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How To Graph Simple Rational Functions
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Feb 20, 2025
In today’s lecture my student and I worked on graphing simple rational functions. There are some easy steps one can take to graph these types of functions and this is what we are going to cover. The very first thing I recommend to my students is finding the domain of the function. But in the assignment my student was given , there were guidelines that were required and that’s what we followed. If you need help in math, be sure to check out http://tayibs.com/get-math-help/
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0:00
okay so we're trying to graph simple
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rational functions okay now in the first
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question you're asked to find the zeros
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of this function that's a rational
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function by the way and we need to find
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the zeros and then the y-intercept and
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then we need to find what kind of
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discontinuity okay and to do that we're
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going to be finding the limit of the
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function this is to find the asymptotes
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all right now the first thing we want to
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do is I guessed it we are asked to find
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the zeros of the function so to find a
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zero that means what are the values for
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which the function is crossing the x
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axis this is the x axis this is the x
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axis so when you are asked to find the
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zeros of a function you're trying to
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find the egg under the point for which
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the function is crossing the x axis okay
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so to do that we're gonna do you're
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gonna first write down the function and
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then we're gonna set it equal to zero
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okay so that I mean where is the
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function crossing the x-axis and then
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we're gonna solve for X all right
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so the first thing I want to do is I'm
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gonna add one and stop me if you know
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again something yeah wait so right so
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we're doing the same thing
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how we use use this equation where to
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find the values for which the function
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is crossing the x-axis all right so we
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have 1 over X plus 2 equals to 1 okay
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now what do I do for me remember where
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we did last with the cross
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no we don't do this I'll cross multiply
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right 1 times 1 is 1 and then X plus 2
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times 1 is X plus 2 okay and then from
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here I can solve for X I'm going to work
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subtract 2 and I'm going to be left with
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1 minus 2 equals negative 1 equals 2x
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okay that's the only zero this one's
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only has one zero so that means this
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function is gonna cross the x-axis only
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once when X is 1 the I mean when X is 1
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Y it's gonna be 0 okay if only if
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there's only one in that goes yeah when
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X is negative 1 Y is gonna be sued I
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mean this is where you're gonna cross
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the x-axis right now to find a
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y-intercept y-intercept is basically the
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same process but now we're doing we want
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to find the values for which the
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function is crossing the y-axis okay so
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the y-intercept is the values for which
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the function is crossing the y-axis so
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to do that we're gonna set y equal to
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zero and solve for no we're gonna set X
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equal to zero and solve for y okay
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so basically what I'm doing is I'm
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finding F of 0 meaning I'm replacing X
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by 0 to find y so anything that's exit
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plug in 0 there you go so it's gonna be
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like this 1 0 to minus 1 you just need
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under the cross multiplication because
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we just want 0 plus 2 is 2 so you do 1/2
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minus 1 right now we need to find a
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common denominator okay we don't have
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you can use your calculator well it's
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better to do just find a common
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denominator right this is like 1 over 1
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so the common denominator is 2 now you
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see we are doing our job right again
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that's why it's good to know this thing
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so the common denominator is going to 2
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so to get to a more more party for my 2
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and this by 2 correct
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so I get 1/2
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- - / - right
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it's 1 - 2 over 2 which is negative 1
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over 2 right so my y-intercept is
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negative 1/2
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all right so so when y is equal to
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negative 1/2 X is 0 now you have your
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two points right so we have X & Y when y
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is 0 X is negative 1 and then when x is
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0 Y is negative 1/2 okay now they want
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to know location and discontinuity okay
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the phone so to find out we need to find
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the domain of the function
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basically we want to know the values for
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which the function is defined okay
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remember we will work on something
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similar when we were working a piecewise
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function I was two weeks ago right so
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look at this function here f of X is 1
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over X plus 2 minus 1 right so I don't
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want this to be equal to 0 you know why
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because you want to see if it's you want
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to see if as the distance away exactly
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because 1 over 0 is undefined right so
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you need a friend when you limit first
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but we need to find the domain of the
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function so we don't want X plus 2 to be
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equal to 0 so I already changed so we
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don't want that right so we're doing the
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same thing so we don't want X to be
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equal to negative 2
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right so when X is negative 2 the
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function will be undefined so this will
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be a vertical asymptote meaning you
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don't want X to be equal to negative 2
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at the moment they'll be discontinuity
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because the function cannot touch this
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this is a limit one of the vertical
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asymptotes this canoe discontinuity
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discontinuity right so that means x
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equals negative B we call it a VA or
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vertical asymptote right
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now we can find a limit of dysfunctions
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all right so as X approaches negative
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infinity right we're gonna replace
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remember what we did in last week or two
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weeks ago plus two minus one right as
