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### Course: Precalculus > Unit 4

Lesson 2: End behavior of rational functions# End behavior of rational functions

Sal analyzes the end behavior of several rational functions, that together cover all cases types of end behavior.

## Want to join the conversation?

- Can anyone explain to me the logic behind "approaching from below" and "approaching from above". I've been doing the exercises and I don't understand the explanation given there. How do I know from where the graph will approach 0?(18 votes)
- 100 -> 10 -> 1 -> .1 -> .01 is approaching 0 from above, or from the positive (positive numbers are 'above' 0)

-100 -> -10 -> -1 -> -.1 -> -.01 is approaching 0 from below, or from the negative (negative numbers are 'below' 0)

As x approaches infinity (as x gets bigger):

1/x approaches 0 from above (smaller and smaller positive values)

-1/x approaches 0 from below (smaller and smaller negative values)(36 votes)

- In the first equation, Sal simplifies the equation to: (7x - 2) / (15 - 5/x). He states at2:32that 7x will approach negative infinity, however he simplified this value from 7x^2 which by definition will always turn negative numbers positive. Shouldn't he have simplified the equation to state: 7|x|?(11 votes)
- Sal simplified the expression down to it's dominate terms: 7x/15.

If you are going to use the 7x^2, then you also need to use the 15x in the denominator. Yes, 7x^2 would be positive, but 15x would be negative. A positive / a negative = a negative.

Hope this helps.(32 votes)

- Did Sal make a mistake in the 1st example? the denominator is supposed to be 15x+5, not 15x-5 (same result nonetheless, but this may confuse others)(16 votes)
- You are correct. The denominator in the first example should be 15x+5 instead of 15x-5. Thank you for bringing this to my attention. It's important to ensure that mathematical expressions are written accurately to avoid confusion and errors.(13 votes)

- Can someone please help me because I am still very confused. How does Sal know that as the numbers approach - infinity they will get to 0 and why is this? What happens as it approaches infinity and why?(14 votes)
- Something approach positive infinity means that it become larger and larger, extremely large. And if something approach negative infinity means it becomes smaller and smaller, as small as you can imagine, and even smaller than that. And if, for example, some number infinity approach 3, means it is so close to 3, extremely close but not yet 3. So, of course, it's not a real number, the symbol of infinity is 8 turn 90 degrees.

And if you want to know which type of approach the question is, you could plug some value in x to figure out what does it approach.

Example 1: What does 7x-2 approach if x approach negative infinity?

What this question means is what number is 7x-2 approach if x become extremely small.

1. If x is -1, 7x-2 is -9

2. If x is -10, 7x-2 is -72

3. If x is -100, 7x-2 = -702

Here's a pattern, as x become smaller and smaller, 7x-2 become smaller and smaller as well. That means when x approach negative infinity, 7x-2 approach negative infinity as well.

Example 2: What does 6x^5/x^9 approach if x approach infinity?

So, this question is similar to the first one, but this question doesn't tell us x is approach to positive infinity or negative infinity. But lucky for us, we don't need to know.

1. If x is 100, 6x^5 is 7.776×10^13, x^9 is 1×10^18, answer is 7.776×10^-5 (it's a very small positive number, but not yet zero)

2. If x is 10, 6x^5 is 777600000, x^9 is 1000000000, answer is 0.7776

3. If x is -10, 6x^5 is -1.29×10^-9, x^9 is -1000000000, answer is 1.29×10^−18

4. If x is -100, 6x^5 is -1.29×10^-14, x^9 is -1×10^18, answer is 1.29×10^-32 (it's a very small positive number, but not yet zero)

So as you can see again, when x become extremely larger or extremely smaller, 6x^5/x^9 extremely approach to 0, but not yet 0.

Okay, hope that helps! :)(9 votes)

- Couldn't you plug in +infinity and -infinity and decide?(5 votes)
- You can't actually plug in infinity, because it is only a hypothetical concept and has no actual value. You can, however, plug in a number that's either very low or very high so that you know what direction it's headed.(9 votes)

- How did Sal know that dividing the numerator and denominator of the function by the highest degree of "x" found in the denominator would make the entire function easier to analyze? In other words, how does he know that the method used in the video will make the function easier to analyze?(6 votes)
- I'm still trying to figure this video out. From reading other comments it seems that dividing by the highest degree in the denominator you make the denominator like a constant as the variable goes to positive or negative infinity. That means that you can concentrate on what is happening in the numerator. I guess whoever came up with it was just trying "algebraic changes which do not change the result" and which could make analysis easier (like using -1.-1 or multiplying by a crazy denominator over that same crazy numerator - those examples both multiply by one but can change the algebra) .(3 votes)

- Why is 5/x going to zero?(3 votes)
- As x approaches infinity, the x will get larger and larger. 5 divided by a very large number is very close to zero. Try it on a calculator to see how it behaves.(7 votes)

- What does end behavior do?(3 votes)
- End behavior is just how the graph behaves far left and far right. Normally you say/ write this like this. as x heads to infinity and as x heads to negative infinity. as x heads to infinity is just saying as you keep going right on the graph, and x going to negative infinity is going left on the graph.

