is going to be the Achilles' heel of rational functions; and also, it is going to be what makes them so interesting to look at.
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Dividing by 0 is not defined--it is never defined; so the zeroes, the roots, the places where it becomes 0,
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of our denominator polynomial, d(x), will break the rational function.
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The zeroes of d(x) are not in the domain of our function; so a rational function, right here--
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wherever it ends up being 0, it is not defined there; that is not in the domain.
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It is not in the set of numbers that we are allowed to use, the domain, because if we plug in a number
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that makes us divide by 0, we don't know what to do; we just blew up the world, so it is no good.
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So, we are not going to be allowed to plug in numbers--we are not going to have those in our domain--
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when d(x) = 0, when we are looking at the zeroes, the roots, of our denominator polynomial.
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Now, other than these zeroes in the denominator, a rational function is defined everywhere else,
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because polynomials have all of the real numbers as their domain.
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Since you can plug any real number into a polynomial, and get something out of it,
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the only place we will have any issues is where we are accidentally dividing by 0.
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So, everything other than these locations where we divide by 0--they are all good.
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The domain of a rational function is all real numbers, except the zeroes of d(x)--all real numbers, with the exceptions of these zeroes in our denominator polynomial.
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Everything will be allowed, with a very few exceptions for that denominator polynomial's zeroes.
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All right, to help us understand rational functions better, let's consider this fundamental rational function, 1/x.
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So, immediately we see that f is not going to be defined at x = 0.
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So, our domain for this will be everything except x = 0; domain is everything where x is not 0--so most numbers.
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We see that we can't divide by 0; but everything else would give us an actual thing.
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So, notice how it behaves near x = 0; it isn't actually going to be defined at x = 0--we will never see that.
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But as it approaches it, f(x) grows very, very large--see how it is shooting up?
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We have it going up and down--it is very, very large.
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What is going on here? Let's look at a different viewing window to get a sense for just how large f(x) manages to become.
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If we look at a very small x width, only going from -0.5 to +0.5, we manage to see a very large amount that we move vertically.
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We can go all the way...as we come from the left, we manage to go from near 0 all the way to -100 on this window.
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And we can go from near 0 all the way up to positive 100 on this window.
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And in fact, it will keep going; it just keeps shooting up.
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It shoots all the way out to...well, not out to infinity, because we can't actually hit infinity...but the idea of shooting out to something arbitrarily large.
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It is going to get as big as we want to talk about it getting out to; so in a way, we say it goes out to infinity.
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We see this behavior of going out to infinity in many rational functions.
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If we look at 1/(x + 2), we get it going out to infinity at -2.
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Notice that, when we plug in -2 into the denominator, we have a 0 showing up; so we see this behavior around zeroes.
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If we look at 5/(x² - 9), we see it at -3 and +3.
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We have these places where it goes out to infinity.
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In a few moments, we will formally define this behavior, and we will name it a **vertical asymptote**.
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But first, let's understand why this strange behavior is occurring.
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To understand that, let's look, once again, at f(x) = 1/x, our fundamental rational function, with what makes the most basic form of it.
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Notice that, while x = 0 is not defined (that would blow up the world, because we are dividing by 0--we are not allowed to do that--
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it just doesn't make sense), everywhere else is going to be defined.
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So, let's see what happens as x goes to 0: this little arrow--we say that x is going to 0.
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We are looking as x gets close to 0--not actually at 0; but as x sort of marches towards 0.
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Let's look at what happens as we go from the negative side.
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As we plug in -2, we get -0.5, 1/-2; if we plug in -1, we get -1.
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But as the x becomes very, very small, we see f(x) become very, very large.
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-0.5 gets us -2; -0.1 gets us -10; -0.01 gets us -100; -0.001 gets us -1000.
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As we divide a number by a very small number, 1 divided by 1/10 is equal to positive 10, because the fraction will flip up.
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We are going to end up having to look at its reciprocal, since it is in the denominator again.
