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### Course: Integrated math 3 > Unit 13

Lesson 4: Graphs of rational functions- Graphing rational functions according to asymptotes
- Graphs of rational functions: y-intercept
- Graphs of rational functions: horizontal asymptote
- Graphs of rational functions: vertical asymptotes
- Graphs of rational functions: zeros
- Graphs of rational functions
- Graphs of rational functions (old example)
- Graphing rational functions 1
- Graphing rational functions 2
- Graphing rational functions 3
- Graphing rational functions 4

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# Graphing rational functions 1

Sal graphs f(x)=(2x+10)/(5x-15). Created by Sal Khan.

## Want to join the conversation?

- How do you find the Horizontal Asymptote if x is multiplied to a higher degree in the denominator than in the numerator? For example: 5x-10/x^2+4x-5(19 votes)
- If the denominator has a higher degree term than the numerator, the horizontal asymptote will be 0. ALWAYS. If the numerator has a higher degree term than the denominator, there is no horizontal asympotote. When the numerator and denominator both have the same highest degree term (the highest in the numerator and denominator) is when you can divide their coefficients (number in front of a variable). There will be just one horizontal asymptote if that is the case. Also, in the video in this section "asymptotes of rational functions" he covers the question you just asked.

http://www.mathops.com/free/a1rf010.php(74 votes)

- How do you find the holes, or as Sal calls them: breaks (see:2:40), in a rational function? And how do you know if a function even has a hole?(11 votes)
- A hole is where a simplified function is undefined. For example, let's say you have:

f(x) = (x + 5)(x + 2) / (x + 5)

You can simplify it by cancelling out the (x + 5) in the numerator and denominator.

f(x) = x + 2

You may think that because this function has no holes at all because there are no points where it is undefined. However, you always have to look at the original equation to find the holes. In the original equation, you had x + 5 in the denominator. To make x + 5 equal to 0, you would need x to be -5. So at -5, f(x) will have a hole there.(24 votes)

- What if f(x) = 1/x, how would you find the asymptotes?(4 votes)
- To find the asymptotes of a function, first recognize that there are three types: vertical, horizontal, and oblique. I can't do as good of a job as Sal, but I can work through this example with you. First, observe that what you have is a rational function. These are the easiest to deal with. Next, recognize what an asymptote actually is: it's a line that the function will get very, very close to, but will never reach.

To find all horizontal asymptotes, observe what happens to y as x gets larger and larger (or more and more negative). If y approaches a specific value, then you have a horizontal asymptote. In your example, As x gets really big, y gets really, really small. Y actually gets infinitely close to zero as x gets infinitely larger. So, you have a horizontal asymptote at y = 0. Applying the same logic to x's very negative, you get the same asymptote of y = 0.

Next, we're going to find the vertical asymptotes of y = 1/x. To do this, just find x values where the denominator is zero and the numerator is non-zero. This clearly happens at x = 0 and nowhere else. So, as we get very close to 0 in x, the y values will approach positive and negative infinity. So, we are actually getting close to forming the vertical line x = 0. This is another asymptote of y = 1/x.

To find oblique asymptotes, the rational function must have the numerator's degree be one more than the denominator's, which it is not. So, there are no oblique asymptotes.

Summing this up, the asymptotes are y = 0 and x = 0. To confirm this, try graphing the function y = 1/x and zooming out very, very far. The graph should look like the lines x = 0 and y = 0.

Sal does a much better job than I do. Take a look at his video here: https://www.khanacademy.org/math/algebra2/polynomial_and_rational/asymptotes-graphing-rational/v/asymptotes-of-rational-functions(19 votes)

- How do you find the horizontal asymptote in a problem like (5x^2-45x+100)/(10x^2-10x-20) ?(4 votes)
- (5x^2-45x+100) and (10x^2-10x-20) both have the same highest degree of x: x^2

Because of that you can divide the coefficients of x^2 to find the horizontal asymptote.

5/10 = 1/2 so, the horizontal asymptote is y = 1/2(4 votes)

- I am asked to graph y=10/x ... how am I supposed to graph that? Is there any tutorial on that?(3 votes)
- 10/x is part of the reciprocal function family and also its an example of inverse variation. well to graph that:

1. make a tale of positive and negative values

2.plot the points from the table

3. (optional) you can also check x/y - intercepts, asymptotes, and end behavior to get a better feel of the graphs shape and continuity/discontinuity(2 votes)

- Hi I have a question about finding the graph of this function: f(x)=((x+1)(x^2-x-2))/((x-1)(x+2))

I've found the x and y-intercept, vertical and horizontal asymptote (in this case there is no horizontal asymptote). What do I need next to be able to graph it fully?(2 votes)- There's no horizontal asymptote, but there is what is called a slant, or oblique asymptote. Basically it just means it's diagonal. how do you determine how diagonal? Let's find out.

First let's factor the entire thing into binomial terms.

((x+1)(x^2-x-2))/((x-1)(x+2))

((x+1)(x+1)(x-2))/((x-1)(x+2))

Now nothing cancels so this is what we hae to work with. The fact nothing cancels means no holes as well.

