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# Graphing rational functions according to asymptotes

CCSS.Math:

## Video transcript

so we have f of X is equal to 3x squared minus 18x minus 81 over 6x squared minus 54 now what I want to do in this video is find the equations for the horizontal and vertical asymptotes and I encourage you to pause the video right now and try to work it out on your own before I try to work through it so I'm assuming you've had a go at it so let's think about each of them so let's first think about the horizontal or is antal asymptote horizontal asymptotes see if there at least is one so the horizontal asymptote is really what is the line the horizontal line that f of X approaches as the absolute value of X approaches as the absolute value of X approaches infinity or you could say what does f of X approach as X approaches infinity and what does X f of X approach as X approaches negative infinity so there's a couple of ways you could think about it let me just rewrite the definition of f of X right over here so it's 3x squared minus 18x minus 81 all of that over 6x squared minus 54 now there's there's two ways you can think about it one you could say okay as X as the absolute value of x becomes larger and larger and larger the highest degree terms in the numerator and the denominator going to dominate are going to Dum the numerator the denominator are going to are going to dominate and so what are the highest degree terms well in the numerator you have 3x squared and in the denominator you have 6x squared so as X approaches so as the absolute value of x approaches infinity these two terms are going to dominate f of X is going to become approximately 3x squared over 6 x squared these other terms are going to matter less obviously minus 54 is I'm going to grow at all and minus 18x is going to grow much slower than the the x-squared the highest degree terms are going to are going to be what dominates and if we look at just those terms then you could think of simplifying it in this way so that's going f of X is going to get closer and closer to 3 6 or 1/2 so you could say that there's a horizontal asymptote at y is equal to Y is equal to 1/2 another way we could have thought about this if you don't like this whole little bit of hand wavy argument that these two terms dominate is that we can divide the numerator and the denominator by the highest degree or X raised to the highest power in the numerator and the denominator so the highest degree term is x squared in the numerator so let's do so let's divide the numerator and the denominator or I guess you could say the highest degree term in the numerator and the denominator is x squared so let's divide both the numerator and denominator by that so if you multiply the numerator times 1 over x squared and the denominator times 1 over x squared notice we're not changing the value of of the entire expression we're just multiplying it times 1 if we assume X does not equal 0 and so we get 2 so in our numerator let's see 3 x squared divided by x squared is going to be 3 minus 18 over X minus 81 over x squared and then all of that over 6 x squared times 1 over x squared that's just going to be 6 and then minus 54 over x squared so what's going to happen what's the if you want to think in terms of if you want to think of limits as something approaches infinity so if you want to say the limit as X approaches infinity here what's going to happen well this this and that are going to approach 0 so you're going to approach 3/6 or 1/2 now if you said as X approaches negative infinity it would be the same thing this this and this approach 0 and once again you approach 1/2 so that's the horizontal asymptote Y is equal to 1/2 now let's think about the vertical asymptotes so let me write that down right over here let me scroll over a little bit so vertical vertical asymptotes or possibly asymptotes vertical maybe there is more than one now it might be very tempting to say okay you hit a vertical asymptote whenever the denominator equals zero which would make this rational expression undefined and as we'll see for this case that is not that is not exactly right just making the denominator equal to zero by itself will not make a vertical asymptote it will definitely be a place where the function is undefined but by itself it does not make a vertical asymptote so let's just think about this denominator right over here so we can factor it out actually let's factor out the numerator and the denominator so we can rewrite this as f of X is equal to the numerator is clearly every term is divisible by three so let's factor out of three it's going to be three times x squared minus 6x minus 27 all of that over the denominator each term is divisible by six six times x squared minus nine and let's see we can we can factor the numerators and denominators denominators out further so this is going to be f of X is equal to three times let's see two numbers their product is negative twenty seven their sum is negative six negative nine and three seemed to work so you can have X minus nine times X plus three just factor the numerator over the denominator this is the difference of squares right over here so this would be X minus three times X plus three so when does the denominator equal when does the denominator equal zero so the denominator denominator equals zero when X is equal to positive three or X is equal to negative 3 now I encourage your pause this video for a second think about are both of these vertical asymptotes well you might realize that the numerator also equals zero when X is equal to negative three so what we can do is actually simplify this a little bit and then it becomes a little bit clearer where our vertical asymptotes are so we could say that f of X we could we could essentially divide the numerator and denominator by X plus 3 and we just have to key if we want the function to be identical we have to keep the caveat that the function itself is not defined when X is equal to negative 3 that definitely did make us divide by 0 so we have to remember that but that will simplify the expression so this exact same function is going to be if we divide the numerator and denominator by X plus 3 it's going to be 3 times X minus 9 over 6 times X minus 3 for X does not equal 4 X does not equal negative 3 notice this is an identical definition to our original function and I have to I have to put this qualifier right over here for X does not equal negative 3 because our original function is undefined at x equals negative 3 x equals negative 3 is not a part of the domain of our original function and so we if we take it out like if we take X plus 3 out of the numerator and denominator we have to remember that if we just put this right over here this wouldn't be the same function because this without the qualifier is defined for x equals negative 3 but we want to have the exact same function so you would actually have a point discontinuity right over here and now we could think about the vertical asymptotes now the vertical asymptotes going to be a point that makes the denominator equals 0 but not the numerator equal to 0 x equals negative 3 made both equal 0 so our vertical asymptotes I'll do this in green just to just a switch or blue our vertical vertical asymptote asymptote is going to be at X is equal to X is equal to positive 3 that's what made the denominator equal zero but not the numerator so let me write that the vertical asymptote is X is equal to 3 and so using these two points of of information or I guess what we just figured out you can start to attempt to sketch sketch the graph this by itself is not going to be enough you might want to also plot a few points to see what happens of I guess around the asymptotes as we approach as we approach the two different asymptotes but if we were to look at if we were to look at a graph actually let's just do it let's just do it for let's just do it for fun here just to complete the picture for ourselves the function is going to look something like this it's going to look something like this and I'm not doing it at scale so if that's 1 and this is 1/2 right over here y equals 1/2 is the horizontal asymptote y equals 1/2 is a horizontal asymptote Y is equal to 1/2 and we have a vertical asymptote at X is equal to positive 3 so we have 1/2 Ln 2 then in blue 1 2 3 once again I didn't draw it to scale or the x and y's aren't on the same scale but we have a vertical asymptote just like that but just looking at this we don't know exactly what the function looks like it could look like something like this it could it and maybe does something like that or it could do something like that or it could do it could do something like that and that or something like that and that so hopefully you get the idea here and to figure out what it does you would actually want to try out some points and the other thing we want to be clear is that the function is also not defined at X is equal to at x is equal to negative 3 so if this is so let me make x equals negative 3 here so x equals negative 3 so 1 2 3 so the function might look and once again I haven't tried out the points it could look something like this it could look something where we're not a whoops where we're not defined at negative 3 we're not defined at negative 3 and then it goes something like this and maybe just something like that or maybe it does maybe it does something like maybe does something like that it's not defined at negative 3 this will be some toad right now so to get closer and closer and it can go something like that or it goes something like that and once again to decide which of these it is you would actually want to try out a few values so I encourage you to - after this video try that out in yourself and try to figure out what the actual graph of this looks like