Let me know if that didn't fully help.(4 votes)

- So all vertical asymptotes are discontinuities, but not all discontinuities are vertical asymptotes, right?(3 votes)
- That's correct. A vertical asymptote is a
**type**of discontinuity, but there are others. For example, a removable discontinuity occurs when there is no real solution for a particular point on the graph.(3 votes)

- there is question in the practice 6x^4-7x/3x^7+18x when x approach postive infinity

what the function approach,i used sal's method, using the highest degree to figure it

out, but the question asked is it approaching 0 from above or below,how would i know is it approaching from above or below?(3 votes)- If f(x) is the function, then as x approaches infinity f(x) approaches 0 from above.

You have established that there is a horizontal asymptote at y=0 [6x^4 / 3x^7 approaches 0 as x is large].

To determine whether f(x) approaches the asymptote from above or below consider the sign of f(x) as x is large.

f(x) = (6x^4 -7x) / (3x^7 +18x)

When x is large both the numerator and denomenator will have +ve signs giving f(x) an overall +ve sign, so f(x) must approach from above.

If the question had asked whether f(x) approaches 0 from above or below as x approaches -ve infinity; as x is large the numerator would be +ve but the denominator would be -ve giving f(x) an overall -ve sign. Then f(x) approaches 0 from below.

I find it helpful to draw a sketch graph to check answers.

You might find the Demos graph drawing calculator helpful. The web address is:

https://www.desmos.com/calculator(3 votes)

