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Current time:0:00Total duration:5:33

Graphs of rational functions: vertical asymptotes

CCSS.Math:

Video transcript

we're told let f of X equal G of X over x squared minus X minus 6 where G of X is a polynomial which of the following is a possible graph of y equals f of X and they give us four choices the fourth choice is off right over here and like always pause the video and see if you can figure it out or if you if you were having trouble with it as I start to think about it with you pause it at any time if you feel inspired it's really important to to try to engage in the problem as opposed to just watch me do it but let's let's start tackling this now together so this is interesting they don't give us a lot of information about f of X in fact we don't know what the numerator is we just know that it is a polynomial well that's somewhat valuable but they do give us the denominator and so we could think about what are the interesting numbers what are the interesting x-values for the denominator in particular what x values will make the denominator equal to 0 and to do that we can factor out the denominator so let's see the coefficient on the first degree term is essentially negative 1 we could write negative 1 there if we want and the constant is negative 6 so if we want to factor that we could say well what two numbers their product is negative 6 and they add up to negative 1 well negative 3 times positive 2 is negative 6 and negative 3 plus 2 is equal to negative 1 so I can rewrite f of X I can say that f of X is equal to G of x over over X minus 3 X minus 3 times X plus 2 so the denominator equals 0 for x equals 3 or x equals negative 2 that's when the denominator is 0 so 0 denominator I'll just write it like that and so if something makes the denominator equal to 0 that tells us that we're either going to have a vertical asymptote at that point or we're going to have a removable discontinuity at that point and the way that that would be a removable discontinuity let's say if we had a removable discontinuity at x equals 3 well that means that G of X could be factored into X minus 3 times a bunch of other stuff if that was the case then x equals 3 would be a removable this continuity if x equals three does not make G of X equal zero so for example if G of three does not equal zero or G of negative two does not equal zero then these would both be vertical asymptotes so let's look at the choices here so in choice a we have one vertical asymptote that vertical asymptote is that x equals negative two so it seems this select this line let me draw this line here this vertical asymptote right over there that is the line X is equal to negative two so at least two B it seems to be consistent with that over there but what about x equals three this one seems completely cool this this graph is defined at x equals three x equals three or is right over there and it seems to be defined there and this f of X is clearly not defined at F at X is equal to three because we go because when x equals three the denominator is zero and dividing by zero is not defined so even though this has one vertical asymptote at an interesting place we're going to rule it out because this graph is defined at x equals three even though f of X is not we would need to see either a vertical asymptote there either a vertical asymptote there or a removable discontinuity alright here we have a vertical asymptote at X is equal to negative two and we have another vertical asymptote at X is equal to positive four so that doesn't make sense either this one just like the last one is actually defined at x equals three we see it x equals three the function is equal to zero but this function f of 3 is not equal to zero f of 3 is undefined we're dividing by zero so we can rule this out once again at x equals 3 we need to see the removable discontinuity or a vertical asymptote because we're not defined there all right let's see choice C we see a vertical asymptote at X is equal to negative two so that looks pretty good and we see a removable discontinuity at x equals three so this function is this graph is not defined for x equals 3 or 4 x equals negative 2 which is good because is not defined at either of those points because at either of those X values our FS denominator is equal to zero so this one looks this one looks quite interesting and this would be consistent this one would be consistent with the width f of X being something of the sort of so the denominator we already know X minus 3 times X plus 2 and in the numerator we would have since X minus 3 is not a vertical ass x equals 3 isn't a vertical asymptote it's a removable discontinuity we must be able to factor for this one G of X into X minus 3 times something else so that's consistent with this one over here so I like this choice now let's look at this choice choice D choice D has two vertical asymptotes one at X is equal to negative 1 this is negative 2 so this x equals negative 1 and this is X is equal to 6 neither of them are what coincide what make our denominator equal 0 so we can rule this out as well so we can feel really good about choice C