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We are going to see this behavior: as we get really, really small--as we get to these really small numbers that we are dividing by--
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we will end up having the whole thing become very, very large.
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We can see this on the positive side, as well, if we go in from the positive side.
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The negative side is us starting from the negative side and us moving towards the 0.
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And the positive one is us starting from the positive side, and us moving towards the 0.
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It is moving from the right, into the left, versus moving from the left into the right.
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Positive: we look at 2; we get one-half, 1/2; we plug in 1; we get 1/1, so 1.
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As we get smaller and smaller, we see it begin to get larger and larger.
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We plug in 0.5, and we will get 2; we plug in 0.1, and we will get 10; we plug in 0.01, and we will get 100; and so on and so forth.
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As we get in smaller and smaller and smaller numbers, we are going to get larger and larger things.
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1/(1 divided by 1 million) is a very small number, but it is one of the things that is defined, because it is not x = 0; it is just very close to x = 0.
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It will cause it to flip up to the top, and we will get 1 million.
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We will get to any size number that we want.
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We can't actually divide by 0; but dividing by very small numbers gives large results.
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This is what is happening: as we get near 0, we go out to infinity, because we are dividing by very, very small numbers.
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And when you divide by something very, very small, you get a very big thing.
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How many times does something very, very small go into a reasonable-sized number?
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It goes in many times; the smaller the chunk that we are trying to see how many times it fits in, the more times it is going to be able to fit in.
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So, the smaller the thing that we are dividing by, the larger the number we will get out in response.
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And that is what is going to cause this behavior of "blowing out."
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With this idea in mind, we will define a vertical asymptote.
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A **vertical asymptote** is a vertical line x = a, where as x gets close to a, the size of f,
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the absolute value of f(x), becomes arbitrarily large--becomes very big.
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Symbolically, we show this as x → a will cause f(x) to go to infinity or f(x) to go to negative infinity.
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Informally, we can see a vertical asymptote as a horizontal location, a, where the function blows out to infinity,
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either positive infinity or negative infinity, as it gets near a.
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So, it is going to be some location a that, as we get close to it, we go out to infinity.
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We manage to blow out to infinity as we get close to it; we get very, very large numbers.
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So, a vertical asymptote is some horizontal location, some vertical line defined by x = a,
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where, as we get close to it, we blow out to infinity.
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Maybe we blow up; maybe we blow down; but we are going to blow out to infinity, one way or the other.
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Vertical asymptotes and graphs; we show the location of a vertical asymptote with a dashed line, as we have been doing previously.
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This aids in drawing the graph and in understanding the graph later.
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So, notice: we just use a dashed line here to show us that this is where the vertical asymptote is.
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So, for the graph of f(x) = 1/(x - 1), we see, right here at x = +1 (because if we plug in x = +1, we would be dividing by 0 there)--
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so near that location--that we will be dividing by very small numbers, which will cause us to blow out to infinity around that location.
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We will see a vertical asymptote there; so as our function approaches that vertical asymptote,
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it ends up having to get very, very large, because it is now dividing by very, very small numbers.
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Notice that the graph gets very close to the asymptote; but it won't be able to touch the asymptote.
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So, it will become arbitrarily close to the line, but it never touches the line.
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Why can't it touch the line? Because it is not defined there.
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We have 1 here; so if we plug in x = 1, then we have 1/(1 - 1), which is 1/0.
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We are not allowed to divide by 0, so we are not allowed to do this, which means we are not allowed to plug in x = 1.
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Our domain is everything where x is not equal to 1; so we are not defined at that thing.
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Our function is not actually defined on the asymptote; it will become very large--it will become very close to this asymptote.
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But it won't actually be allowed to get onto the asymptote, because to be on that point,
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to be on that horizontal location, would require it to be defined there.
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And it is not defined there, because that would require us to divide by 0, which is not allowed.
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So, we are not allowed to do that, which means that we are not actually going to be able to get on it.
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We will never touch the vertical asymptote, because it is not defined.
