Graphs of rational functions: vertical asymptotes | High School Math | Khan Academy

• [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. In 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 numbers, their 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 tells 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, then x equals three would be 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 me, 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 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 see that x is equal to 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 asymptote, 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 would coincide with what makes our denominator equal zero, so we could rule this out as well. So we could feel really good about choice C.