## Video transcript

- [Voiceover] So, we're
given this function, f of x, and it equals this rational
expression over here and we're asked "What does f of x approach "as x approaches negative infinity?" So, as x becomes more and more
and more and more negative, what does f of x approach? And, like always, pause the video and see if you can think
about that on your own. Well, one thing that I like to do when I'm trying to consider
the behavior of a function as x gets really positive
or really negative is to rewrite it. So, f of x, I'm just rewriting it once, is equal to 7x-squared, minus 2x over 15x minus five. Now, an interesting technique
to think about what happens to the different terms
as x gets very positive or x gets very negative, is to divide both the
numerator and the denominator by the highest degree term
of x in the denominator. And the highest degree term
of x in the denominator is the first-degree term. We have just a single x there. So, let's multiply both the
numerator and the denominator by one over x, or another
way of thinking about it is we're dividing both the
numerator and the denominator by x. And if we're doing the
same thing to the numerator and the denominator, if we're multiplying or
dividing them by the same value, I should say, well then, I'm just really
just multiplying it by one. So, I'm not changing its value. This will make it a little
bit more interesting, and a little bit easier for
us to think about what happens when x becomes very, very, very negative. So, 7x-squared divided by x, or being multiplied by one
over x, is going to be 7x. 2x times one over x, or 2x
divided by x, is just two. And then all of that over 15x divided by x, or 15x over x, is just going to be 15. And then you have five over x. Five times one over x
is equal to five over x. Minus five over x. Now, this is equivalent, for our purposes, to what we started with but it
makes it a little bit easier to think about what happens when x gets very, very, very, very negative. Well, when x gets very,
very, very, very, very, very, very negative, this is going to become a
very large negative number. You subtract two from it,
it really won't matter much. You divide that by 15, well,
that's not gonna matter much. And this is just going to
become very, very, very small. You're taking five and
you're dividing it by ever-larger negative numbers, or more and more negative numbers. So, this right over here
is gonna go to zero. This thing over here is
gonna go towards infinity. Or, I should say, it's gonna
go towards negative infinity. Seven times a negative trillion, seven times a negative googol, seven times a negative googolplex, we're getting more and
more negative numbers, this is gonna get, this is going to approach negative infinity. Doesn't matter that you're
subtracting two from that. In fact, that'll get even more negative. And it doesn't matter if
you then divide that by 15, you're still approaching
negative infinity. If you had a arbitrarily negative number, you divide it by 15, you still have an arbitrarily negative number. And, so, you could say
that this is going to go to negative infinity. Now, another way that you
could've thought about it. This is actually how I do think about it when I'm trying to, when I see these types of problems. I say, well which terms in the numerator and the denominator are going to dominate? And what do I mean by "dominate"? Well, as x gets very positive
or x gets very negative, another way to think about
it is the magnitude of x gets large, the absolute
value of x gets large. The higher degree terms are
going to grow much faster than the lesser degree terms. And so, we could say that for large x, for large x, and when I say "large" I
mean high absolute value. High absolute value. And if we're going to negative infinity, that's high absolute value. So, f of x is going to
be approximately equal to the highest degree term on the top, which is 7x-squared, divided by the highest degree term on the bottom. 15x is going to grow, in
fact, this is right over here, this constant. So, as this becomes larger
and larger and larger, this is going to matter a lot, lot less. So, it's going to be approximately that. Which is equal to 7x over 15. Well, even here, if you
think about what happens when x becomes very, very negative here. Well, you're just gonna get larger, you're gonna get more and more and more negative values for f of x. So, once again, f of x
itself is going to approach, is going to go to, negative infinity as x goes to negative infinity. Let's do another one of these. So, here they're telling us to find the horizontal asymptote of q. A horizontal asymptote,
you can think about it as what is the function
approaching as x becomes, as x approaches infinity, or as x approaches negative infinity. And just as a couple of examples here. It's not necessarily the q
of x that we're focused on. But you could imagine a function, let's say it has a horizontal asymptote at y is equal to two, so that's y is equal to two there. Let me draw that line. So, let's say it has a
horizontal asymptote like that. Well then the graph could
look something like this. It could look, let me
draw a couple of them that have horizontal asymptotes. So, maybe it's over
here, it does some stuff, but as x gets really large,
it starts approaching, the function starts
approaching that y equals two without ever quite getting there. And it could do that on this side as well. As x becomes more and more negative. As it gets more negative, it approaches it without ever getting there. Or, it could do something like this. You could have, if it has
a vertical asymptote, too, it could look something like this. Where it approaches the
horizontal asymptote from below, as x becomes more negative, and from above, as x becomes more positive. Or vice versa. Or vice versa. So, this is just a sense of
what a horizontal asymptote is. It'll show you what's
the behavior, what value is this function approaching,
as x becomes really positive or x becomes really negative. Well, let's just think about it. We could essentially do what we just did in that last example. What happens if we were to, if we were to divide all of these terms by the highest degree
term in the denominator? Well, if we divide, so q of
x is going to be equal to, the highest degree term in the denominator is x to the ninth power. So, we could say six, 6x to the fifth divided by x to the ninth is going to be six over x to the fourth. And then minus two times x to the ninth. All of that over three over, I'm gonna divide this by x to the ninth, x to the seventh, plus one. Well, if x approaches
positive or negative infinity, six divided by arbitrarily large numbers, that's gonna go to zero. Two divided by arbitrarily large numbers, whether they are positive or negative, that's going to go to zero. So your numerator's
clearly gonna go to zero. This term of the
denominator, three divided by arbitrarily large numbers, whether we're going in the positive or the negative direction, it is gonna approach zero. It'll approach zero from
the negative direction, or we could say from below. If we're dealing with very negative x's. If we're dealing with very positive x's, then we're going to
approach zero from above. We're gonna get smaller and
smaller positive values. So, all of these things go to zero and this right over here is
going to be, would stay at, one. And so if you're approaching
zero in your numerator and approaching one in your denominator, the whole thing is going to approach zero. So, in the case of q of x, you have a horizontal asymptote at y is equal to zero. I don't know exactly
what the graph looks like but we could draw a horizontal
line at y equals zero and it would approach it. It would approach it from above or below. Let's do one more. What does f of x approach as x
approaches negative infinity? Well, let's divide all of these terms by the highest degree that
we see in the denominator. We see an x to the fourth. So, 3x to the fourth divided
by x to the fourth is three. Minus seven over x-squared,
I'm just dividing by x to the fourth, minus
one over x to the fourth, over, x to the fourth divided
by x to the fourth is one, minus two over x, plus
three x to the fourth. This is an equivalent,
this right over here is, for our purposes, for thinking
about what's happening on a kind of an end
behavior as x approaches negative infinity, this will do. I've just divided everything
by x to the fourth. And so what's gonna happen as x approaches negative infinity? This is going to approach zero. This is going to approach zero. This is going to approach zero. And this is going to approach zero. And so, as all of that
stuff approaches zero, what we're left with is
we're going to approach, we're going to approach three over one, or we could say just three. Another way you could
think about doing these is look at the highest degree terms. 3x to the fourth, x to the fourth. Ignore everything else because they're going to be overwhelmed by
these higher degree terms. So, you could say f of
x is approximately equal to 3x to the fourth over x to the fourth for large magnitude x. Magnitude x. And very negative is still
a very large magnitude, large absolute value. And so, 3x to the fourth
divided by x to the fourth, f of x is going to be
approximately equal to three. Or it's going to approach three. So, that's another way that
you could think about it.