Now, to find horizontal/ slant asymptotes we look at the degree of the numerator over the degree of the denominator. I assume you know but just to keep things ordered as I would do it to solve the problem, equal degrees would have the horizontal asymptote be the ratio of the leading coefficients, if the degree of the denominator is larger then the HA is 0 and finally we have the degree of the numerator greater.

To find the slant asymptotes you perform the division depicted in the rational function To do this I am going to expand both the numerator and denominator, if you have trouble with polynomial long division let me know.

(x+1)(x^2-x-2) = x^3-x^2-2x+x^2-x-2 = x^3-3x-2

(x-1)(x+2) = x^2+2x-x+2 = x^2+x+2

So now we have (x^3-3x-2)/(x^2+x+2)

Again, if you need help with the division let me know, but in the end you wind up with x-1 - 4x/(x^2+x+2) Now we use this to find the oblique asymptote. Essentially we look at what this heads to as x goes to infinity. The answer is x-1, as the last term heads to 0.

So there is no horizontal asymptote, but the function does head toward the line x-1 as x heads toward infinity.

I hope this helpped, if not let me know though.(4 votes)

- when will there be more than one horizontal asymptote?(3 votes)
- Rational functions will never have more than one horizontal asymptote. However there are functions that exist that have more than one horizontal asymptote like the logistic function.(2 votes)

- In a rational expression, when I have a numerator that is equal to zero (0) but the denominator is not equal to zero (0) what do I have. The expression is equal to zero (0) but what do I have? Example, what is the expression then called?

At 0.59 seconds Sal talks about the function equaling zero (0).(2 votes)- 1) 0 divided by any value (except 0) = 0

2) 0 divided by 0 is indeterminate. It can be made to equal anything. There is no single agreed upon solution.

3) Any non-zero value divided by 0 is undefined.

Hope this helps.(2 votes)

- 9:13Sal said because we still have the +10 and -15, shouldn't the fact that we can never reach infinity another reason we can't get to 2/5 ?(2 votes)
- While that's a good way to thing, and how many rational functions do work, in this case you have (2x+10)/(5x-15). Well, what would happen if you just had 2x/5x?

It would just be a line equivalent to y = 2/5 with a hole at x=0, so the only reason we get the asymptotes is the face we have more than just monomial terms in the numerator and denominator.(2 votes)

- When trying to find the x-intercept and there is not an x in the numerator, what should be done?(1 vote)
- If there is no x in the numerator and there is no other term added to the fraction, then the x intercept will be undefined.(3 votes)

