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
Sal graphs y=(x)/(x^2-x-6). Created by Sal Khan and CK-12 Foundation.
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- Can an asymptote have an asymptote or a point where it doesn't exist?(4 votes)
- Interesting question, but I can't think of any way for either of those to happen.
The asymptotes you will typically see at this level are all lines (horizontal, vertical, or oblique/slant). Curved asymptotes do exist, but they are 'simple' polynomials not rational equations. This means that there is no denominator and so no way for division by zero to occur.
If you are interested in more about this subject, the following page should be helpful:
- The rational function that I'm graphing has an expression with three terms in the numerator and the denominator. I probably won't get a response in time, but for future reference; how would I go about with graphing this sort of rational function?(2 votes)
- Well, first see if you can get into factored form. If you can combine the three terms into factors of terms, you should be all set as you know the locations of the asymptotes and everything. If you can't factor anything, you can at least determine the asymptotes of the function by applying the rational roots theorem to the denominator. You can also find the horizontal/slant asymptote by looking at the degrees.(3 votes)
- how can the graph cross 0 if the horizontal asymptote is at 0?(2 votes)
- Graphs of rational functions can cross the horizontal asymptote because it may not be undefined at that value But, the graph will never cross a vertical asymptote. The vertical asymptotes occur where the function is undefined.(2 votes)
- How do I find the horizontal asymptote if the numerator does not contain an x?(2 votes)
- If the numerator is a constant and the denominator is a polynomial then the asymptote will always be at
- Why can the parts of the function at the far negative and far positive portions of the graph not pass through the horizontal asymmtote, while the part in the middle can? Also, how do you know whether a part of the function will pass through the horizontal asymmtote?
- The horizontal asymptote is not much like a vertical one, It's caused by trends as x gets very large, not by /0. So before |x| gets large things can be very different.
Just plot the graph according to the methods described so far and see where the points lie. Whether or not a function passes through a horizontal asymptote depends on the function.(2 votes)
- If a point crosses an asymptote, is that a point of discontinuity?(2 votes)
- I'm not sure what you mean by a point crossing an asymptote – do you mean a line?
The graph of a function can't cross a vertical asymptote, and thus vertical asymptotes are a type of discontinuity.
The graph of a function can cross a horizontal asymptote – no discontinuity.
Except for piecewise functions, you only get discontinuities when there is division by zero.(1 vote)
- If y=0 is an asymptote, how is there a value for x at y=0? Shouldn't an asymptote mean the graph is not touching that line at all?(2 votes)
- In all of the videos you explained how to find asymptotes when we have "something / something",but what"s the approach if i have "something +(something/something)",maybe something like this y=2x+(1/x-1),how to find the asymptotes of that? If the problem was 1/x-1 OK,similar to your examples , but how to approach it now?(1 vote)
- You just get the common denominator;
2x+(1/(x-1)) = (2x^2 - 2x)/(x-1)
Now you set the denominator equals zero;
x-1 = 0 -> x=1
Now, you set the limit to x->infinity;
lim y where x -> ± infinity =
lim x -> ± infinity = 2x + (1/(x-1)) = 2x
- Because, if you imagine that 1/x-1, and x was infinite large, it would be almost 0, right?
There you have it;
Diag. asymptote: Y= 2x
vertical asymptote: x=1(2 votes)
- What happens when the numerator has higher degree term than the denominator ?(1 vote)
- You have to divide the numerator by the denominator. Either long division or synthetic division, there won't be a horizontal asymptote, the asymptote will be oblique. Note, that there is an oblique asymptote only if the numerator degree is 1 greater than the degree of the denominator, if the degree is greater then 1, then there is no asymptote.(2 votes)
- i have the same question as "Malsonasojoshua7". in my home work i got a problem of "(3x+21)/((x^2)-x-2)" and i put it into Desmos Graphing calculator online and it gave me a line that crossed the asymptote, but if the x value where 0' then the whole top of the equation would become 0 and wouldnt that make the function undefined?(1 vote)
- The function becomes undefined only if the denominator (the bottom) becomes 0.