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Now, not all 0's give asymptotes; so far, all of the vertical asymptotes we have seen have happened at the location of a 0 in the denominator.
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And that is going to be the case: all vertical asymptotes will require there to be a 0 in the denominator.
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But a 0 in the denominator does not always imply an asymptote.
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We are not going to always see an asymptote.
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Consider f(x) = x/x; now, normally, x/x is a very boring function; the x's cancel, and we are left with this effectively being just 1,
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except at one special place: when we look at x = 0, we are given 0/0.
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Now, 0/0 is definitely not defined, so we are not allowed to do anything there.
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We have a pretty good sense that it would be going to 1.
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But when we actually look at that place, it is not defined, because we are looking at 0/0.
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0/0 is definitely not defined; so there is going to be a hole in the graph.
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f is a rational function, x/x, a polynomial divided by a polynomial; they are simple linear polynomials, but they are both polynomials.
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So, f is a rational function; and it is not defined at x = 0; but it still has no asymptote there,
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because normally, we just have x/x; it is just really a constant function, 1.
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We see it here; it is going to be 1 all the time, forever, always, except when we look at x = 0,
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at which point it blips out of existence, because it is not allowed to actually have anything at 0/0, because that is not defined.
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So, formally, we can't give it a location; so instead, we have to use this hole.
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We show that there is a missing place there with this empty circle.
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As opposed to a filled-in circle to show that there is something there, we use an empty circle to show that we are actually missing a location there.
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It has a hole at that height there, because it would be going there, but it doesn't actually show up there.
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There is no asymptote there; why does it happen--why do we lack this asymptote?
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It is because the numerator and the denominator go to 0 at the same time, so the asymptote never happens.
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Because they are both going to 0, there is nothing to divide out to blow up to a very large number--to blow down to a very large negative number.
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The asymptote can't happen, because normally they just cancel out to 1/1...well, not 1/1; they cancel out to something over something.
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But since it is the same something, it becomes just 1.
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So, if we have .1 divided by .1, well, they cancel out, and we just get 1.
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If we have .0001 divided by .0001, well, once again, they cancel out, and we have just 1.
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So, we don't end up seeing this asymptote behavior, because we have to have something divided by very small numbers.
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But if we have a very small number divided by a very small number, they end up going at the same rate; and so we don't get this asymptote.
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To find an asymptote, then, that means we first need to cancel out common factors.
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We need to be able to say, "Oh, x/x means I am going to cancel these out; and I will get it be equivalent to being just 1."
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We will have to remember that we are forbidden to actually plug in x = 0; but other than that, it is just going to be 1 all the time, forever.
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So, with our newfound understanding of all these different things that make up rational functions,
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we can create a step-by-step guide for finding vertical asymptotes.
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So, given a rational function in the form n(x)/d(x), where n and d are both polynomials,
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the first thing we want to do is figure out all of the x-values that we aren't going to be allowed to have in our domain--
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all of our forbidden x-values that would break our function.
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These are the zeroes of d(x), because they would cause us to divide by 0.
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Since dividing by 0 is not allowed, then our domain is everything where we won't divide by 0.
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So, the zeroes of d(x) are not in our domain.
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Next, we want to determine if n(x) and d(x) share any common factors.
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So, to do this, you might have to factor the two polynomials.
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If they share common factors, cancel out those factors.
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Alternatively, this step can be done by checking that the zeroes to d(x) are not also zeroes to n(x),
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because for them to have a common factor, where they would both be going to 0 at the same time,
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then a common factor means that they have the same root.
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So, if they are both going to be zeroes at the location, that means we will see this effect.
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So, we can either cancel out the factors, or we can just and make sure that the roots to d(x) are not also roots to n(x).
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But the easiest way to find roots in the first place is usually to factor.
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So, you will probably want to factor, for the most part.
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But occasionally, it will be easier to just see, "If we plug in this number, does it come out to be 0?"
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The next step: once you cancel out the common factors, you determine any zeroes that are left in the denominator.
