Algebra (all content)
- 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
Graphs of rational functions: vertical asymptotes
Sal picks the graph that matches f(x)=g(x)/(x²-x-6) (where g(x) is a polynomial) based on its discontinuities.
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- How can we be sure of what the question requires?like isn't there a way to figure out whether a function leads to the formation of a vertical asymptote or when it would lead to a discontinuity in the graph? For eg. in this ques. you said it could either be a vertical asymptote or a discontinuity.Isn't there a definite way out...so that we can look out for that particular thing itself.(6 votes)
- This example is a question about interpreting the parts of expressions. At first, rational functions seem wildly complicated. It isn't like the equation of a line, (linear function), f(x) = mx +b, where you just have slope and intercept to worry about. With a little practice, though, you can figure out a lot about a graph by looking at the parts of these rational functions.
In this case, the
only detailwe have is that there is a quadratic in the denominator. We have no information about the numerator. The denominator of a rational function can't tell you about the horizontal asymptote, but it CAN tell you about possible vertical asymptotes. What Sal is saying is that the factored denominator (x-3)(x+2) tells us that either one of these would force the denominator to become zero -- if x = +3 or x = -2. If the denominator becomes zero then the function is undefined at THAT point -- that value of x.
So, what happens when the denominator of a rational function becomes zero? It either just has a gap at that point (indicated by an empty dot, or little circle), or it has a vertical asymptote at that point which means that the function swoops toward positive or negative infinity, never crossing that point.
IFwe could see the numerator, we could find out whether each of the binomial factors cancels out with a corresponding binomial factor in the numerator. If so, it would be an empty circle on the graph (shows up as ERROR on table list of calculator), which means it is a
removable discontinuity. If the binomial factor remains in the denominator because it cannot be cancelled, it will show up as a
vertical asymptoteon the graph at the value of x that would be undefined.
So, to answer your final question, in this specific example, we cannot tell
whichwould happen without seeing the numerator. Instead we should look for
eithertwo asymptotes at the correct spots
ortwo empty dots at the correct spots
orone of each.(24 votes)
- How did he determine that 3 was the removable discontinuity?(6 votes)
- Sal checked what was happening at x = -2 and at x = 3.
At x = -2, the dotted line indicates an asymptote,, a line that the graph does not cross. But a removable discontinuity is a single point that cannot be included.
The open circle at x = 3 indicates that that specific point is not included. Therefore, that is the removable discontinuity.(4 votes)
- Can we consider rational function as a quotient of two functions ? somewhat draw their graphs through the intersection of the functions in the numerator and the denominator ?(4 votes)
- A graph that is a quotient of two functions is slightly different than just a function, because a quotient of two functions creates a removable discontinuity. For example, the lines y=x and y=x²/x are the exact same, except at the x-value of 0. It results in 0 for the first function but it is undefined in the second function(5 votes)
- Around2:09, Sal talks about x=3 possibly being a removable discontinuity. What about x=-2? We do not know the numerator ( g(x) ). So why doesn't Sal mention x=-2 possibly being a removable discontinuity? Could a possible graph show x=-2 as a removable discontinuity and x=3 as a vertical asymptote?(4 votes)
- I've never come across "removable discontinuities" before, but I think I grasp the basic concept. However, why is there one only at (x-3)(...)/(x-3)(x+2), x cannot equal 3, and not one at (x+2)(...)/(x+2)(x-3), x cannot equal -2? Could someone please explain this concept to me better, or direct me to useful material? Thanks!!(3 votes)
x+2was a factor of
x=-2would be removable discontinuity – there is nothing in this video that says otherwise.
In the situation Sal shows where
x+2is not a factor of the numerator (i.e. g(x)) then you get an infinite discontinuity.
Khan Academy material on types of discontinuities (including removable ones):
- Around2:15, Sal mentions "removable discontinuity." I watched that part 5 times but I still don't get it. Can someone explain using the example Sal provided?(3 votes)
- Removable discontinuities are defined in the prior section of videos. Here's a link: https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions#discontinuities-of-rational-functions
Sal's function has discontinuities at x=3 and x= - 2.
If the numerator had been defined and you we able to cancel out one of the factors in the denominator, then it would remove that discontinuity.
For example: f(x) = (x-3)(x+1) / [(x-3)(x+2)]
The factor (x-3) could be cancelled out. This would remove x=3 as a discontinuity.
Hope this helps.(2 votes)
- why the removable discontinuity is the specific x that makes both the denominator and numerator equal to zero?(2 votes)
- It's a discontinuity because plugging that value in doesn't give a number, it gives 0/0. It's a removable discontinuity because, at any point around there, whatever will make the numerator equal to 0 will cancel with whatever makes the denominator 0, and so we don't get asymptotic behavior or anything else weird.(3 votes)
- Why f(x) = (( x^(2)-x)) / (x^(2)-1) function has a vertical asymptote at x = -1 but not at x = 1?(2 votes)
- x=1 is a removable discontinuity.