For example: 0/5 just = 0; but 5/0 = undefined.
The denominator becomes 0 if x=-2 or x=1.(2 votes)
Let's graph another rational function, because you really can't get enough practice here. So let's say we have y is equal to x over x squared minus x minus 6. So the first thing we might want to do is just factor this denominator so we can identify our vertical asymptotes, if there are any. So what two numbers when I take their product I get negative 6 and if I add them up I get negative 1? So they have to be of different signs. So one's going to be a plus and one-- let me write my x's a little bit neater than that-- so one is going to be a positive and one is going to be a negative. A 2 and a 3 seem to be pretty close, because they're one apart, and I'm going to subtract the larger number because when I add them, I get a negative. So x minus 3 times x plus 2 seems to work. That gets negative 6. Negative 3x plus 2x. negative 3 times x plus 2 times x is negative x, so that works. So this is equal to x over x plus 2 times x minus 3. And like we saw in the last video, since these expressions, since the x plus 2 doesn't cancel out with anything in the numerator and the x minus 3 doesn't cancel out with anything in the numerator, we know that these can be used to find our vertical asymptotes. The vertical asymptotes are when either that term is equal to zero or when that term is equal to zero, because at those points, our equation is undefined. So this is equal to zero when x is equal to negative 2, and this is equal to zero when x is equal to positive 3. And you could try it out here. If x is equal to negative 2 or positive 3, you're going to get a zero in the denonminator, y will be undefined. So vertical asymptotes at x is equal to negative 2. So there's a vertical asymptote, a vertical asymptote right there. Another vertical asymptote is x is equal to 3. One, two, three. There is our other vertical asymptote. Now let's think about horizontal asymptotes, or if there are any. So what happens as x gets super-positive or super-negative? And as we said before, you just have to look at the highest degree term on the numerator and the highest degree term on the denominator. Now, notice the highest degree term on the denominator is x squared, while the highest degree term on the numerator is only an x. So when x gets really large, what's going to happen? You could imagine, this is going to be like a million over a million squared, which is still one over a million. These terms over here don't matter much. But this term right here is going to grow faster than everything. This is an x squared term. As x gets large, it's going to get way larger than everything, including this term on the top, so it's essentially going to go to zero. When the denominator just gets bigger, faster than the numerator as you're going to approach zero. So we have a horizontal asymptote at y is equal to 0. I could draw it as a dotted line over our x-axis. So that right there is the line, y is equal to 0. Once again, we identify that looking at the highest degree term there. The denominator has a higher-degree term, so it's going to grow faster than the numerator. You could try it out on your calculator. And that's true whether you go in the super-negative direction or the super-positive direction. This thing is going to overwhelm this thing up here, the denominator grows faster than the numerator, which essentially we're going to approach zero. You're going to get smaller and smaller fractions. Just remember, 1/10 and then-- let me actually just-- as x gets larger and larger and larger, what's going to happen? Let me just show you on my calculator. Let's say x is equal to 10. 10 divided by 10 squared minus 10, and normally you wouldn't have to do this. I just really want to show you the intuition. Whoops! I'm not trying to graph. Let me exit from here. So if we have 10 over 10 squared minus 10, once again, you normally wouldn't have to do this. I just want to show you, give you the intuition. Let me put some parentheses there. Let me put some parentheses over here. So let me insert the parentheses there and put a parentheses over here. You get a small number. What happens if x gets even larger? Let me make, instead of a 10, let me make it all 100. Let me make these tens into hundreds, into 100. Insert 100 there, what do we get? We get even a smaller number. And if you try x is equal to 1,000, it's going to be be even smaller than that. That's because this term right here is growing faster than every single other term. That's why our horizontal asymptote is y is equal to 0. Now, the last