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These are the vertical asymptotes, because once we have canceled out common factors,
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we don't have to worry about both the top and the bottom going to 0 at the same time.
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So, any roots that are left are going to have to make very small numbers to blow everything out to infinity.
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So, anything that is left after canceling out (it is crucial that it is after you cancel out common factors)--
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whatever is left as zeroes in the denominator after canceling out will give you vertical asymptotes.
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Finally, an optional step: if we are graphing the function, you want to figure out which way the graph goes:
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positive infinity or negative infinity on either side of each asymptote.
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You can do this by using test values that are very close to the asymptote.
00:18:18.100 --> 00:18:24.700
For example, if we had an asymptote at x = 2, we might want to check out f at 1.99 and f at 2.01.
00:18:24.700 --> 00:18:27.900
They are very close to the asymptote, so we will have an idea of where it is going.
00:18:27.900 --> 00:18:33.700
Alternatively, you can also do this in your head by thinking in terms of positive versus negative,
00:18:33.700 --> 00:18:38.300
and just thinking, "If I was very slightly more or very slightly less," and we will discuss this in the examples;
00:18:38.300 --> 00:18:43.500
you will see it specifically in Example 3; we will see this idea of "If I was just a little bit over;
00:18:43.500 --> 00:18:47.200
if I was just a little bit under," and it will make a little bit more sense when we actually do it in practice.
00:18:47.200 --> 00:18:52.100
But you can always just test a number, use a calculator, and figure out, "Oh, that is what that value would be";
00:18:52.100 --> 00:18:56.400
"I am clearly going to be positive on this side; I am clearly going to be negative on this side."
00:18:56.400 --> 00:18:59.200
All right, let's look at some examples.
00:18:59.200 --> 00:19:04.900
f(x) = (x - 1)/(2x² + 8x - 10); what is the domain of f?
00:19:04.900 --> 00:19:15.000
To figure out the domain of f, we need to figure out where our zeroes on the bottom thing are.
00:19:15.000 --> 00:19:21.600
So, our very first step, where we do factoring: 2x² + 8x - 10--how are we going to factor this?
00:19:21.600 --> 00:19:25.000
Well, let's start out by pulling out that 2 that is in the front; it is just kind of getting in our way.
00:19:25.000 --> 00:19:31.400
So, x² +...8x, pulling out a 2, becomes 4x; -10, pulling out a 2, becomes -5.
00:19:31.400 --> 00:19:33.900
We can check this if we multiply back out; we see that we have the same thing.
00:19:33.900 --> 00:19:38.200
So, 2(x² + 4x - 5); could we factor this further? Yes, we can.
00:19:38.200 --> 00:19:46.500
We see that, if we have +5 and -1, that would factor; so 2(x + 5)(x - 1)...sure enough, that checks out.
00:19:46.500 --> 00:19:53.600
x² - x + 5x...that changes to 4x...plus 5 times -1...-5...great; that checks out.
00:19:53.600 --> 00:20:00.800
So, that means we can rewrite (x - 1)/(2x² + 8x - 10).
00:20:00.800 --> 00:20:09.900
It is equivalent to saying (x - 1) over the factored version of the denominator, 2(x + 5)(x - 1).
00:20:09.900 --> 00:20:21.800
So, all of the places to figure out the domain of f, all of the places where our denominator will go to 0, are going to be x at -5 and x at +1.
00:20:21.800 --> 00:20:28.100
Now, we don't want to allow those; so when x is not -5, and x is not 1, then we are in the domain.
00:20:28.100 --> 00:20:35.500
We are allowed those values; so our domain is everything where x is not -5 and not positive 1.
00:20:35.500 --> 00:20:41.200
Great; the next question is where the vertical asymptotes are.
00:20:41.200 --> 00:20:44.400
To find the vertical asymptotes, we need to cancel any common factors,
00:20:44.400 --> 00:20:47.700
so that the numerator and the denominator don't go to 0 at the same time.
