Integrated math 3
Sal analyzes the behavior of q(x)=(x²+3x+2)/(x+3) around its vertical asymptote at x=-3.
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- Wait a minute, q approaches +infinity when x-values are between -3 and -2. If x -> 3^+ means "values over -3," then why do we limit our values?(9 votes)
- It does not mean values over -3. It means the limit is approaching from the right side of 3. Most of the times, they are just same as when you put 3, but sometimes they are not the same.(8 votes)
- is simplifying the expressions eqivalent when i search for the asymptotes and their behaviour?(8 votes)
- When you simplify by cancelling, "(x-a)/(x-a)" = 1 for all x, so cancelling doesn't affect the graph - it won't affect any asymptotes that are left after the cancelling, or remove any asymptotes.(6 votes)
- I didn't quite get the answer of the first question. Shouldn't q(x) be approaching to positive infinity? I mean even Sal points that out on the numberline in blue color with the arrow pointing to the right side. What's happening... I'm confused.(7 votes)
- He starts to explain around3:30, as q(x) approaches the vertical asymptote of -3, the function goes down and approaches negative infinity.
Try substituting any value less than -3 for x, and you'll find the function always comes out as a negative. If we look at x = -4, for example, the numerator simplifies to (-3)(-2) = 6. The denominator simplifies to -4+3 = -1. The function as a whole then simplifies to q(x) = -6 for x = -4. You can try this for any x value smaller than -3 and you'll find the function approaches negative infinity the closer x gets to it's vertical asymptote of -3.(4 votes)
- I would definitely start this video by considering the interval from -3 to -2 first (approaching from positive side), because if we start considering approaching from the negative, there is a temptation to use whole numbers such as (-5) and (-4) and by substituting these numbers (instead of -3.1 and -3.01) function DOES NOT appear to approach negative infinity. There is a sign change at x=-4.4.(6 votes)
- So why do you have to restrict the -3^+ but not -3^-?(2 votes)
- You don't. You can approach -3 from either side. Just get close enough that y changes rapidly.(2 votes)
- Why do we care about the intervals -3<x<-2? What does Sal mean at5:10when he says "I don't want any kind of weird sign changes"?(2 votes)
- Because, at x=-2, that would be a zero, and that has no sign. Also, if you stray too far from -3, you might get some wrong answers. Usually, I would check for numbers about 0.1 away from the asymptote maximum. The closer to the asymptote, the more accurate.(2 votes)
- How can we differentiate between horizontal and vertical asymptotes? Sometimes I hear Sal refer to horizontal asymptotes but like in this video, it focuses on vertical.(1 vote)
- Horizontal asymptotes are when a function's y value starts to converge toward something as its x value goes toward positive or negative infinity. This is the end behavior of the function.
Vertical asymptotes are when a function's y value goes to positive or negative infinity as the x value goes toward something finite.
Let's say you have the function
a(x) = (2x+1)/(x-1).
As x → 1 from the negative direction, a(x) → -∞. As x → 1 from the positive direction, a(x) → +∞. This is your vertical asymptote, because as x approaches something finite, a(x) approaches something infinite.
As x gets bigger and bigger (you can think of this as x → ∞, I don't know if you have done end behavior at this point in the course), a(x) goes to 2. You can confirm this by plugging in really big values, for example:
(2(1000000000)+1)/((1000000000)-1) = 2.000000003
Anyway, the point is that as x approaches something infinite, a(x) approaches something finite, so this is your horizontal asymptote.
This is your bottom line:
Vertical asymptote: As x → finite, y → infinite
Horizontal asymptote: As y → finite, x → infinite
Hope this clarified a bit.(4 votes)
- In the example in the video, the line x = -3 is the vertical asymptote because it is the denominator and as x approaches -3 from either side, the denominator gets smaller at a faster rate than the rate at which the numerator is getting bigger.
If the factor (x+3) is also in the numerator, would this mean that x = -3 is no longer the vertical asymptote? Because now the numerator is changing at a pace that is just as fast as the denominator?(2 votes)
- Yes. (x+3)/(x+3) is always 1. At x = -3, this quotient 0/0 = 1, according to mathematicians, is undefined (because division by 0 is undefined in general), so there the function is undefined.
