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Then differentiation rule is?
u times the coefficient by it
$ax^{a-1}$
and add one to power
Minλ
You are mixing it with integration
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test
To type in the advanced section please read #get-advanced-access. It contains the instructions to gain access.
!volunteers
Helpers are just people volunteering their time to help you. Be polite and patient.
!trigs
#geometry-and-trigonometry - euclidean geometry, coordinate geometry, trigonometry
is !trigs new or sth
!trig
#geometry-and-trigonometry - euclidean geometry, coordinate geometry, trigonometry
they are the same
ive never seen either of those
talks and γ dont work
!elliptic curve meme
🍎 = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
🍌 = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
🍍 = 4373612677928697257861252602371390152816537558161613618621437993378423467772036
!cohomology meme
H*(🌭, 🍔) ≅ 🍔[🍉]/(🍉ⁿ⁺¹)
.close this was fun
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can someone tell me where i went wrong
question: find x+2y
on the left side, you effectively only multiplied the second term by 2sqrt(2)
you need to multiply the WHOLE side
also why did the = sign turn into an arrow
oh i thought so
habit
i should have just took whole term in common denominator
and if the end goal is to isolate
x + 2y,
you shouild only multiply using 2
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how do i know which one to put as u and dv? I used LIATE and it didn't work!
this is a "go with your gut" kinda question tbh, but lemme see which combination yields the least calculation
i'm pretty sure it's making xe^2x as u
since I did the opposite and it led me nowhere
yea e^(2x)/(2x+1)^2 is nasty to integrate
and reducing the (2x+1)^2 to (2x+1) could help out
actually you could sub and do some nice algebra but ill let you do the ibp for now
awesome, so im just kinda supposed to try and see with ibp which works out better
basically
liate works on simpler problems
you should visit the bprp video on liate (and why he dislikes it like i do haha)
tabular just helps me out more
@alpine igloo Has your question been resolved?
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i've got an issue with diverging and converging sequences; the traduction is pretty easy
ik know it but I can't proove it
C'est un vrai/faux je suppose?
Bha faut y aller un par un de tte façon
oui
T'as 0 idée des réponses là ?
bah pr moi la 1ere est vraie
et pareil pr la 3eme
mais je sais pas si les autres le sont ou pas
Bha la 4eme c'est l'inverse de la 3eme en vrai
Ok ils ont échangé un et vn mais ça reste suite convergente + suite divergente
oe pas faux
Ouais donc la 2 alors
Pas forcément tu peux avoir des trucs à la (-1)^n aussi qui oscillent
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this isn't exactly a math question but I am looking for any insights on how to study for a math accuplacer
I do have study sheets but it seems underwhelming for an engineering school to have pre-algrebra questions
am i tripping or did a certain discord server pop up
might wanna go in #study-discussion there's more ppl looking in there
At least they're automodded now
So many fkin spammers these days
yeah
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hi i wanna know the idea of solving this derivative problem
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
Let the point of tangency on the graph (C) be P(x_0, y_0), since P is on the curve y = x^3 + 3x^2. You can use this to find an equation of the tangent line at x_0
once you get that, you can substiute M(m,0) as well
Also, what is the relation between the slope of two perpendicular lines?
k₁.k₂=-1
Quick question for MxRgD, kind of a spoiler: ||Arent there infinite solutions?||
||I don't think so?||
it would be
Δ: y = (3x₀²+6x₀)(x-x₀) + x₀³ + 3x₀²
Δ is the equation of the tangent to (C) at x₀
right..