00:20:47.700 --> 00:20:58.200
We notice that we have x - 1 here and x - 1 here; so we can cancel these out, and we get 1/[2(x + 5)].
00:20:58.200 --> 00:21:02.900
So, now we are looking for where the denominator goes to 0.
00:21:02.900 --> 00:21:08.800
That is going to happen at...when is x + 5 equal to 0? We get x = -5.
00:21:08.800 --> 00:21:15.000
So, we have a vertical asymptote when the denominator in our new form with the common factors canceled out.
00:21:15.000 --> 00:21:19.700
So, we will have a vertical asymptote after we have canceled out common factors,
00:21:19.700 --> 00:21:27.500
when we can figure out that the denominator still goes to 0, at x = -5; great.
00:21:27.500 --> 00:21:36.900
The second example: Given that the graph below is of f, our function, find a and b if our function f is 3x/[(x - a)(x - b)].
00:21:36.900 --> 00:21:42.600
So, we notice, first, that we have a vertical asymptote here, and a vertical asymptote here, off of this graph.
00:21:42.600 --> 00:21:48.100
So, we see that one of our vertical asymptotes happens at -3; the other one happens at positive 2.
00:21:48.100 --> 00:21:55.500
So, x = -3; x = +2; we have vertical asymptotes.
00:21:55.500 --> 00:22:04.400
OK, so if x = -3 and x = 2 mean vertical asymptotes, then that means we need to have a factor that will cause a 0 in the denominator at those locations.
00:22:04.400 --> 00:22:12.700
So, we want a 0 in the denominator at -3; so x + 3 would give us a 0 when we plug in -3.
00:22:12.700 --> 00:22:18.000
And x - 2 would give us a 0 when we plug in positive 2.
00:22:18.000 --> 00:22:22.600
Plus 3...we plug in -3 + 3; that gives us a 0 in our denominator.
00:22:22.600 --> 00:22:26.500
We plug in +2; 2 - 2...that gives us a 0 in our denominator; great.
00:22:26.500 --> 00:22:32.500
So, that means that (x + 3)(x - 2)...we need those factors in our denominator to be able to have vertical asymptotes.
00:22:32.500 --> 00:22:38.400
We don't have to worry about it interfering with our numerator, because 3x doesn't have any common factors with (x + 3)(x - 2).
00:22:38.400 --> 00:22:40.800
They are very different linear factors.
00:22:40.800 --> 00:22:51.300
At this point, we are ready...we want -a to be + 3, so that means that a must be equal to -3, because it comes out to be positive 3.
00:22:51.300 --> 00:23:00.600
There is our a; and b...minus b is -2, so it must be that b is equal to positive 2, so it is still actually subtracting.
00:23:00.600 --> 00:23:03.200
And there are our answers.
00:23:03.200 --> 00:23:08.900
All right, the third example: Draw the graph of f(x) = 1/(3 - x).
00:23:08.900 --> 00:23:12.500
It is a pretty simple one, so we don't have to worry about factoring; it has already been factored.
00:23:12.500 --> 00:23:18.300
We don't have to worry about canceling any common factors; we see that there are very clearly no common factors.
00:23:18.300 --> 00:23:20.800
So, we can just get right to figuring this out.
00:23:20.800 --> 00:23:29.200
We have a vertical asymptote; first, we notice that we have a vertical asymptote at x = 3,
00:23:29.200 --> 00:23:36.600
because when we plug in 3, 3 minus x...we have 3 minus 3, so we would have a 0, so we would be dividing by very small numbers;
00:23:36.600 --> 00:23:41.000
so we would be dividing 1 by very small numbers at x = 3.
00:23:41.000 --> 00:23:46.400
That means we will blow out when we are near x = 3; so we have a vertical asymptote there.
00:23:46.400 --> 00:23:56.100
Let's draw in our graph axes: let's put this at...we will need more to the right, because the interesting thing happens out here.
00:23:56.100 --> 00:24:10.200
So, here is 1, 2, 3, 4, 5; I will extend that a little bit further: 6, 1, 2, 3, 4, 5, 6.