So therefore there is a location on the x axis (x = -3), with no extent (length), where the function is undefined. Everywhere else the quotient is 1.(1 vote)
- Is this something like finding limits? I think this is what calculus is all about. Am I right? I am kinda confused why it´s here in algebra II.(1 vote)
- It is similar to limits, but it is not. It is in Algebra II material because, for one, you are not actually using limits, and because you need it to graph rationals.(3 votes)
- Could someone give me an example of a function where the asymptote approaches positive infinity from both directions or negative infinity from both directions?(2 votes)
- [Voiceover] We're asked to describe the behavior of the function q around its vertical asymptote at x = -3, and like always, if you're familiar with this, I encourage you to pause it and see if you can get some practice, and if you're not, well, I'm about to do it with you. All right, so this is q of x, it's defined by a rational expression, and whenever I'm dealing with asymptotes, I like to factor the numerators and the denominators so I can make more sense of things. So the numerator here, what two numbers if I were... Their product is two and their sum is three, well that's two and one, so I can factor this as x + 1 times x + 2, if that's unfamiliar to you, I recommend you watch the videos on Khan Academy about factoring quadratics, and over x + 3, and when x = -1, or x = -2, it would make the numerator equal zero without making the denominator equal zero, so those are points where the function is equal to zero, but when x = -3, the denominator equals zero, while the numerator is not equal to zero, so we're dividing by zero, and so that's a pretty good sign of a vertical asymptote, that our function as we approach that value is either going to pop up like that, or it's going to pop down, or it's going to pop down like that, or maybe, or either way, or could pop up like that, or it could go down something like that, but we're going to have a vertical asymptote that the function as you approach x = -3, that the function is going to approach either positive infinity or negative infinity or it might do positive infinity from one direction or negative infinity from another direction and if this idea of directionality is a little bit confusing, well that's where we're, that's what we're about to address in this video. Let me just draw a number line here that focuses on these interesting values. So we care about x = -3, and then the other interesting values may be -2, - 1, it's all there, now what does it mean to be approaching x, what does it mean for x to be approaching -3 from the negative direction? And just to be clear, this little superscript right over here, that means we're approaching from the negative direction. So that means we're approaching from values more negative than -3, so those are these values right over here. We are approaching, we are approaching from that direction. Another way to think about it is we approach from the negative direction where on the interval x is less than -3, so let's think about what the sign of q of x is going to be as we're approaching -3 from that negative direction, from the left. Well, if we have something less than -3, and you add one, this is going to be negative. If you have something less than -3, and you add two, that's going to be negative as well, and if you have something less than -3, and you add three, well that's going to be negative as well. So a negative times a negative is a positive, but then you divide by a negative, it's going to be a negative, so q of x is going to be negative on that interval. So as we approach, we have our vertical asymptote. As we approach -3 from the left hand side, well q of x is going to be negative and so it's going to approach negative infinity. So as x approaches -3 from the negative direction, q of x is going to approach negative infinity. So at least this accurate, this is accurate, and this is accurate right over here, and you can validate that, try some values out. Try negative, let's see, if you did negative three point... q of negative three point, - 3.1, once again, that's on the left side, it might be like right over there. Actually, it'd probably be a little bit further. It might be something like this, like the scale that I've drawn it on. q of -3.1, if you wanted to verify it, it's going to be -3.1 + 1, which is -2.1, times -3.1 + 2, which is negative, - 1.1, 1.1, I could put that in parentheses just to make it clear, and then all of that is going to be over, well -3.1 + 3 is going to be, is going to be -0.1. So notice whatever we get up here that's a positive value, we're essentially going to, if we're dividing it by -0.1, that's like multiplying it by -10, so it's going to become a very negative value, and if instead of it being -3.1, imagine if it was -3.01, and this would be a 01 here. This would be a 01 here, and this would be a 01 here. And so this denominator, you're dividing by -0.01, it's going to be even larger negative value, so you're going to approach negative infinity. So it's going to be one of these two choices. Now let's think about what happens as we approach x from the positive direction, and that's what this notation over there means, this superscript on the right hand side, the positive direction, so we're going to approach x from the positive direction, and I'm going to pay attention in particular to the interval between -2 and -3, because then we know we don't have any weird sign changes going on in the numerator. So I care about the interval -3<x<-2. So I can draw an open circle here to say we're not considering when we're at -2, and of course we're not going to include -3 because our function isn't defined there, but over this interval, so x + 1 is still going to be negative, x +1 is still going to be negative. If you took a -2.5 + 1, it's going to be -1.5. x + 2 is still going to be negative, you're taking values that are more negative than -2, that are less than -2, so you add two to that and you're still going to be negative, and then let's see, if you add three, if you add three to these values, remember, they're greater than -3, or you could say they're less negative than -3, well then this is going to give you a positive value. This is going to give you a positive value, and think about it, q of, q of, I don't know, -2.99, what is that going to be equal to? If you add one to that, that is -1.99, and if you add two to that, times, times - 0.99 all of that over -2.99 + 3, well that's going to be 0.01, so you're going to get a positive value on top and then you're going to divide it by 0.01, that's the same thing as multiplying by 100, so you're going to get larger and larger values. You're going to approach infinity as you get closer and closer to it from the right hand side. So q of x is going to approach positive infinity. So this is the choice that is correct. This one is wrong, this says we're going to approach negative infinity, so that's incorrect, and we will go with that choice.