Integrated math 3
- 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
Graphing rational functions 2
Sal graphs y=(2x)/(x+1). Created by Sal Khan and CK-12 Foundation.
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- What if the denominator doesn't factor out? In my example that I am working on, the numerator is "2" and the denominator is "x^5 - 3x^3"
And thanks Salman, your videos are saving me!!(10 votes)
- If the denominator doesn't factor out in the case of finding asymptotes, then you wouldn't have a vertical asymptote or a horizontal asymptote. You would still be able to find points to plot but you wouldn't have any asymptotes.
Hope this helps!(1 vote)
- From y = (x+1)/(x+1) - no vertical asymptote. Since a vertical asymptote is when f(x) is undefined and that is undefined when the denominator = 0. Since 0/0 has no vertical asymptote - can someone explain the difference between x/0 and 0/0?
- The reason why y = (x + 1)/(x + 1) does not have a vertical asymptote is because the equation simplifies to y = 1. Instead of an asymptote, however, there must be a hole at -1.(4 votes)
- If I don't have a horizontal asymptote, but I have a slant asymptote how do I find where, and if, it crosses the x axis?(4 votes)
- Yes, it will cross the x-axis if you have a slant asymptote (oblique asymptote). I am assuming here that you are talking about oblique asymptotes of rational functions and not the oblique asymptotes of hyperbolas (which are not functions).
The rules for determining that you have an oblique asymptote are complicated and then there are some steps for finding the x-intercept. First you determine whether you have a proper rational function or improper one. Only improper rational functions will have an oblique asymptote (and not all of those). A proper one has the degree of the numerator smaller than the degree of the denominator and it will have a horizontal asymptote. An improper rational function has either the same degree or a larger degree in the numerator.
If the degree of the polynomial in the numerator is the same as the degree of the polynomial in the denominator, you will have a horizontal asymptote but no oblique asymptote.
If the degree of the numerator is greater by two or more, then you do not have a horizontal asymptote --nor an oblique one.
Finally! If the degree of the numerator is greater by ONE than the degree of the denominator, then you have an oblique asymptote. You can find the equation of that asymptote by dividing your polynomials to find the quotient, which should be of the form y = mx+b -- the familiar equation of a straight line with slope m. You can find the x-intercept by substituting in a value of zero for the y and solving for x. That is where the oblique asymptote crosses the x-axis.
Note that an oblique asymptote shows how the function behaves, but the function may cross the oblique asymptote before it settles down unlike vertical and horizontal asymptotes which the function approaches but never touches.(5 votes)
- why isnt 0 over 0 just 0?(2 votes)
- It helps to think of division as the inverse of multiplication. In that case, 0/0 would be 0 *(1/0). Before you can complete that multiplication, think about what 1/0 is: the number that, when multiplied by 0, yields a product of 1. Since there is no such number (0*x = 0 for all x, so nothing can be multiplied by 0 to get 1,) 0 has no multiplicative inverse, and therefore 0/0 can not exist. It's undefined.(5 votes)
- Is there any sort of explanation why something like (x+1) in the denominator fails to be a vertical asymptote when it gets canceled out with the same expression in the nominator? I didn
t actually get it from the video. Cant it still be an asymptote? The graph still approaches it, but never touches.(1 vote)
- A vertical asymptote happens because as you get closer and closer to the point where you'd divide by zero - in this case, x=-1 - your result is going to keep getting larger and larger (or smaller, if the number is negative. The absolute value becomes huge). That creates the vertical asymptote. The exception is when it cancels out, meaning there's another (x+1) in the nominator... no matter how tiny x+1 is, (x+1)/(x+1) is still going to be 1, so you only get the "hole" in the graph.(8 votes)
- so exactly how do you find the horizontal asymptote?(2 votes)
- Another solution, as Sal mentioned in his first video, is to look at the degree of the numerator and denominator.
