Up next in 10
How To Find A Slant Asymptote Of A Rational Function & Graph It
35 views
Feb 20, 2025
In today’s tutoring session, the focus will be on finding slant asymptotes for more complex rational functions. Last week we covered simple rational functions and our main objective was finding the vertical and the horizontal asymptote. We will also be introducing the “ long division “ as it is an important part of This process. I hope this helps
View Video Transcript
0:00
in today's lecture you're trying to find
0:01
about slant asymptotes last week we
0:03
talked about the vertical asymptotes and
0:06
in the horizontal asymptotes and to find
0:08
the vertical asymptotes we usually use a
0:10
domain of the function right which is
0:12
the denominator and to find the
0:15
horizontal asymptote we look at their
0:16
numerators and the denominator right now
0:19
you notice in this case the denominator
0:22
is X and the numerator the highest
0:26
degree in the denominator is one right
0:29
which is X to the power of one and then
0:31
the highest degree in the numerator is X
0:33
to the power of two - right so well I'll
0:39
take the one with the highest degree all
0:40
right degree means when you look at this
0:42
like 3x squared minus 2x plus 2 right
0:45
the variable with the highest number is
0:47
well x squared right and in the
0:51
denominator the the one with the highest
0:53
is X and the degree is one that's the
0:57
highest part this denominator has here X
0:59
or one right now whenever you have this
1:03
type of function where the degree in the
1:04
denominator is less than the degree in
1:07
the numerator when you compare the
1:09
highest you know that there is
1:11
possibility of a slant asymptote okay
1:15
and to find the slant asymptote we do
1:18
the long division we're gonna divide
1:19
this fraction here so let's pull this
1:22
aside and we're gonna divide that on a
1:24
piece of paper so we made a function
1:26
again the denominator is 3x squared
1:34
minus 2x minus 1 so remember the first
1:42
thing is this degree right of the new
1:45
denominator is 1 right because X to the
1:49
power of 1 okay
1:51
and the numerator is x square right
1:54
degree is 2 right so since the degree in
1:59
the numerator is higher than the
2:00
denominator you know that there is a
2:02
slant asymptote so we're gonna find it
2:05
so we're gonna do our division have you
2:08
ever done a long division before yeah
2:10
you forgot
2:12
it'll come back okay I set it up so it's
2:15
gonna be X minus one here ding didn't
2:18
hold the paper down okay thank thank you
2:19
X minus 1 and then we have on top three
2:23
x squared minus 2x plus you do move your
2:29
finger for one second - right okay and
2:32
this is I'm gonna do I just said I'll be
2:40
right away - you'll be ended at the
2:41
bottom here I don't know what I was
2:43
thinking
2:43
are you watching this this is that what
2:45
I just did it that was worst thing I
2:47
could have done three x squared so then
2:58
yes we do how many tank and X minus 1
3:02
goes into 3 x squared three x times
3:04
right oh because you take any X here you
3:09
put it into the 3x you left for what 3 X
3:12
- 3 x times X is 3 X square right and 3
3:16
x times negative 1 is negative 3 X okay
3:20
and then we're gonna do the subtraction
3:21
yeah that's the long division I'm sure
3:23
you don't you don't and then negative 2
3:37
X minus negative 3 X will be X is X
3:41
because that's what a negative and in
3:43
any negative gives you so they'll be 3x
3:46
minus 2x you're left with X right and
3:48
then you bring down - so this is not X
3:52
plus 2 correct now how many times as X
3:55
goes into X it's one time good job so
3:59
you surprise one okay so 1 times X is X
4:04
right and one times negative 1 is
4:09
negative 1 right negative 1 okay I mean
4:14
you do this X minus X 0 2 minus negative
4:18
1 3 2
4:22
twisted asha remainder right now this
4:25
piece that we have here that would be
4:28
your slinked or your horizontal
4:31
asymptote so we're gonna call it y
4:33
equals 3x plus one is the horizontal
4:38
asymptote but they all know this length
4:40
or the oblique asymptote okay so why is
4:43
it so you've been three X plus one we're
4:50
gonna graph it okay three X plus one and
4:53
then this is how we write it I'm gonna
4:55
I'm gonna show you also Croatian I tried
4:59
to find where the slant asymptote is by
5:01
using long division yes so first I saw
5:05
how many times X goes into three x three
5:08
x squared
5:09
mm-hm and I found that it only goes into
5:12
three like three exercises and then
5:16
after that I did one mm-hmm
5:19
I didn't wanna how many times is no well
5:21
then when you did the 3x time what did
5:22
you set up here sure I'll do it when I
5:25
got to X I put it down here but how did
5:27
you get that you did 3x times 3x times X
5:31
gives you what 3x squared and then 3x
5:34
times not a 2x and negative one negative
5:37
one that gave me negative 2x but I
5:39
probably made a positive soccer remember
5:41
because you have to switch off the
5:43
