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Analyzing vertical asymptotes of rational functions | High School Math | Khan Academy


5m read
·Nov 11, 2024

We're as to describe the behavior of the function Q around its vertical asymptote at x = -3. Like always, if you're familiar with this, I encourage you to pause it and see if you can get some practice. 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. 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)(x + 2). If that's unfamiliar to you, I recommend you watch the videos on Khan Academy on factoring quadratics.

And over x + 3, and x when x = -1 or x = -2 would make the numerator equal 0 without making the denominator equal 0. So those are points where the function is equal to zero. But when x equals -3, the denominator equals 0 while the numerator is not equal to 0. 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 it 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, 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 what we're about to address in this video. So 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 and 1. It's all there. Now, what does it mean to be approaching x from the negative direction? And just to be clear, this little superscript right over here 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 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, this, 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 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, and we have our vertical asymptote, 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 is accurate, this is accurate, and this is accurate right over here. You can validate that, try some values out. Try negative, let's see if you did -3.1. Once again, that's on the left side; it might be like right over there. Actually, it would 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 want to verify it, it's going to be -3.1 + 1, which is -2.1 times -3.1 + 2, which is -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, which is going to be 0.1. So notice whatever we get up here, this 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 get, it's going to become a very negative value.

And if instead of -3.1, imagine if it was -3.01. Then this would be a 0.01 here, this would be a 0.01 here, and this would be a 0.01 here. So this denominator, you're dividing by -0.01, is going to be an 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, the 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 could draw an open circle here to say we're not considering where 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 -2.5 + 1, it's going to be -1.5. x + 2 is 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; 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 are 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 -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 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.

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