00:24:10.200 --> 00:24:19.100
I will go for roughly the same scale; OK.
00:24:19.100 --> 00:24:27.100
And 1, 2, 3, -1, -2, -3, -1, -2; great, we have drawn out our graph.
00:24:27.100 --> 00:24:35.400
We know that we have a vertical asymptote at 3; so let's draw in a vertical line.
00:24:35.400 --> 00:24:41.300
OK, we see our asymptote on our graph; and now we need to draw it in.
00:24:41.300 --> 00:24:45.700
So, at this point, we realize that we actually probably should have some points; so let's try some points.
00:24:45.700 --> 00:25:00.200
If we try x =...oh, let's say 4, just one unit to the right of our asymptote; when we plug that in, we will get 1/(3 - 4), which gets us 1/-1.
00:25:00.200 --> 00:25:05.600
1/-1 is -1; so at 4, we are at -1; so we plot in that point.
00:25:05.600 --> 00:25:15.200
Let's try one unit to the left, x = 2: we will get 1/(3 - 2), so we will get 1 over positive 1 (3 - 2 is positive 1).
00:25:15.200 --> 00:25:17.700
We will be at positive 1 when we are at 2.
00:25:17.700 --> 00:25:25.300
We might still want a little bit more information, so maybe we will also try x = 0.
00:25:25.300 --> 00:25:33.000
At x = 0, we will get 1/(3 - 0), so we will be at 1/3; so at 0, we are just 1/3 of the way up to our 1.
00:25:33.000 --> 00:25:39.700
So, we will put in that point there at x = 6; let's try that one.
00:25:39.700 --> 00:25:50.600
We will be at 1/(3 - 6), so 1/-3; so we will be at -1/3, so once again, we will be just a little bit...one-third of the way out there.
00:25:50.600 --> 00:25:56.300
All right, now we are starting to get some sense; what happens is that we actually get near that asymptote.
00:25:56.300 --> 00:25:59.700
Are we going to go positive on the left side, or are we going to go negative on the left side?
00:25:59.700 --> 00:26:04.600
Are we going to go positive on the right side, or are we going to go negative on the right side?
00:26:04.600 --> 00:26:14.900
What we can think about is: we can consider just a hair to the left of 3: consider 2.99999.
00:26:14.900 --> 00:26:18.500
Consider this number; now, we could actually plug it into a calculator, and we could get a value.
00:26:18.500 --> 00:26:30.700
But we can also just think about this in our head, and say, "1/(3 - 2.99999)...if we have a bunch of 9's,
00:26:30.700 --> 00:26:35.200
3 - 2.99999, well, is it going to be a positive number, or is it going to be a negative number?"
00:26:35.200 --> 00:26:41.000
Well, 3 - 2.99999...2.99999 is still smaller than 3, so it is going to come out as a positive thing.
00:26:41.000 --> 00:26:48.500
We have 1, a positive number, divided by a positive number; we can just think of it as dividing by a small positive number.
00:26:48.500 --> 00:26:53.300
That means that the whole thing will come out to be a positive thing, when we are on the left side,
00:26:53.300 --> 00:27:00.400
because we are subtracting by something, but it ends up being just under; so we stay positive.
00:27:00.400 --> 00:27:03.600
So, we can think in terms of "Are we being positive, or are we being negative?"
00:27:03.600 --> 00:27:07.000
We are being positive; so we know we are going to go up on this side.
00:27:07.000 --> 00:27:22.600
If we consider a very small number above 3, 3.00001--if we consider that, it will be 1/(3 - 3.00001).
00:27:22.600 --> 00:27:26.800
Well, 3 minus something that is just a little bit above it is going to end up being negative.
00:27:26.800 --> 00:27:29.500
It is going to be a very small number, but it is going to be negative, and that is key.
00:27:29.500 --> 00:27:35.600
1 over a negative number on the bottom: since it is a positive divided by a negative, the whole thing will stay negative.
