Discontinuities of rational functions

so we have this function f of X expressed as a rational expression here or defined with the rational expression and we're told at each of the following values of X select whether F has a zero a vertical asymptote or or a removable discontinuity and before even looking at the choices what I'm going to do because you're not always going to have these choices here sometimes you might just have to identify the vertical asymptotes or the zeros or the removable discontinuities is I'm going to factor this out and hopefully or I'm going to factor the numerator and the denominator to hopefully make things a little bit more clear and we're going to think about what X values make the numerator and/or the denominator equal to zero so the numerator can i factor that well let's see what two numbers their product is negative 24 and they add to negative 2 let's see that would be negative 6 and positive 4 so we could write x minus 6 times X plus 4 did I do that right negative 6 times 4 is negative 24 negative 6x plus 4x is negative 2x if that's right all right and now in the denominator let's see 6 times 4 is 24 6 plus 4 is 10 so we could say 6x plus 6 times X plus 4 so let's look at the numerator first the numerator numerator 0 if X is equal to positive 6 or X is equal to negative 4 the denominator denominator is 0 if X is equal to negative 6 or X is equal to negative 4 so let's think about it can we could we simplify this what we have up here a little bit and I and we're just in keep in mind what I just wrote down well it might be just you know jumping out it you have an X plus 4 divided by an X before why can't we simplify this expression a little bit and just write this whole thing right our original original expression as being X minus 6 over X plus 6 where X if we want to be algebraically equivalent we have to constrain constrain the domain where X cannot be equal to negative 4 well this is interesting because before when x equals negative 4 we had all this weird 0 over 0 stuff but we were able to remove it and we were able to make it just kind of this extra condition and if you were to graph it it would show up is a little bit of a removable discontinuity a little bit of a point discontinuity it's just one point where the function is not actually is actually not defined so in this situation and this is typical where there's a thing that you can factor out it was making the numerator and the denominator equals zero but you can factor that out so it no longer does it that's going to be a removable discontinuity so x equals negative 4 is a removable discontinuity and once you factor it out all of the things that would make it a removable discontinuity then you can think about what's going to be a zero and what's going to be a vertical asymptote if it makes once you've factored it out everything that's common to the numerator and the denominator if of what's left something makes the numerator equal to zero well then it's going to make this whole expression equal to zero and so you're dealing with a zero so at x equals 6 6 would make the numerator equal to 0 6 minus 6 is 0 so there you're dealing with a zero and two can make the denominator equal to 0 you get x equals negative 6 that would make the denominator equal to zero so that's a vertical that's a vertical asymptote and it's called a vertical asymptote because as you approach this value is approach negative 6 either from values smaller than negative 6 or larger than negative 6 the denominator is going to become either very very very it's either going to become a very very very small positive number or a very very small negative number it's going to approach 0 either from above or below and so when you divide by that you're going to get very large positive values are very large negative values and so your graph your graph does stuff like that so you have a vertical you have a vertical asymptote it might either do something like that or it might do something it might do something like this where this is your vertical asymptote and then when you approach from here goes bumps up like that so either way so that's why that's a vertical asymptote let's do another one all right so same drill and like before let's see if you can work it out on your own let me factor it so see two numbers the product is negative 32 they're gonna have to have different signs 8 & 4 and then we want the larger one to be positive since they have to add up to positive for X so X plus 8 times X minus 4 yep I think that's right and then divided by this is 4 times 4 is 16 4 negative 4 plus negative 4 is negative 8 so this is X minus 4 times X minus 4 now this is really really interesting because you might say oh I have an X minus 4 X minus 4 maybe x equals 4 is a removable discontinuity and it would have been unless you had if you didn't have this over here because x equals 4 even after you factor this you still can't it's the fact the expression the function is still not going to be defined when x equals 4 and so this this function is actually equivalent to X plus 8 over X minus 4 I don't have to put that extra little constraint like I did before I don't have to put I don't have to put this little constraint that describes this removable discontinuity because that constraint is still there after I've factored out after I've cancelled out this X minus 4 with this X minus 4 and so here we've we've this is actually an algebraically equivalent expression to this one right over here and so now we can think about what the what the zeros or the vertical asymptotes or the removable discontinuities are going to be well something that makes the numerator 0 without doing it to the denominator is going to be a 0 and x equals negative 8 makes the numerator equal to 0 without doing anything to the without making the denominator equal to zero it would make the denominator equal to negative 12 so you can literally just evaluate H of negative 8 is equal to 0 over negative 12 Oh which is equal to which is equal to 0 so that's why we call it a zero what about x equals 4 well x equals 4 is going to make the denominator only equal to 0 and so that's going to give us a vertical asymptote and we're all done