switch up the orders of are very good
5:45
didn't window signs in it yeah and then
5:47
I subtracted I subtracted in this I got
5:51
3x minus 3x squared and I got zero
5:54
that's wrapped in negative 2 plus 3x and
5:57
I got 1x and I brought up the to then I
6:01
did the same thing and we're also how
6:03
many times does X going to do when a 1x
6:07
uh-huh and I can't one in one time one
6:09
time good yeah and then I saw
6:13
multiplication again now then how many
6:16
times 1x you did one times tax way yes
6:23
you did
6:24
1 times X Y and one times X is negative
6:28
no he's uh negative is 1x right 1x and
6:31
wanted to have negative one is negative
6:32
one day we switched oh yeah I switched
6:35
order it was negative imagine any
6:38
negative sign here so he has to be
6:39
switched in there yeah and then as I
6:43
didn't 1 X minus 1 X 0 netted 2 plus 1
6:49
and then once your slant asymptote 3 X
6:52
plus 1 was that easy over yeah we're
6:56
gonna do is trying to rewrite the
6:58
function giving the asymptotes and you
7:01
on divine and the dividend and remainder
7:04
correct yeah so this is the equation 3x
7:07
squared minus 2x plus 2x minus 1 that's
7:10
a function and when we did our long
7:12
division we came up with this over here
7:14
right this is the asymptotes and this is
7:17
your denominator and this is your
7:18
remainder so to rewrite this function
7:20
it's gonna look like this so I just do
7:23
now I'm gonna write it down and you see
7:25
I can hold on the paper from you can
7:29
work over there ok f of X so first thing
7:32
we need is the asymptotes all right
7:33
absolutely 3 X plus 1 correct
7:36
so it's gonna be 3 X plus 1 and then the
7:41
remainder plus oppose the remainder day
7:43
3 over the denominator X minus 1 and
7:48
this is how this function decomposed
7:50
would look like okay so this is your
7:54
essence asymptote and this is your
7:56
remainder and this is your denominator
7:58
so we're gonna write this this is
8:00
asymptote was slant asymptote right and
8:06
this is your denominator right and this
8:11
is your remainder so this is how you
8:14
break down the function once you're done
8:15
and this is gonna be your asymptotes
8:17
here correct correct
8:19
the limit as X approaches negative
8:23
infinity and positive infinity right so
8:26
we have a function and what we need to
8:28
do is just break it down into this here
8:31
we have
8:33
limit of the entire function which is 3x
8:35
square negative 2x plus 2 and X minus 1
8:38
we talked about simple rational function
8:40
last week and what we do is we figure
8:44
out the the coefficient with a highest
8:49
degree over here right which is or in
8:52
this case is 3x squared I didn't fashion
8:54
the use coefficient I mean the variable
8:56
in this case and we do the same thing
8:57
with the denominator which is X or 3x
8:59
over X we get to this point and as I X
9:03
approaches negative infinity what's
9:05
gonna happen to the time it's gonna go
9:07
to negative infinity so it's gonna be
9:09
negative infinity same thing with a
9:11
positive side so this is how you find a
9:14
limit you take the highest degree on the
9:16
numerator and the highest degree in the
9:18
denominator and then you find a limit of
9:20
the function as X varies from negative
9:23
infinity to positive infinity
9:24
all right last step to graph the
9:26
function is meaning to figure out you
9:28
need to draw the asymptotes and then get
9:32
the X and y intercept and then we can go
9:34
ahead and draw the function for the
9:36
first thing the one thing that we were
9:38
missing was the vertical asymptote and
9:40
to find a vertical asymptote we have to
9:42
look at the denominator we worked on the
9:44
last week so we have to set it up this
9:46
way and X cannot equal 1 so x equals 1
9:49
will be your vertical asymptotes and
9:52
then the next thing is graphing the
9:54
slant and we chose some random point and
9:58
when x is 0 well with one you just plug
10:01
it in here when X is 1 well before so we
10:04
drew that over here this length so when
10:05
x is 0 Y is 1 when X is 1 Y is 4 and if
10:09
you draw it as a dotted line and then it
10:11
either thing will be to find it to draw
10:13
the vertical asymptote which is x equals
10:15
1 right here we just broke one and draw
10:19
vertical line and then the last step is
10:21
we found
10:23
y-intercept in this case when X to 0
10:27
will be negative 2 now to find X and
10:29
said we have to use the quadratic
10:30
formula we don't and I'm not going to
10:33
use that because I was gonna fracture
10:35
this function but it's not factorable so
10:38
we have to use a quadratic but once you
10:40
have this is pretty much a done deal you
10:43
can draw a function given your
10:44
asymptotes and your vertical asymptotes
10:47
in your sláinte
10:48
so the functions got look something like
10:49
this is gonna go to negative 1 and it's
10:51
gonna be symmetric over here so this is
10:53
the function make sense yeah okay now
10:56
we're gonna work on the next one
#Mathematics
#Other
#Teaching & Classroom Resources