00:27:35.600 --> 00:27:39.400
And so, we will get very large negative numbers; so we will be going down on the right side.
00:27:39.400 --> 00:27:47.900
So, at 3.00001, we see that we end up being still negative; so we will have to be negative when we are on the right side of our asymptote.
00:27:47.900 --> 00:27:54.800
On the left side, 2.99999, just to the left of our asymptote at x = 3...we see that we are going to end up being positive.
00:27:54.800 --> 00:27:58.800
And so, that guides us on how our asymptote should grow.
00:27:58.800 --> 00:28:03.900
So, at this point, we can draw in this curve; it is going to be growing and growing and growing.
00:28:03.900 --> 00:28:08.900
And as it gets to the asymptote, it is going to get larger and larger and larger and larger and larger.
00:28:08.900 --> 00:28:13.900
It will never actually touch the asymptote; it will end up just getting larger and larger and larger and going up and up and up.
00:28:13.900 --> 00:28:20.400
It never becomes perfectly vertical; it never touches the asymptote; but it is going to blow out to infinity, slowly but surely.
00:28:20.400 --> 00:28:26.000
It won't ever actually touch infinity, because we can't touch infinity--infinity is just an idea of ever-larger numbers.
00:28:26.000 --> 00:28:28.600
But it will go out to ever-larger numbers.
00:28:28.600 --> 00:28:37.400
On the other side, we see a very similar thing; sorry, that should be curved just a little bit more--not quite straight.
00:28:37.400 --> 00:28:41.100
And it is going to get very, very large, and once again, going to curve out.
00:28:41.100 --> 00:28:45.500
It never quite touches that vertical asymptote, but we will get very, very, very close.
00:28:45.500 --> 00:28:48.800
My right side is a little bit closer to the vertical asymptote than it probably should be.
00:28:48.800 --> 00:28:53.900
But we have a good idea, and that would be acceptable when turning that in on a test or homework.
00:28:53.900 --> 00:28:56.000
What happens as we go very, very far to the left?
00:28:56.000 --> 00:29:00.800
Well, let's think about as we plug in some really big negative number, like, say, -100.
00:29:00.800 --> 00:29:08.700
Well, x = -100...we are going to end up having 1/(3 - -100); those cancel out, and we get a positive.
00:29:08.700 --> 00:29:12.300
So, we get 1/103, which is a very small number.
00:29:12.300 --> 00:29:19.900
So, since it is 3 - x, as we plug in very large negative numbers, we are going to effectively get very large positive numbers in our denominator.
00:29:19.900 --> 00:29:24.700
And so, we will be dividing by very large things; so we will sort of crush down to nothing.
00:29:24.700 --> 00:29:28.800
As we get larger and larger and larger, we are going slowly approach 0.
00:29:28.800 --> 00:29:33.200
We will never quite hit 0 (we will talk about this more when we talk about horizontal asymptotes in the next lesson),
00:29:33.200 --> 00:29:36.000
but we will get very, very close to the 0 height.
00:29:36.000 --> 00:29:40.900
The same thing happens as we go out to the right: but we will be coming from the negative side,
00:29:40.900 --> 00:29:49.000
because if we look at a very, very large positive number, like 1/(3 - 100), a positive 100 plugged in for x,
00:29:49.000 --> 00:29:52.900
we will get 1/-97, so we will be a very small number,
00:29:52.900 --> 00:29:56.700
but we will be a very small negative number, because we have that negative on the bottom.
00:29:56.700 --> 00:30:00.400
And so, we will be on the bottom side; we will be below the x-axis.
00:30:00.400 --> 00:30:06.900
Great; the final example, Example 4: Give a rational function where the graph goes up on both sides of an asymptote.
00:30:06.900 --> 00:30:09.300
Also give one where it goes down on both sides.
00:30:09.300 --> 00:30:14.800
So, so far, we have always seen our asymptotes where on one side it goes up, and on the other side it goes down (or maybe the other way).
