okay this video I'll show you how to find the horizontal and also the vertical asymptotes of
this function first for the horizontal acid hope remember we have to take two limits the
first one is the limit as X approaches positive infinity of this function and if we do get a
finite valued right here then y equals 2 that a finite value will be the horizontal slope and we
also have to work out the limit as X approaches negative infinity of the function because this
value might be different than that if this is also finite then we will have another horizontal slope
so for example just like the inverse tangent X it has two different horizontal as tops but of
course these two files could be the same so in that case you still have one and if the
limit don't exist what limit is like infinity or negative infinity then we don't have horizontal
Stokes so we're just going to work this out first and let's get going first we take the limit as X
approaches positive infinity of this function 3x minus 6 over x squared plus X minus 6 and notice
this one here is really nice because we have a polynomial over another polynomial so to take
the limit as 8 approaching infinity X approaching infinity you can just compare the highest power
of X on the top and the highest power of X on the bottom and you see that this is the same as a the
limit as X approaches passed team Finity on the top is just 3x on the bottom it's just x squared
remember we can do this if X is approaching infinity or negative infinity right and the
purpose of this is that we are just trying to compare the ratio of the top and the bottom
the dominating part of the top over the dominating part of the bottom and you see that in this case X
x-square they can cancel so we have only one of it and this is saying the limit as X approaches
infinity we get 3 over X and plugging infinities who here we get 3 over infinity and notice that
this is just a none infinity it's just a finite number over infinity we get to a conclusion right
here which is 0 and in this case we see that this is finite so we have y equals to 0 as one of the
horizontal obstacles if this is different that we have to if this is the same was to have won
by anyway that's pretty much it for that part and now for this part we do the same thing the
limit as X approaches negative infinity of the same function of course and we can still do the
same thing right when X is approaching negative infinity on the top we can just care about the
3 X on the bottom we can just care about the x squared once again we are just trying to
compare the dominating part of the top and also the bottom and this is the limit as X approaches
negative infinity of 3x over X square and I know some of you guys noticed this is pretty much the
same as that and you cancel things out you get the limit as X approaches negative infinity of 3 over
X and you can plug in negative infinity here and you get 3 over negative infinity right here even
though you have positive 3 over negative infinity this is still 0 if you want to put on negative 0
that's still 0 so this is 0 and notice we have y equals to 0 so if they are the same that means
we only have one horizontal acetal and that is y equals to 0 and this right here is it right and
now let's talk about how we can find the vertical asymptotes and in this situation when we have a
rational function like this what we want to worry about us we have to have a noun serious situation
on the top over zero like that well sometimes you may end up with zero over zero in that case
it might be an open circle on the graph namely a removable discontinuity but when you have a non
zero over zero situation most likely the function will have a vertical stop there so with this set
we are just going to worry about the bottom for now so we can find a candidate of the particular
slope that means I will just have to set x squared plus X minus 6 I will have to set this to be zero
and I can solve this by factoring we get X plus 3 times X minus 2 from the first part we see X plus
3 has to be 0 that means X has to be negative 3 for the second part we see that X minus 2 has
to be 0 that means X has to be 2 and from this point on you have to be extra careful yes these
two numbers are going to make the denominator 0 so you have to get rid of these numbers from the
domain but if you're talking about the vertical a sort of situation pay attention to this we
also have to make sure that the top cannot be 0 so what do I mean by that let me demonstrate we
will have to plug in negative 3 into the original right into the original and see if we have a non
zero over zero situation or not let's check that out plugging negative 2 into here we get 3 times
negative 3 and then minus 6 over over what over zero because we know if you plug in negative 3
into the pattern it will be zero things with just theta right and now you see on the top 3 times
negative 3 is negative 9 minus 6 is negative 15 over this dorota already half so you know if it's
this condition so yes there will be a vertical asset hope at negative 3 so I'll just put this
down the a right here right so checks on the other hand though if you plug in 2 into the original
function we see that that's going to be 3 times 2 minus 6 over what over the 0 why because you know
it already because you just thought that but on the top you seeing this time 3 times 2 is 6 minus
6 is 0 over this rule that we already have in this situation you actually don't have a vertical asset
or you can do a little bit more investigation and investigation is you take the limit as X
approaches 2 of the function and you see that you get 3x minus 6 over that x squared plus X minus 6
and to do this limit you know that if you plug in 2 into all the X you and os3 over 0 so you
simplify the expression you see that on the top is 3 times X minus 2 on the bottom you factor it you
get that X plus 3 times X minus 2 and of course you see X minus 2 can be cancelled it and then
from here you get the limit as X approaches 2 of 3 over X plus 3 and plug in 2 now you get 3 over
2 plus 3 namely you get 3 over 5 this right here tells you when X is approaching to the Y value is
approaching 3 over 5 and when X is exactly to the functions actually undefined that means you will
have an open circle when X is 2 then this right here it's actually the removable discontinuity
right so you'll see the open circle you have a nice graph but there's an open circle there's no
jump because limit you to have a left rudder mapping equal to 3 over 5 so this right here
it's a removable there's continuity right this continuity like that it's movable it's not jump
even though you have an open circle but anyway this it's not a vertical as though so be sure
it's just to all this kind of check and Indian will just write this down for you guys right
here for the 30 call us at hope we only have that and you write it down the formula x equals 2y but
this is so this X is equal to negative 3 this is this is it and in addition if the question
is asking you for removable discontinuity this is what you do so I can tell you every movable
and she's ready down removable removable this continuity right is add x equals to 2 so yeah
that's what you have to answer if the questions also asking you for that by anyway this is it