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# Graphs of rational functions (old example)

CCSS.Math:

## Video transcript

so we've got three functions graphed here the function f in magenta we have the function G and 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 and I'll I encourage you to pause this video and try to think about it on your own before I work through it so 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 functions graphs look like that or 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 not you're going to do 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 5 let's think about which of these would have a vertical asymptote at x equals negative 5 and or to have a vertical asymptote it can't be defined there so that's the first way I would think about it and then even if it 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 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 zero 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 zero so that's not F this one when X is equal to negative 5 this denominator does become zero so this seems if I just use purely deductive reasoning this looks like my best candidate for candidate for f of X plus 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 one now is that the true 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 one as X gets very very very large then the negative two and the subtracting the two of the numerator and adding 5 in the denominator we're 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 when the new this 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 at X is equal to 5 so the vertical asymptote isn't what's going to help us resolve between G and H 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 both of them are undefined for it when x equals 5 both of their denominators are 0 so let's see if the 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 this top one well if we expand out the numerator this is equal to 2x 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 and so this is for very very large X's this is going to be approximately 2x over X let me make it very clear for maybe I'll write as as X approaches infinity and 2x 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 is asymptote H is asymptotes 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 one would H of x equals 0 well the numerator equals 0 when x equals 6 and we see it right over here now that might have not helped us that much because G also equals 0 G also equals 0 at 6 but at least the horizontal asymptotes are what cluing is what is cluing us in as X gets really really really large then the negative 2 subtracting the 12 in the numerator subtracting the final denominator we're going to matter less and less and less we want to look at the highest degree terms it's going to approach 2x over X which is 2 and that we see that for H now G if we 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 2x over X minus 5 but as which is approximately going to be equal to we look at the highest degree terms negative 2x over X negative 2x over X as 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 this X gets very very large we approach negative 2 or frankly as X gets very very small where 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