## Video transcript

right over here I have the graph of f of
X and what I want to think about in this video is whether we could have sketched
this graph just by looking at the definition of our function which is
defined as a rational as a rational expression we have 2x plus 10 over 5x
minus 15 so there's a couple of ways to do this first you might just want to
pick out any numbers that are really easy to calculate so for example what
happens when X is equal to zero so we could say F of 0 F of 0 is going to be
equal to well all the X terms are going to be 0 so you're going to be left with
10 over negative 15 which is negative 10 15 which is negative 2/3 so you could
plot that when x equals 0 f of x or y equals f of X is negative 2/3 and we see
that point let me do that in a darker color you see that point right over
there so we could have plotted that point we could have also said when does
this function equal zero well the function is equal to zero when the only
way to get the function equal to zero is if you get this numerator equal to zero
so you could try to solve 2x plus 10 is equal to zero well that's going to
happen when 2x is equal to negative 10 I just subtracted 10 from both sides and
if I divide both sides by 2 that's going to happen when X is equal to negative 5
and you see that you see this right over here when X is equal to negative 5 the
function intersects the x-axis so that's just two points but that still doesn't
give us enough to really form this interesting shape over here and you
could think about what other functions have this type of this type of shape so
now what I want to think about is the behavior of the function at different
points so first I want to think about when this function is undefined and what
type of behavior we might expect for that function when it's undefined so
this function is going to be undefined the only way I can think of to make this
undefined is if I make the denominator equal to zero we don't know what it
means to divide by zero that is undefined so this the function is going
to be undefined when five x this in blue when five X minus 15 is equal to zero or
adding 15 to both sides when five X is equal to 15 or dividing both sides by
five five when X is equal to three f is
undefined now there's a couple of ways for a function to be undefined at a
point you could have something like this so let's let me draw some axes right
over here so let's say that this is three you could have your function it
could look like this it could be defined it might kind of approach something but
just not be defined right at three and just keep on going like that or the
other possibility is it might have a vertical asymptote there and if it has a
vertical asymptote it's going to look something like this it might it might be
approach it might just pop up to infinity and then it might pop down from
infinity on this side or it might go from negative infinity right over here
that says that's what a vertical asymptote would look like that as we
approach from the left the graph is approaching a vertical but it never it
never quite gets two x equals three I guess one we could say it or the
function is defined not defined at x equals three and as you approach from
the right the same thing the function just in this case drops down it almost
becomes vertical its approaching negative infinity as X approaches three
from the positive direction so how would we know obviously when you look at here
when you look we know the graph ahead of time and if you say okay this is e this
is x equals ten so let's see how many this is one two three four five so each
of these are two so x equals three is right over here
when you look at the graph if you had the graph in front of you you'd see oh
look this is indeed a vertical asymptote just looking at the graph you see that
you have a vertical asymptote at x equals three and let me write that down
vertical asymptotes vertical asymptotes asymptote at x equals three at x equals
three but how would you have known that how would you have known that if you
didn't have the graph here if you just had this we know it's not defined at
three but how do we know it's not a point discontinuity and not a instead of
a vertical asymptote well there's a couple of ways to do it one ways you
could try values near three and see what happens
so for example you could get your calculator out and you could try let's
say three point zero one so if you say two times three point zero one plus ten
X G well that's the numerator and then I'm going to divide that the numerator
by five times three point zero one minus fifteen it gets us a fairly large number
it's kind of exploding on us and if we've got even closer so if we did two
times three point zero zero one plus let me plus ten plus ten divided by five
times three point zero zero one so now I'm just now I'm trying X is equal to
three point zero zero one minus 15 minus 15 we see we get even a larger number so
as X gets closer and closer to three f of X seems to be exploding it seems to
be approaching positive infinity so that's one way to say okay this looks
like at least from this side we are approaching we are approaching positive
infinity so we would have been able to draw we would have been able to draw
something like that and then you could have tried values below you could have
tried values below so you could have said I actually let me just let me just
put the last entry here and let me just change the 3.00 ones to two point nine
nine nine two point nine nine nine and whoops let me go over here we have two
point nine nine nine and we get notice we're going really negative now we're
approaching negative infinity so if you just try that out that would give you a
pretty good indication that the graph is looking something like that right over
here which also seems to match connecting these two points that we've
already thought about but now let's see what's going on as X approaches really
large values or really negative really positive values are really negative
values it looks like there is a horizontal asymptote here just looking
at the graph just looking at the graph it seems like there's some value that as
X approaches really large values really positive values f of X is going to be
approaching that value that asymptotes from above
and as X becomes really negative it looks like f of X is approaching that
from below but how would we be able to figure that out just by looking at this
well one thought experiment is just to say well what happens to f of X as X
approaches infinity so let me write that down as X approaches infinity then then
f of X is going to approach what well as X approaches larger and larger values
the positive 10 and the negative 15 start to matter a lot less that the
highest degree term in the numerator denominator start to dominate so we
could say as X approaches infinity f of X is getting closer and closer to 2 x
over 5 X which is which is 2/5 which is equal to 2/5 so you could say f of X is
approaching 2/5 and if you really want to kind of see that a little bit more a
little bit more concretely let's imagine different values for X as X gets larger
and larger and larger so if we have so X f of X if X is 1 that f of X is just
going to be 2 plus 10 over 5 minus 15 so here the 10 and the negative and the
subtracting the 15 matter a lot but if X was if X were a thousand then f of X
would be 2,000 plus 10 over 5,000 minus 15 so now the 2,000 and 5,000 are really
kind of setting the agenda and then if X were if X were let's say 1 million 1
million and I just used blue for a more contrast then f of X would be 2 million
2 million plus 10 we move over to the right a little bit 2 million plus 10
over 5 million 5 million minus 15 so here the 10 and the 15 are almost
inconsequential you could imagine if X were a billion or trillion or googol
then these then the 10 and the negative 15 starting to matter a lot and lot less
so as X approaches infinity these matter less the higher degree terms the
degree terms matter so f of X is going to approach 2x over 5x which is 2/5 so f
of X is approaching 2/5 and that's what this line looks like 2/5 is the same
thing as 0.4 so f of X and we see that in the graph f of X is approaching that
but it's not quite getting close to it it's it was it's not quite getting there
it's getting closer and closer to it as X goes to infinity but it's not quite
getting there because you're always going to have that plus 10 and that
minus 15 there's you're never going to be exactly at 2/5 and the same thing is
happening as X gets more and more and more negative you can make all of these
negative values so if if this were negative 1 then it would be negative 2
negative 5 this was negative 1000 would be negative 2000 over negative 5000 if
there were negative 1 million it would be negative 2 million over negative 5
million but you see even in this case f of X is approaching 2x over 5x which is
approaching 2/5 or you could say it's approaching negative 2 over negative 5
which is still 2/5 and you see that right over here so we would say that
this function has a horizontal asymptote horizontal horizontal asymptote
asymptote at y equals the horizontal line right over here is y is equal to
2/5 so hopefully this graph here is helping us appreciate what these
vertical and horizontal asymptotes actually are but if we didn't have the
graph we could have said okay we're undefined at x equals 3 we could test
some values around it and so we could say okay look it does look like we're
approaching negative infinity as X approaches negative 3 from the left it
looks like we're approaching positive infinity as X approaches negative 3 from
the right so we could we could put we could draw that kind of blue point right
there we could graph these two points when does F equal 0
what happens to F when x equals 0 and then we could think about the behavior
as X approaches infinity or negative infinity as X approaches negative as X
approaches infinity or negative infinity and draw this horizontal asymptote and
between all of those that would have been a pretty good way to be able to
sketch this actual graph