If the degrees are equal, the horizontal asymptote is the quotient of the leading coefficient of the numerator and the leading coefficient of the denominator.
If the degree is higher in the denominator, the horizontal asymptote is y = 0 (the x-axis).
If the degree is higher in the numerator, there is no horizontal asymptote.(4 votes)
- Can someone give me an example where there's NO asymptote AT ALL?(2 votes)
- f(x)=(x+3)/(x+3) There are no asymptotes. Just a hole.(4 votes)
- Shouldn't curves be in quadrant 1 and 3 because the graph is positive?I don't understand why the curves became to be on the quadrants 2 and 4...(2 votes)
- Just because "the graph is positive" doesn't mean it needs to be in quadrants 1 and 3. All it will mean is that the curve rises as you move from left to right. You can see that this is the case -- as we move to the right along the top-left-hand curve, the y-value rises, and as we move to the right along the bottom-left-hand curve, the y-value also rises.(4 votes)
- At4:34, is there no horizontal asymptote in this expression? Can anyone explain why? I thought a rational expression must have an asymptote.(3 votes)
- No, not all rational expressions have vertical or horizontal asymptotes. Some have oblique asymptotes instead.
For example, (x³+5x +1) / (x²+2x + 7) has no vertical asymptote because the denominator has no rational zeros. But, because the numerator has a higher degree than the denominator, it does not have a horizontal asymptote either. It does have an oblique asymptote of x − 2.
So, rational expressions may have, but do not necessarily have vertical or horizontal asymptotes. Here is the general pattern:
If there are no rational zeros of the denominator, there are no vertical asymptotes.
If there are rational zeros in the denominator, then those are possible locations of vertical asymptotes, but they must be checked to make sure.
If the degree of the denominator is greater than the degree of the numerator, then the horizontal asymptote is the x axis.
If the degree of the numerator and denominator are equal, then there is a non-zero horizontal asymptote.
If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote but instead there is an oblique asymptote.(1 vote)
- is horizontal line is undefined slope and vertical slope has zero slope?(1 vote)
- Slope is defined as change in y divided by change in x.
A horizontal line has no change in y, so its slope is 0 / change in x
which equals zero.
A vertical line has no change in x, so its slope is change in y / 0.
Since division by 0 is not allowed, that means that the slope of the vertical line is undefined.(5 votes)
Let's do a couple more examples graphing rational functions. So let's say I have y is equal to 2x over x plus 1. So the first thing we might want to do is identify our horizontal asymptotes, if there are any. And I said before, all you have to do is look at the highest degree term in the numerator and the denominator. The highest degree term here, there's only one term. It is 2x. And the highest degree term here is x. They're both first-degree terms. So you can say that as x approaches infinity, y is going to be-- as x gets super-large values, these two terms are going to dominate. This isn't going to matter so much. So then our expression, then y is going to be approximately equal to 2x over x, which is just equal to 2. That actually would also be true as x approaches negative infinity. So as x gets really large or super-negative, this is going to approach 2. This term won't matter much. So let's graph that horizontal asymptote. So it's y is equal to 2. Let's graph it. So this is our horizontal asymptote right there. y is equal to 2, that right there-- let me write it down-- horizontal asymptote. That is what our graph approaches but never quite touches as we get to more and more positive values of x or more and more negative values of x. Now, do we have any vertical asymptotes here? Well, sure. We have when x is equal to negative 1, this equation or this function is undefined. So we say y undefined when x is equal to negative 1. That's definitely true because when x is equal to negative 1, the denominator becomes zero. We don't know what 1/0 is. It's not defined. And this is a vertical asymptote because the x doesn't cancel out. The x plus 1-- sorry-- doesn't cancel out with something else. Let me give you a quick example right here. Let's say I have the equation y is equal to x plus 1 over x plus 1. In this circumstance, you might say, hey, when x is equal to negative 1, my graph is undefined. And you would be right because if you put a negative 1 here, you get a 