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CCSS.Math:

so we're told let f of X equal to x squared minus 18 over G of X where G of X is a polynomial and then they tell us which of the following is a possible graph of y is equal to f of X and they give us four choices here and like always I encourage you to pause the video and see if you could give a go at it look at our f of X and then think about which of these graphs actually match up or could match up to that f of X all right now let's let's work through this together and they don't give us a lot of information they don't tell us anything about the denominator of this rational expression but they do tell us this numerator and like we've seen before it's useful to factor the numerator and see at what X values do interesting things happen in particular at what X values does the numerator equal zero so if we factor the numerator up here we could rewrite f of X as being equal to let's see we could factor out a 2 out of the numerator so it's 2 times x squared minus 9 and that's all going to be over G of X we don't know what the denominator is we just know that it's a polynomial now let's see here in the numerator x squared minus 9 you might recognize that as a difference of squares so we can factor that further so we still have that original two and it's going to be X plus 3 times X minus 3 we've seen that at multiple times if that looks unfamiliar to you I encourage you to watch the video on differences of squares or factoring polynomials x squared minus 9 is the same thing as x squared minus 3 squared so it's x plus 3 times X minus 3 and then all of that is still going to be over G of X so the first thing that we might realize is okay when does our numerator equal 0 well when when X is equal to negative 3 or when X is equal to positive 3 if X is equal to negative 3 this expression is going to be 0 if X is equal to positive 3 this expression is equal to 0 so you might just say well maybe we have zeros at plus or minus 3 so maybe at X is maybe maybe F of negative 3 is equal to 0 and F of positive 3 is equal to zero those values sure look like they make the numerator equal to zero and then we look at our choices and when we look at our choices this choice a does seem to have a zero at positive three but it doesn't have one at negative three it has a vertical asymptote at negative three so that seems that seems a little bit confusing this choice B does have a zero at at positive 3 but it has nothing going on here at or nothing interesting going on at negative 3 it defines negative 3 it doesn't even have a vertical asymptote there so once again this looks a little bit perplexing choice C choice C has a removable discontinuity at positive 3 and then it has a vertical asymptote at negative 2 so once again this doesn't have anything interesting going on it at X is equal to negative 3 so still a little perplexing and this one has zeros at positive 6 and negative 6 so none of the choices have zeros at both x equals positive 3 and x equals negative 3 so what's going on well what we need to realize is just because something makes the numerator equal to 0 doesn't mean that it's definitely going to be a 0 for that function and you might say well how can that be well think about situations in which those values would also make the denominator equal to 0 so let me write out some potential f of X is here so we just know that G of X is a polynomial so f of X could be we know the numerator 2 times X plus 3 times X minus 3 over well let's just say G of X is I'm just going to make up something G of X is equal to X plus 1 well in this situation none of the values that make the numerator equal 0 make the denominator equal 0 so this is a situation where you would have two zeros at x equals positive 3 and x equals negative 3 so this would be the two two zeros to two zeros and so let's look at another sitov situation let's look at a situation where f of X is equal to we know the numerator X plus 3 times X minus 3 and let's say that we do have let's say that one of those X values positive or negative 3 do make the denominator equal to zero so let's say X plus 3 and then say times X plus 1 well you see here now since X plus 3 can is both in the numerator and the denominator you could divide X plus 3 divided by X plus 3 they cancel out and here x equals negative 3 would be a removable discontinuity so this would have this would have 0 0 at x equals 3 and a removable removable discontinuity at x equals negative 3 and so those values that make the numerator equal to 0 we now see it could be a zero or it could it could represent a removable discontinuity and here I just picked a removable discontinuity at negative 3 it could be or it could be the other way around or it could be at both values if this was X plus 3 times X minus 3 over X plus 3 times X minus 3 then you would have a removable discontinuity at both X is positive 3 and negative 3 and then you could go even further f of X could look like this it could be 2 times X plus 3 times X minus 3 over over X plus 3 squared and I'll just make up some other another expression here X plus 1 so what's going to happen here even if you divide the numerator and the denominator by X plus 3 you're still going to have 1 X plus 3 left over in the denominator that could cancel with one of the X plus 3s but you're still going to have an X plus 3 and so in this case you would have a vertical asymptote so in this case you would have a 0 0 at X is equal to 3 and you would have a vertical asymptote I'll just shorten it shorthand you'd have a vertical I'll just write out a vertical asymptote at X is equal to negative 3 so these particular examples that I just showed you showed you that any value that makes the numerator equal 0 aren't necessarily zeros for the function they could be zeros they could be removable discontinuities or they could be vertical asymptotes but they would all occur at X is equal to positive or negative 3 so what that lends now let's look at the choices again so choice a has zero at x equals positive three and it has a vertical asymptote at x equals negative three so that's actually very consistent with this this circum this this situation that i just described so choice a actually is looking is looking pretty good choice B has a 0 at x equals 3 but its vertical asymptote looks like it's at x equals 2 and nothing interesting is happening at x equals negative 3 so we can rule that out now you look at choice C you have a removable discontinuity at x equals positive 3 which is completely possible we've seen that situation where something that makes the numerator equals zero could be a removable discontinuity if you have that same expression in the denominator but then that vertical asymptote isn't at x equals negative 3 it's at x equals negative 2 so that rules it out once again nothing interesting happening at x equals negative 3 and here you have two zeros but they're not at x equals positive or negative 3 they're at x equals positive or negative 6 so we can definitely rule that one out so we should feel pretty good about we should feel pretty good about choice a