If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:11:31

CCSS.Math:

let's graph another rational function because you really can't get enough practice here so let's say we have y is equal to x over x squared minus X minus 6 so the first thing we might want to do is just factor this denominator so we can identify our vertical asymptotes if there are any and so what two numbers when I take their product I get negative 6 and if I add them up I get negative 1 so they have to be of different signs so one's going to be a plus and 1 let me write my X is a little bit cleaner than that so 1 is going to be a positive and one is going to be a negative so it's going to be a 2 and a 3 seem to be pretty close because they're 1 apart and I'm going to subtract the larger number because they when I add them I get a negative so X minus 3 times X plus 2 seems to work that gets to negative 6 negative 3x plus 2x negative 3 times X plus 2 times X is negative x so that works so this is equal to x over X plus 2 times X minus 3 and like we saw in the last video since these expressions since the X plus 2 doesn't cancel out with anything in the numerator and the X minus 3 doesn't cancel out with anything in the numerator we know that these can be used to find our vertical asymptotes the vertical asymptotes are when either that term is equal to 0 or when that term is equal to 0 because at those points our equation is undefined so this is equal to 0 when X is equal to negative 2 and this is equal to 0 when X is equal to positive 3 and you could try it out here if UX is equal to negative 2 or positive 3 you're going to get a 0 in the denominator Y will be undefined so vertical asymptotes at X is equal to negative 2 so there's a vertical asymptote vertical asymptote right there and other vertical asymptotes at X is equal to 3 1 2 3 there is our other our other vertical asymptotes now let's think about horizontal asymptotes or if there are any so what happens as X gets super positive or super negative and as we said before you just have to look at the highest degree term on the numerator and the highest degree term on the denominator now notice the highest degree term of the denominator is x squared well the highest degree term on the numerator is only an X so when X gets really large what's going to happen you can imagine this is going to be like a million over a million squared right which is still 1 over a million right these two these terms over here don't matter much but this term right here is going to grow faster than everything this is an x squared term as X gets large it's going to get way larger than everything including this term on the top so it's essentially going to go to 0 when the denominator just gets bigger faster than the numerator is you're going to approach 0 so we have a horizontal asymptote at Y is equal to 0 at Y is equal to 0 I could draw it as a dotted line over our x-axis so that right there is the line y is equal to 0 once again we identify that looking at the the highest degree term there the denominator has a higher degree term so it's going to grow faster than the numerator and you could try it out on your calculator and that's true whether you go into the super negative direction or the super positive direction this thing is going to overwhelm this thing up here denominator grows faster in the numerator which essentially you're going to approach the 0 you're going to get smaller and smaller fractions just remember you know if 1 over 10 and then let me let me actually just as X gets larger and larger and larger what's going to happen let me just show you on my calculator so as X let's say X is equal to 10 10 divided by divided by 10 squared minus 10 and normally you wouldn't have to do this I just really want to show you the intuition oh whoops I'm not trying to graph let me I'm not trying to graph let me exit from here exit so if we have 10 over 10 squared minus 10 once again you normal you wouldn't have to do this I just want to show you give you the intuition on we put some parentheses there let me put some parentheses over here so let me insert the parentheses there and then put a parentheses over here you get a small number that what happens if X gets even larger if X gets even larger let me make a set of a 10 let me make it all 100 let me make these tens into hundreds into a hundred let me insert a hundred there what do we get we get even a smaller number if you try X is equal to thousands going to be even smaller than that and that's because this term right here is going faster than every single other term that's why our horizontal asymptote is y is equal to zero now the last thing we want to do we've drawn all of our asymptotes let's just try out some points so let's try out let's draw it like a little table here there's our table when x is equal to 0 what is y x is 0 we have 0 over all the 0 minus 6 0 over negative 6 is just 0 when X is equal to I don't let's just try when X is equal to 1 what do we have we have 1 over I'll write it here 1 over 1 squared minus 1 that's just 0 so we have negative 6 when X is equal to negative 1 what do we have when X is equal to negative 1 X is equal to negative 1 we have negative 1 over negative 1 squared which is 1 minus negative 1 so that's plus 1 right minus negative 1 minus 6 so what is this right here this is negative 1 so this is going to be negative 1 over 2 minus 6 over negative 4 this is going to be equal to 1/4 so we're going to get a positive value so we have let me draw this negative 1 we get we're at 1/4 right here that's about right there I'll do it in a darker color we had the point zero zero and then it at X is equal to one we had negative one-sixth so you could you could keep you could keep graphing more and more points but it looks like as we approach this vertical asymptote from the right we go to positive infinity we go to positive infinity and that should make sense let's see if we put if we were to put in we're approaching negative 2 from the right so if you were to put in negative one point nine nine nine nine nine nine nine this term is going to be a very small positive number this term is going to be a negative number this term is going to be a negative number the negatives cancel out you have a very small positive number the denominator one over that gives you a very positive number now as we go as we approach the other vertical asymptote from the left we're going to go super negative and my my gut tells me that because when I tried X is equal to one I already went to a negative value but you could imagine if you did two point nine nine nine nine nine right two point nine nine nine let me draw that a little bit better if you you get the idea if X is equal to two point nine nine nine so we get really close to the asymptote this is going to be positive this is going to be negative that's going to be positive so you're going to and this is going to be a small number so you're going to have one over a very small number which very small negative number which is a very I guess negative number right you could you know it's it's the negative of a one over very small numbers are going to proach negative infinity now let's try some points out here see what happens so what happens when X is equal to four when X is equal to four you have four over over 16 minus four minus six what is that that's 16 minus 10 that is six so this is equal to 4 over 4 over 6 which is equal to two-thirds so the point four two-thirds is here so one two three four two-thirds just like that so that gives me the sense look I have to approach this horizontal asymptote as we go further and further out we're going to probably approach neg positive infinity like this like let me draw a little neater than that like that you get the idea and then over here we're going to get closer and closer to our horizontal asymptote as we approach infinity this should be a smoother looking curve right around there I'm making a mess here this should be a smoother looking curve you get the idea I think and let's see what happens when X is equal to negative three X is equal to negative three so when X is equal to negative three we have negative 3 over negative 3 squared which is 9 minus negative 3 so that's plus 3 minus 6 so what is this equal to this is equal to negative 3 over not this is 12 minus 6 over 6 right which is equal to negative 1/2 so negative 3 negative 3 negative 1/2 negative 1/2 right there so we're going to approach this asymptote as we get really negative and we're probably going to go straight down like that as we approach this vertical asymptote right there so let's and you could try more points if you don't you know believe me but let's let us graph it just to verify it for ourselves so our equation our equation is X divided by X divided by x squared minus X minus 6 and let's graph it and they go there you go all right looks pretty good looks pretty good we went we start our asymptote is 0 we go down vertical asymptote BAM go up there then we go back down here and then we go like just like that again so once again this looks just about exactly what we got obviously the graphing calculator it kind of Pitters out as you get close to these values and it does we things but it has the same general shape we could actually we could actually close the range a little bit if we want if we want to graph it let's make our X minimum value let's make it 5 let's make it and the spec R will X maximum value let's make that 5 as well we're kind of zooming in a little bit so let's graph it now's bam bam there you go what is the same shape as we graphed right here hopefully you found that satisfying