Do you not have any information on graph C?
noo, the problem only provides these info, i cannot use web to create and see the graph c
looks right
sooo M(m;0) ∈ Δ
0 = (3x₀²+6x₀)(m-x₀) + x₀³ + 3x₀²
i understand this, what should we do next
expand the bracket part
0 = 3mx₀² - 2x₀³ +6mx₀ - 3x₀²
0 = -2x₀³ + (3m-3)x₀² + 6mx₀
now you can factor out the x_0
which gives x_0 = 0 being one of the solutions. The other two points of tangency are roots of the quadratic
how can i factor out the x₀
$x_0 [-2x_0^2 + (3 - 3)x_0 + 6m] = 0$ like this
MxRgD
Since one tangent is at x_0. What would the slope at x_0 = 0 be
slope = 0
Alright so if the slope is 0, what does that imply?
it's a horizontal line
Yep, but the two other tangents could be perpendicular to it
What does that also imply?
im confused
what are you confused about
like the other two are perpendicular to the first one?
yeah, the one at x_0
since the question says two tangents being perpendicular
okay if it's a horizontal line, and you have another line being perpendicular to it. What does it look like?
it is a vertical line
vertical... or two of them are perpendicular to each other.. i dont know how to answer 😭
Yeah they are vertical
or two of them are perpendicular to each other
ig you meant 3-3m here
but if it's vertical, what does that imply for the slope?
hmmm, the slope would go to infinity i guess
eh sort of? But a better answer would have been it's undefined basically
the slope is undefined
but if the slope is undefined then its derivative is also undefined
why though, my teacher didnt talk much about these things
oooohhhh x₂-x₁ will always be 0 right
yeah
y = f(x) = x³ + 3x²
okay
alright I'm back
Yeah and it's derivative is 3x^2 + 6x
but we said here the derivative has to be undefined
see the contradiction here?
yea f'(x) = 3x² + 6x
so then which one of these options is right
the 2nd one
but im still so confused 🥹 cannot load
i still cant connect all of them together to fully understand here
Which part confuses you
f(x) = x³ + 3x²
f'(x) = 3x² + 6x
Δ: y = (3x₀²+6x₀)(x-x₀) + x₀³ + 3x₀²
M(m,0) ∈ Δ
-> 0 = x₀.[-2x₀² + (3m-3)x₀ + 6m]
hmmm
from the contradiction, x₀=0, where the slope is unidentified, all the things after, kinda new to me and hard to understand
Yeah I'm basically explaining why the 2nd option has to be true
which you would maybe need to do in a question like this (justifying why you rejected the 1st option)

so from here
I'll rewrite this as $x_0 [2x_0^2 + 3(1 - m)x_0 - 6m] = 0$
MxRgD
okay
since we know that x_0 = 0 implies the two perpendicular tangents cannot include the one at x_0
the other two points of tangency must come from the roots of this quadratic $$2x_0^2 + 3(1 - m)x_0 - 6m = 0$$
MxRgD
ohhhh alright read it again and somewhat understand this
not fully but better now
cool
yeaaa
Let these two other points of tangency be x_1 and x_2
alright
what would the slopes of both of these tangents be
the slope would just be the derivative right...
i find the derive of this
That's incorrect since your original function is this
and this is your derivative needed for tangents with point x_1 and x_2
so 3x₁² + 6x₁ and 3x₂² + 6x₂
yep
I'll rewrite this as $$k_2 = 3x_1^2 + 6x_1 = 3x_1(x_1 + 2)$$
$$k_3 = 3x_2^2 + 6x_2 = 3x_2(x_2 + 2)$$
MxRgD
so you know k_2 and k_3 are perpendicular
we can use this
ooooo
so apply that and you get...?
9x₁x₂(x₁x₂+2(x₁+x₂)+4) = -1
cool
do i have to use viete here
MxRgD
yep
you'll get everything in terms of m
and then you can solve for m
and that's your answer
yeah that's right
It's pretty long winded the problem 
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,rccw
what's the original question?
sorry, I was sending in the question
the part I am stuck on is c
no worries 
@chilly crane Has your question been resolved?
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Help my math or get terminated
A - 7B + 3C = 2
2A - 6B + 4C = 8
A + B + C = ?
No decimals or fractions accepted in this
Only whole numbers
I hate this level
subtract
I've been trying to solve by substitution for over 30 minutes
Elaborate bro
subtract equations
Use elimination.