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this approaches negative infinity what's
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gonna happen over infinity infinity is
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born like it's gonna be when your
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purchase no no don't worry about it look
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let's look at it right replace this
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number here by the biggest number you
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can think of in a negative side negative
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mainly negative meaning right if you do
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1 over negative the minute
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what are you gonna get you're gonna get
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zero or something close to zero okay
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what's one of them any zero point one
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zero zero zero zero zero zero are close
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that to dislike like 9 no no can you do
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can you divide one take one biscuit and
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try to divide it into 1 million people
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you can't what are you gonna get the
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crane nothing basically a so X which is
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negative infinity this is gonna be
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getting close down to 0 so it's gonna go
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small smaller so therefore you're gonna
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have negative 1
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the limit will be negative 1 all
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Colossus policy this this one here we
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get close to zero right
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it goes 1 over X plus 2 we tend to go
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close to 0 and negative so therefore all
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you have left is what is your next
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negative 1
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nothing so all you have left is negative
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1 you see what I mean yes uh Bobby yeah
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you only mean will be negative 1 and
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then from there you can grab your
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function right so now to graph the
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function first we need to graph the
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vertical asymptote right it's negative 1
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what was our introduction to negative 2
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right so negative 2 we have vertical
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asymptotes here
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right and then we have how many zeros or
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ones you have negative 1/2 right
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actually no that was nothing and then we
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also have zero and negative 1/2 this
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would be the function here right and
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then we can find some out of point so
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you can hang it up like we have every
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piece of information that we need right
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we have negative 1 and 0 for 0 0 and
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negative 1 half our y-intercept we have
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x equals negative 2 us our vertical
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asymptotes and as X approaches negative
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infinity we have negative and say the
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same thing for infinity right now on
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this portion we've graphed our vertical
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asymptotes and then we have our
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horizontal asymptote right and the
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function is gonna look like this when we
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uh get this curve you see yeah that
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makes sense like it doesn't touch it
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doesn't talk nothing to this
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discontinuity there so you what you want
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so what we're trying to do is find the
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two points like these two points yeah so
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that's what we want to find and then if
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you have to do it but this is only this
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stuff is only to determine this did you
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see how the function is behavior okay
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and most likely you're gonna have
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something here and you're gonna have
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something here too like this on the
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bottom right because this is a simple
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rational function all reflect reflect
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exactly right yeah it's a little
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different from the last one the last one
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had a integers enumerator right and this
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one we went to the same process find
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found the zeroes found a y-intercept we
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found a vertical asymptote and now we're
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trying to find a limit as X approaches
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negative infinity of the function now if
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you see here there's a difference right
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here here you have X was fallen in X
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minus 3 the easiest way to learn this is
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this the highest degree of the
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polynomial as of under the numerator is
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1 right
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and they asked you three of the kamo
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meters in the denominator is one as well
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so what you do is this you have X plus 4
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and X minus 3 y it is regard the
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integers for negative to disregard so I
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don't care about don't care about their
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like this moment so what you do is this
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divide x over X right you get what
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Thanks X over X gives you and no X of X
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gives you 1 what's 1 over 1 verse 2 over
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2 so what's to loyalty which one you
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think in to her again what's X over X
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its 1 1 Y X X is 1 X over X is 1 1
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that's it
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ok now to find it limit of this function
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right you're doing this limit as X
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approaches infinity negative infinity
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right of X over X right which is one
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meaning no matter how far you go all is
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all be one so that'd be the name that
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would be your horizontal asymptote so
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for this for this equals 1 for this the
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first will be y equals 1 yeah because
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whatever you do okay so anytime it's any
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time you just disregard this the
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integers and so on especially if it
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looks like this before something like
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this and this will always be the magic
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right so any number the whole whole
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number yeah but it's not disregard it no
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b1u not always 1 because if this was
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like this let me write down functioning
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at like 3 2 X plus 5 over 3x minus 1
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disregard these integers right let's
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take these two here okay in your ass
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do not that it's over dude so what's 2x
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or 3x ago just look this will go away
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right yes what you left with
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so earth day 2/3 two things same thing
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no matter where you go infinity is
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always going to be 2/3 for this function
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so it's just a matter of trying to make
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the X's go away you take the highest
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degree okay and this is what happens
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most of the time that exists yeah
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