thing we want to do, we've drawn all of our asymptotes, is just try out some points. So let's draw like a little table here. There's our table. When x is equal to 0, what is y? x is 0, we have 0 over all of this. 0 minus 6, 0 over negative 6 is just 0. When x is equal to-- I don't know, let's just try when x is equal to 1, what do we have? We have 1 over-- I'll write it here. 1 over 1 squared minus 1. Now that's just 0, so we have negative 6. When x is equal to negative 1, what do we have? When x is equal to negative 1, we have negative 1 over negative 1 squared, which is 1 minus negative 1. So that's plus 1-- right, minus negative 1-- minus 6. So what is this right here? This is negative 1, so this is going to be negative 1 over 2 minus 6 over negative 4. This is going to be equal to 1/4, so we're going to get a positive value. So we have-- let me draw this-- negative 1, we're at 1/4 right here. That's about right there. I'll do it in a darker color. We had the point 0, 0, and then at x is equal to 1, we had negative 1/6. So you could keep graphing more and more points, but it looks like as we approach this vertical asymptote from the right, we go to positive infinity. And that should make sense. Let's see, if we were to put in-- we're approaching negative 2 from the right. So if you were to put in negative 1.9999999, this term is going to be a very small positive number. This term's going to be a negative number. This term's going to be negative number. The negatives cancel out. You have a very small positive number in the denominator. 1 over that gives you a very positive number. Now, as we approach the other vertical asymptote from the left, we're going to go super-negative. My gut tells me that because when I tried x is equal to 1, I already went to a negative value. But you could imagine if you did 2.99999, right? Let me draw that a little bit better. You get the idea. If x is equal to 2.999, so we get really close to the asymptote, this is going to be positive, this is going to be negative, that's going to be positive, and this is going to be a small number. So you're going to have 1 over a very small negative number, which is a very, I guess, negative number. It's a negative of 1 over a very small number, so you're going to approach negative infinity. Now, let's try some point out here to see what happens. So what happens when x is equal to 4? When x is equal to 4, you have 4 over 16 minus 4 minus 6. What is that? That's 16 minus 10. That is 6. So this is equal to 4/6, which is equal to 2/3. So the point 4, 2/3 is here, so one, two, three, four. 2/3, just like that. So that gives me the sense, look, I have to approach this horizontal asymptote as we go further and further out. We're going to probably approach positive infinity like this. Let me draw it a little neater than that. Like that. You get the idea. Then over here, we're going to get closer and closer to our horizontal asymptote as we approach infinity. This should be a smoother-looking curve right around there. I'm making a mess here. This should be a smoother-looking curve. You get the idea, I think. Now, let's see what happens when x is equal to negative 3. So when x is equal to negative 3, we have negative 3 over negative 3 squared, which is 9 minus negative 3, so that's plus 3 minus 6. So what is this equal to? This is equal to negative 3 over-- this is 12 minus 6 over 6, right, which is equal to negative 1/2, So negative 3, negative 1/2. Negative 1/2 is right there. So we're going to approach this asymptote as we get really negative. And we're probably going to go straight down like that as we approach this vertical asymptote right there. And you could try more points if you don't believe me. But let us graph it just to verify it for ourselves. So our equation is x divided by x squared minus x minus 6. And let's graph it. And there you go. There you go. All right, looks pretty good. Our asymptote is zero, we go down. Vertical asymptote, bam! Go up there, then we go back down here, then we go just like that again. So once again, this looks just about exactly what we got. Obviously, the graphing calculator, it kind of a pitters out as you get close to these values and it does weird things, but it has the same general shape. We could actually close the range a little bit if we want, if we want to graph it. Let's make our x minimum value, let's make it 5. And let's take our x maximum value, let's make that 5 as well. We're kind of zooming in a little bit. So let's graph it now. Bam! bam! There you go! It's the same shape as we graphed right here. Hopefully, you found that satisfying.