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Sal matches three graphs of rational functions to three formulas of such functions by considering asymptotes and intercepts. Created by Sal Khan.
Video transcript
Voiceover:So, we've got three functions graphed here. The function f in magenta. We have the function g in this green color. And we have the function h in this dotted purple color. And then we have three potential expressions, or three expressions that could be potential definitions, for f, g, and h. And what I want to do in this video, is try to match them, try to match the function to a definition. I encourage you to pause this video and try to think about it on your own before I work through it. There's a couple of ways that we could approach this. One, we could think about what the graphs of each of these look like, and then think about which of these function's graphs look like that. Or, we could we could look at the function graph and think about the vertical and horizontal asymptotes, and think about which of these expressions would have a vertical or horizontal asymptotes at that point. So, I'm actually going to do it the second way. Let's look at the graphs. I tend to be a little bit more visual, or I like to look at graphs. So it looks like - f, I'll start with f. f looks like it has a vertical asymptote at x equals 5. Or, sorry, at x equals negative 5. We have a vertical asymptote right over here, at x is equal to negative five. Let's think about which of these would have a vertical asymptote at x equals negative 5. In order to have a vertical asymptote, it can't be defined there. So that'sthe first way I would think about it. And then even if isn't defined there, we need to make sure that it's an actual vertical asymptote, and not just a hole at that point, not just a point discontinuity. So, let's think about it. So, this first expression is actually defined at x equals negative 5. The only reason why it wouldn't be defined, if somehow we got a 0 in the denominator. But if you have negative 5 minus 5, that's negative 10. So this one's defined at that point, so that's not f. This one's also defined at x equals negative 5. The denominator does not become 0, so that's not f. This one, when x is equal to negative 5, this denominator does become 0. So this seems, I just used purely deductive reasoning, this looks like my best candidate for f of x. But let's confirm that it's consistent with the other things that we're seeing over here. So, let's look at f's horizontal asymptote. If I look at the graph, it looks like there's a horizontal asymptote. Especially as x gets to larger and larger values, it looks like f of x is approaching 1. f of x is approaching 1. Now, is that the same case over here? As x gets larger and larger and larger, as x approaches infinity, well, then the negative 2 and the plus 5 don't matter. As x approaches infinity, this is going to approximate, for very large x's, it's going to be x over x. We look at the highest degree terms, and so that is going to approach 1, as x gets very very very large. Then the negative two, subtracting the 2 in the numerator and adding 5 in the denominator are going to matter less and less and less, because the x gets so big. So, this will approach 1. So, that seems consistent from that point of view. And let's see, is there anything else interesting? Well, when does this equal 0? Well, the numerator equals 0 when x is equal to 2. And we see that's indeed the case for f right over there. So, I'm feeling really good that this is our f of x. Now, let's look at g of x. g of x has, and actually they're trying to trip us up here, because it looks like both h and g of x, have a vertical asymptote at x is equal to 5. So the vertical asymptote isn't what's going to help us resolve between g and h. Did I just say g and f? g and h both have a vertical asymptote at x equals 5. And you see that over here. At x equals 5, both of them are undefined. When x equals 5, both of their denominators are 0. So, let's see if the horizontal asymptotes could help us. So, it looks like g has a horizontal asymptote at y equals negative 2. y equals negative 2. if x becomes very positive or very negative, it looks like y is approaching negative 2. So, what's happening up here, for this top one? Well, if we expand out the numerator, this is equal to 2 x minus 12, over x minus 5. For very large x's the negative 12 and the negative 5 aren't going to matter as much. So this is, for very very large x's, this is going to be approximately 2 x over x. Let me make it very clear, for, maybe I'll write, "as x approaches infinity." And 2 x over x would be 2, would be approximately equal to 2 as x approaches infinity. g's asymptote isn't at 2, it's at negative 2. h's asymptote's the one that looks like it's at 2. At y equals 2. So this top one looks like h of x. This top one looks like at h of x. And we can confirm that. When would h of x equal 0? Well, the numerator equals 0 when x equals 6, and we see it right over here. Now, that might not have helped us that much because g also equals 0 at six, but at least the horizontal asymptotes is what is cluing us in. As x gets really really really large, then the negative, subtracting the 12 in the numerator, subtracting the 5 in the denominator, are going to matter less and less and less. We want to look at the highest degree terms. It's going to approach 2 x over x, which is 2. And we see that for h. Now, g, if we just use our deductive reasoning, we'd say "okay, well, hopefully this is our g of x." Now does this make sense? g of x is equal to 12 minus 2 x, over x minus 5. Which is approximately going to be equal to, we look at the highest degree terms, negative 2 x over x, as x approaches infinity. Which is going to be equal to, or we're going to approach negative 2. And that is indeed where g's horizontal asymptote is. As x gets very very large, we approach negative 2, or frankly as x gets very very small, we're also going to approach negative 2. Negative 2 times the negative billion over a negative billion is also going to be negative 2. So, we can feel pretty good that this is our g of x.