00:30:14.800 --> 00:30:20.900
But is it possible to create something where it is going to go up on both sides or going to go down on both sides?
00:30:20.900 --> 00:30:24.900
Let's look at our first one that we think of when we think of a rational function.
00:30:24.900 --> 00:30:29.900
We immediately think, "Oh, the fundamental, basic thing that makes up rational functions is 1/x."
00:30:29.900 --> 00:30:34.300
This is the simplest form we can think of that really shows these ideas we are looking for.
00:30:34.300 --> 00:30:45.600
1/x ends up giving us a graph that gets very, very large, that gets to the right side, but goes down on the right.
00:30:45.600 --> 00:30:49.100
And then, the same thing blows down.
00:30:49.100 --> 00:31:04.800
The issue here with 1/x is: on one side, it is negative; on the left side, we were dividing by negative numbers, so we have negative.
00:31:04.800 --> 00:31:12.500
On the other side, it is positive; that is why we go up and down.
00:31:12.500 --> 00:31:19.200
That is what causes up and down effect--well, down and up, if we go negative to positive.
00:31:19.200 --> 00:31:22.300
But that is why we are seeing this split, where we are going in two opposite ways,
00:31:22.300 --> 00:31:25.000
because on one side of the vertical asymptote, we are dividing by a negative number.
00:31:25.000 --> 00:31:27.900
On the other side of the vertical asymptote, we are dividing by a positive number.
00:31:27.900 --> 00:31:34.600
They are both very small numbers, but that negative versus positive causes this incredible difference in whether you go to -∞ or +∞.
00:31:34.600 --> 00:31:42.700
So, what we want is positive on both sides, if we are going to go up on both sides.
00:31:42.700 --> 00:31:48.900
So, what could we create that would be similar to 1/x, but always putting out positive things on the bottom?
00:31:48.900 --> 00:31:52.100
Let's keep it as 1 over; but what is always positive?
00:31:52.100 --> 00:31:56.100
Well, instead of dividing by x, let's divide by x².
00:31:56.100 --> 00:32:02.400
We can force it to always be positive: 1/x²...x² is still a polynomial, so it is still a rational function.
00:32:02.400 --> 00:32:07.600
And if we draw that one in, we are going to see it behaving very similarly on the right side.
00:32:07.600 --> 00:32:12.500
It will end up being here and here and here as we get to small numbers, as we get below 1.
00:32:12.500 --> 00:32:18.800
We are going to be getting to dividing by very small numbers, so it is going to blow out to infinity.
00:32:18.800 --> 00:32:20.900
And we will get crushed down to 0 as it goes to the right.
00:32:20.900 --> 00:32:30.400
On the left side, though, when we square a negative number, it becomes positive; so we are going to see a mirror image on the left side, as well.
00:32:30.400 --> 00:32:33.400
So, we are going to see it going up on both sides.
00:32:33.400 --> 00:32:38.500
If we want to get one that is going to go down on both sides, if we want it to be down on both sides--
00:32:38.500 --> 00:32:44.200
well, we see 1/x²; we just want it to go down; so let's make it as easy as just flipping it down.
00:32:44.200 --> 00:32:48.800
We flip it down by making everything negative, so it is -1/x².
00:32:48.800 --> 00:32:57.900
And we will see a graph that looks like this; we have managed to make it go down on both sides.
00:32:57.900 --> 00:33:06.200
So, a simple way to make both sides go in the same direction is by having it be squared, or by having it be negative and then squared...
00:33:06.200 --> 00:33:09.500
not negative and then squared, because that would be positive; but a negative square,
00:33:09.500 --> 00:33:12.600
because then it is a square number, times a negative; cool.
00:33:12.600 --> 00:33:14.500
All right, we will see you at Educator.com later.
00:33:14.500 --> 00:33:18.600
Next time, we will talk about horizontal asymptotes and have an understanding of why it is getting crushed to 0,
00:33:18.600 --> 00:33:22.000
and see that it can, in fact, be crushed to value other than 0--pretty interesting; goodbye!