- 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
Sal picks the graph that matches f(x)=(2x²-18)/g(x) (where g(x) is a polynomial) based on its zeros.
Want to join the conversation?
- The example Sal gives at5:51,
(2(x+3)(x-1))/ (x+3)^2(x+1) I don't understand why x at -3 isn't a removable discontinuity. If you put -3 in the original equation without simplying, you get 0/0.
Thank you.(7 votes)
- A removable discontinuity would divide COMPLETELY out of the denominator.
In this case, after the division there is still a factor of ( x + 3 ) left in the denominator,
so that takes precedence over the fact that a single ( x + 3 ) was able to be divided out
and indicates an asymptote at x = -3 instead of a removable discontinuity.(4 votes)
- So I'm still confused about how you would solve a problem like this on your own, without knowing which random examples to pick?(4 votes)
- g(x) can be anything, so Sal is choosing factors that cancel out factors in the numerator or create asymptotes.(3 votes)
- Hello all, could someone please explain how the answer choice is A? Thanks :)(3 votes)
- The zeros of the numerator are -3 and 3. So, at x = -3 and x = 3, the function should have either a zero or a removable discontinuity, or a vertical asymptote (depending on what the denominator is, which we do not know), but it must have either of these three “interesting” behaviours at x = -3 and x = 3. The answer A is the only one that has “interesting” behaviour at both x = -3 and x = 3.(5 votes)
- By the end of the video, Sal disregard the choice letter c. Isn't it possible to have a discontinuity at x = 3 since we don't exactly know what g(x) is? It could have and (x-3) or (x+3) so it's possible to have a discontinuity at 3 right?(1 vote)
- It is certainly possible to have a discontinuity at x = 3, but the problem with choice c is that its asymptote is not at x = -3.
Listen again carefully from about6:20to6:36where Sal sums up various situations that may be indicated by the numerator.(7 votes)
- at6:42why sal is pertaining to x-intercepts when he is talking about zeros of the function?(2 votes)
- A zero of a function is the value of x that makes the function value zero.
All points with a function value of zero lie on the X-axis.
In other words, the function intercepts the X-axis at those points.(1 vote)
- Why can't 3 be a vertical asymptote?(2 votes)
- Sorry for necroposting but I wanted to clarify this for future readers.
Since g(x) can be any polynomial (no other info is given about it), it actually can if g(x) = (x - 3)². If that were the case, then f(x) would simplify to 2(x + 3) / (x - 3), which has a zero at x = -3 and a vertical asymptote at x = 3. But the choices don't include that which is fine since they only want you to pick one possibility.
The only thing you need to note is something interesting must happen at both x = -3 and x = 3. As a pair, they need to be a combination of zeros, vertical asymptotes, and/or removable discontinuities.(1 vote)
- sal seid We've seen that situation where something
that makes the numerator equals zero
could be a removable discontinuity
if you have that same expression in the denominator.
why is that?(1 vote)
- Consider a function
f(x)=(x-1)/(x-1). The numerator will be
x=1, but then you will also have
0in the denominator, and division by zero is still an undefined operation. So the graph of this function will have a "gap", or removable discontinuity at
- why -1 is not vertical asymtote?(1 vote)
- Because nothing about this function suggests that it is. You're not given any information about g(x), so there's no reason to assume that there has to be a vertical asymptote at this x value or any other.
What Sal written to the right are just examples meant to show that the zeros you found in the numerator could also be removable discontinuities or asymptotes.(2 votes)
- Using GeoGebra I generated graph C with the function, f(x) = 2(x+3)(x-3)/(x+2)(x-3), on the other hand, graph A is reversed. Although it does fits the conditions...(1 vote)
- 𝑓(𝑥) = 2(𝑥 + 3)(𝑥 − 3)∕((𝑥 + 2)(𝑥 − 3)) has a horizontal asymptote 𝑦 = 2.
Also, 𝑓(−3) = 0 and 𝑓(0) = 3.
None of those features are present in Graph C.
– – –
Graph A has a vertical asymptote at 𝑥 = −3,
which tells us that 𝑔(𝑥) must have a factor (𝑥 + 3)².
Also, graph A has a horizontal asymptote 𝑦 = −2,
which tells us that 𝑔(𝑥) must be of degree 2 and have a leading coefficient of −1.