0 down here. In fact, you'll also get a 0 on top. You'll get a 0 over a 0. It's undefined. But as you can see, if you assume that x does not equal negative 1, if you assume that this term and that term are not equal to zero, you can divide the numerator and the denominator by x plus 1, or you could say, well, that over that, if it was anything else over itself, it would be equal to 1. You would say this would be equal to 1 when x does not equal negative 1 or when these terms don't equal zero. It equals 0/0, which we don't know what that is, when x is equal to negative 1. So in this situation, you would not have a vertical asymptote. So this graph right here, no vertical asymptote. And actually, you're probably curious, what does this graph look like? I'll take a little aside here to draw it for you. This graph right here, if I had to graph this right there, what this would be is this would be y is equal to 1 for all the values except for x is equal to negative 1. So in this situation the graph, it would be y is equal to 1 everywhere, except for y is equal to negative 1. And y is equal to negative 1, it's undefined. So we actually have a hole there. We actually draw a little circle around there, a little hollowed-out circle, so that we don't know what y is when x is equal to negative 1. So this looks like that right there. It looks like that horizontal line. No vertical asymptote. And that's because this term and that term cancel out when they're not equal to zero, when x is not equal to negative 1. So when your identifying vertical asymptotes-- let me clear this out a little bit. when you're identifying vertical asymptotes, you want to be sure that this expression right here isn't canceling out with something in the numerator. And in this case, it's not. In this case, it did, so you don't have a vertical asymptote. In this case, you aren't canceling out, so this will define a vertical asymptote. x is equal to negative 1 is a vertical asymptote for this graph right here. So x is equal to negative 1-- let me draw the vertical asymptote-- will look like that. And then to figure out what the graph is doing, we could try out a couple of values. So what happens when x is equal to 0? So when x is equal to 0 we have 2 times 0, which is 0 over 0 plus 1. So it's 0/1, which is 0. So the point 0, 0 is on our curve. What happens when x is equal to 1? We have 2 times 1, which is 2 over 1 plus 1. So it's 2/2. So it's 1, 1 is also on our curve. So that's on our curve right there. So we could keep plotting points, but the curve is going to look something like this. It looks like it's going approach negative infinity as it approaches the vertical asymptote from the right. So as you go this way it, goes to negative infinity. And then it'll approach our horizontal asymptote from the negative direction. So it's going to look something like that. And then, let's see, what happens when x is equal to-- let me do this in a darker color. I'll do it in this red color. What happens when x is equal to negative 2? We have negative 2 times 2 is negative 4. And then we have negative-- so it's going to be negative 4 over negative 2 plus 1, which is negative 1, which is just 4. So it's just equal to negative 2, 4. So negative 2-- 1, 2, 3, 4. Negative 2, 4 is on our line. And what about-- well, let's just do one more point. What about negative 3? So the point negative 3-- on the numerator, we're going to get 2 times negative 3 is negative 6 over negative 3 plus 1, which is negative 2. Negative 6 over negative 2 is positive 3. So negative 3, 3. 1, 2, 3. 1, 2, 3. So that's also there. So the graph is going to look something like that. So as we approach negative infinity, we're going to approach our horizontal asymptote from above. As we approach negative 1, x is equal to negative 1, we're going to pop up to positive infinity. So let's verify that once again this is indeed the graph of our equation. Let's get our graphing calculator out. We're going to define y as 2x divided by x plus 1 is equal to-- delete all of that out-- and then we want to graph it. And there we go. It looks just like what we drew. And that vertical asymptote, it connected the dots, but we know that it's not defined there. It just tried to connect the super-positive value all the way down. Because it's just trying out-- all the graphing calculator's doing is actually just making a very detailed table of values and then just connecting all the dots. So it doesn't know that this is an asymptote, so it actually tried to connect the dots. But there should be no connection right there. Hopefully, you found this example useful.