If you factor the polynomials, you can see that the factor of (x-1) can be divided out which is why it is a removable discontinuity.(3 votes)
- So... the numerator can't be zero? Why is that?(1 vote)
- I'm assuming you meant why the denominator can't be zero – there is no reason why the numerator can't be zero ...
If so, this is because dividing by zero is incompatible with real numbers.
Ask yourself, what is the value of 1/0?
Let's assume that there is an answer and say it equals
Take the reciprocal of both sides:
Multiply both sides by a:
But since anything times zero is still zero:
0 = 1
This obviously can't be true! Therefore our original assumption that
1/0=amust be invalid ...(5 votes)
- So in what ways can an asymptote be represented. I've seen a dashed line so far and now I see an empty dot or a "hole". Is this "hole" another way of representing an asymptote/the excluded value of the graph which is defined by the horizontal/vertical asymptote?(2 votes)
- The asymptote is the dotted line. You will see the graph go to infinity or negative infinty as it approaches this line.
The empty dot or "hole" is called a discontinuity. The graph approaches that point directly, but can't ever equal that point.
Hope this helps.(3 votes)
- [Voiceover] We're told, let f of x equal g of x over x squared minus x minus six, where g of x is a polynomial. Which of the following is a possible graph of y equals f of x? And they give us four choices. The fourth choice is off right over here. And like always, pause the video, and see if you can figure it out or if you were having trouble with it as I start to think about it with you, pause it anytime, if you feel inspired. It's really important to try to engage in the problem as opposed to just watch me do it. But let's start tackling this now together. So, this is interesting. They don't give us a lot of information about f of x. If fact, we don't know what the numerator is. We just know that it is a polynomial. Well, that's somewhat valuable. But they do give us the denominator and so, we can think about what are the interesting numbers, what are the interesting x-values for the denominator. In particular, what x-values will make the denominator equal to zero? And to do that, we can factor out the denominator. So let's see, the coefficient on the first degree term is essentially negative one. We can write negative one there if we want. And the constant is negative six. So if we want to factor that, we can say, well, what two number,s they're product is negative six and they add up to negative one? Well, negative three times positive two is negative six. And negative three plus two is equal to negative one. So I can rewrite f of x. I can say that f of x is equal to g of x over x minus three times x plus two. So the denominator equals zero for x equals three or x equals negative two. That's when the denominator is zero. So zero denominator. I'll just write it like that. And so something makes the denominator equal to zero. That tell us that we're either going to have a vertical asymptote at that point or we're going to have a removable discontinuity at that point. And the way that that would be a removable discontinuity, let's say, if we had a removable discontinuity at x equals three, well that means that g of x could be factored into x minus three times a bunch of other stuff. If that was the case, the x equals three would a removable discontinuity. If x equals three does not make g of x equal zero. So, for example, if g of three does not equal zero, or g of negative two does not equal zero, then these would both be vertical asymptotes. So let's look at the choices here. So choice A, we have one vertical asymptote. That vertical asymptote is at x equals negative two. So it seems, this line, let me draw this line here. This vertical asymptote, right over there, that is a line, x is equal to negative two. So at least to be, it seems to be consistent with that over there but what about x equals three? This one seems completely cool. This graph is defined at x equals three. X equals three is right over there and it seems to be defined there. And this, f of x, is clearly not defined at f, at x is equal to three because when x equals three, the denominator is zero and dividing by zero is not defined. So even though this has one vertical asymptote at an interesting place, we're going to rule it out because this graph is defined at x is equals three, even though f of x is not. We would need to see either a vertical asymptote there or a removable discontinuity. Alright, here we have a vertical asymptote at x is equal to negative two and we have another vertical asymptote at x is equal to positive four. So that doesn't make sense either. This one, just like the last one, is actually defined at x equals three. We that x is equals three, the function is equal to zero. But this function? F of three is not equal to zero. F of three is undefined. We're dividing by zero. So we could rule this out. Once again, at x equals three, we need to see the removable discontinuity or a vertical asymptote, because we're not defined there. Alright, let's see choice C. We see a vertical asymptote at x is equal to negative two. So that looks pretty good. And we see a removable discontinuity at x equals three. So this function is, this graph is not defined for x equals three or for x equals negative two, which is good because f is not defined at either of those points because at either of those x-values, our f's denominator is equal to zero. So this one looks quite interesting. And this would be consistent. This one would be consistent with the, with f of x being something of the sort of, so the denominator, we already know. X minus three times x plus two. And in the numerator, we would have, since x minus three is not a vertical as-, since x equals three isn't a vertical asymptote, it's a removable discontinuity, we must be able to factor, for this one, g of x into x minus three times something else. So that's consistent with this one over here. So I like this choice. Now let's look at this choice, choice D. Choice D has two vertical asymptotes. One at x is equal to negative one. This is negative two, so this, x equals negative one. And this x is equal to six. Neither of them are, would coincide with what make our denominator equal zero, so we could rule this out as well. So we could feel really good about choice C.