Thus, our only choice is 𝑔(𝑥) = −(𝑥 + 3)²
⇒ 𝑓(𝑥) = (2𝑥² − 18)∕(−(𝑥 + 3)²)
Luckily, 𝑓(3) = 0 and 𝑓(0) = 2, just like graph A.(2 votes)
- [Voiceover] So we're told let f of x equal two x squared minus 18 over g of x where g of x is a polynomial. And then they tell us, which of the following is a possible graph of y is equal to f of x? And they gave us four choices here. And like always, I encourage you to pause the video and see if you could give a go at it. Look at our f of x and think about which of these graphs actually match up or could match up to that f of x? All right, now let's work through this together. And they don't give us a lot of information. They don't tell us anything about the denominator of this rational expression. They do tell us it's numerator. And like we've seen before, it's useful to factor the numerator and see, and what x values do interesting things happen. And particular, at what x values does the numerator equal zero? So if we factor the numerator up here, we could rewrite f of x as being equal to, let's see we could factor out a two out of the numerator. So it's two times x squared minus nine. And that's all going to be over g of x. We don't know what the denominator is. We just know that it's a polynomial. Now see, here are the numerator x squared minus nine. You might recognize that. It's a difference of squares so we can factor that further. So we still have that original two and it's going to be x plus three times x minus three. We've seen that multiple times. If that looks unfamiliar to you, I encourage you to watch the video on differences of squares or factoring polynomials. X squared minus nine is the same thing as x squared minus three squared. So it's x plus three times x minus three. And then all of that is still going to be over g of x. So the first thing that we might realize, okay, when is our numerator equal zero? Well, when x is equal to negative three or when x is equal to positive three. If x is equal to negative three, this expression is going to be zero. If x is equal to positive three, this expression is equal to zero. So you might just say, "Oh, maybe we have zeros "at plus or minus three." So maybe at x is maybe, maybe f of negative three is equal to zero and f of positive three is equal to zero. Those values sure look they made the numerator equal to zero. And then we look at our choices. And when we look at our choices, this choice A does seem to have a zero at positive three but it doesn't have one at negative three. It has a vertical asymptote at negative three. So that seems a little bit confusing. This choice B does have a zero at positive three but it has nothing going on here at, or nothing interesting on at negative three. It defines negative three. It doesn't even have a vertical asymptote there. So once again, this looks a little bit perplexing. Choice C has a removable discontinuity at positive three and then it has a vertical asymptote at negative two. So once again, this doesn't have anything interesting going on at x is equal to negative three. So still a little perplexing. And this one has a zero at positive six and negative six. So none of the choices had zeros at both x equals positive three and x equals negative three. So, what's going on? Well, what we need to realize is just because something makes the numerator equal to zero doesn't mean that it's definitely going to be a zero for that function. And you might say, "Well, how can that be?" Well, think about situations in which those values would also make the denominator equal to zero. So let me write out some potential f of x is here. So we just know that g of x is a polynomial. So f of x could be, we know the numerator, two times x plus three times x minus three, over. Well, let's just say g of x is, I'm just gonna make up something, g of x is equal to x plus one. Well, in this situation, none of the values that make the numerator equal zero make the denominator equal zero. So this is a situation where you would have two zeros at x equals positive three and x equals negative three. So this would be the two zeros. Two zeros. And so let's look at another similar situation. Let's look at a situation where f of x is equal to, you know, the numerator, x plus three times x minus three. And let's say that we do have, let's say that one of those x values, positive or negative three, do make the denominator equal to zero. So let's say, x plus three and then, say times x plus one. Well, you see here, now since x plus three is both the numerator and the denominator, you could divide x plus three divided by x plus three. They cancel out. And here x equals negative three would be a removable discontinuity. So this would have zero at x equals three and a removable discontinuity at x equals negative three. And so those values that make the numerator equal to zero we now see it could be a zero or it could represent a removable discontinuity. And here I just pick a removable discontinuity be at negative three or it could be the other way around or it could be at both values. If this was x plus three times x minus three over x plus three times x minus three, then you would have a removable discontinuity at both x is positive three and negative three. And then you could go even further. F of x could look like this. It could be two times x plus three times x minus three over x plus three squared and I'll just make up some other, another expression here, x plus one. So, what's going to happen here? Even if you divide the numerator and the denominator by x plus three, you're still going to have one x plus three left over the denominator. That could cancel with one of the x plus threes but you're still going to have an x plus three. And so in this case, you would have a vertical asymptote. So in this case, you would have a zero at x is equal to three. And you would have a vertical asymptote. I'll just shorten it shorthand. You would have a vertical, I'll just write that out, vertical asymptote at x is equal to negative three. So these particular examples that I just showed you showed you that any value that makes the numerator equals zero aren't necessarily zeros for the function. They could be zeros. They could be removable discontinuities or they could be vertical asymptotes. But they would all occur at x is equal to positive or negative three. So if that lands, now let's look at the choices again. So choice A has a zero at x equals positive three and it has a vertical asymptote at x equals negative three. So that's actually very consistent with this situation that I just described. So choice A actually is looking pretty good. Choice B has a zero at x equals three but its vertical asymptote looks like it's an x equals two and nothing interesting is happening at x equals negative three. So we could rule that out. If I look at choice C, you have a removable discontinuity at x equals positive three which is completely possible. We've seen that situation where something that makes the numerator equals zero could be a removable discontinuity if you have that same expression in the denominator. But then the vertical asymptote isn't that x equals negative three. It's an x equals negative two. So that rules it out. Once again, nothing interesting happening at x equals negative three. And here you have two zeros but they're not at x equals positive or negative three. They're x equals positive or negative six. We can definitely rule that one out. So we should feel pretty good about, we should feel pretty good about choice A.