A - 7B + 3C = 2
-(2A - 6B + 4C = 8)
u can subtract eqn 1 and 3, and the A's will cancel out
You're better off the other way around
Do this, but subtract the other way around.
oh nvm there is no eqn 3
Should end up the same either way
I ain't scared of negative numbers
That's okay then. You can keep going.
after you do that you get −8B+2C=−4
then divide by 2
you get -4b+C=-2
solve for C
-A -B -C = -6
C = -2 + 4b
This is overkill, the equations are set up so that you get A+B+C if you subtract one from the other
Bruh but now there's no coefficients left
Wdym
A+B+C=(8−5B)+B+(4B−2)
💀
The coefficients are -1, you can just multiply through by -1
The Bs cancel out
Yeah
and you get 6
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how do i solve this
it says: the value of the expression 18-(-30):(-2)+(-2)*(-1)
(El valor de la expresion)
a) 9
b) -5
c) 5
d) 9
i thought it was simple arithmetic's but when i try to do it i notice is dividing from 0
is on my diagnosis to university : (
isnt multiply first?
parentesis, power, division/multiplication, add/sub
then (-2) * (-1)
minus * minus = +
so is 2
-2 + 2 on the numerator
mejor, por que me cuestan algunos terminos
18-(-30) = 48 y ese es el numerador
Si, el orden es
- Resolver todo lo que este en parentesis
- Potencias
- Multiplicaciones y Divisiones
- Suma y Resta.
Estas seguro?
Estas empezando con sumas, y ya se ven varias multiplicaciones y divisiones.
deberia empezar con (-30):-(2)?
Si, y te queda tambíen pendiente la multiplicacion.
Listo
eso gracias : DDD
Si te tengo que dar una explicacion facil
Podes hacer esto
,tex $18 - (-30) : (-2) + (-2) \cdot (-1)$\par
$\underline{18} - \underline{(-30):(-2)} + \underline{(-2) \cdot (-1)}$
Fijate que dividis los terminos en cualquier lugar que veas un mas o un menos cuando no este entre parentesis.
que es que este subrayado abajo?
Para que sepas cuales terminos se separan
18 es un termino
-(-30):(-2) es otro
+(-2) * (-1) es otro
entiendo
ya entendi, voy a separar los terminos para hacer este tipo de ejercicios gracias : DD
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How do I find the horizontal intercept here?
do you know what defines a horizontal intercept?
the position where the line touches the X axis?
right. and at the x-axis, what is necessarily true?
mhm. so set y = 0 and find x.
6/5?
looks correct.
um.
(5, 6) implies your horizontal intercept is at x = 5, y = 6.
didn't we agree earlier that the horizontal intercept occurs at y = 0?
have a look at how you wrote the coordinates for the vertical intercept.
follow that method, but change it to fit the concept of a horizontal intercept.
sure. all the best.
@dark cairn Has your question been resolved?
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Okay. I do not know how to prove linear independence or dependence. I have no familariaty with complex numbers.
I tried with the usual definition of linear dependence
i am about to do that
uhm, uh. I deleted it
for a), $a_1, a_2 \in \mathbb{R}$ such that $a_1(1 + i) + a_2(1 - i) = 0$ with not all $a_i = 0$
andre
$a_1(1 + i) + a_2(1 - i) = 0 \implies a_1(1+i) = -a_2(1 - i)$
andre
what do you want to show ?
(a)
first
sure but here what is your goal ?
Show that 1+i, 1-i is linearly independent
$a_1 + a_1 \cdot i = a_2 \cdot i - a_2$
andre
so if you have an equality of two complex numbers like this, is there any equalities you can extract from it?
i'm not sure where you got that from?
I don't know
well if you had a + bi = 3 + 4i, what are a and b?
a = 3 and bi = 4i
yes, so the real and imaginary parts must separately be equal
why
how is a = 3
and bi = 4i
Is there something I can see that proves it?
Normal algebraic manipulation would mess with me, I think
Lin Xia
your textbook defines complex numbers as being pairs of real numbers (i.e. a + bi is notation for (a,b)), and pairs of numbers can be equal only if each member of the pair is equal
I did not see that
LADR books do that??
it's the first thing in the entire book
Omg
I am dying
I must have forgotten about it
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I dont even know where to start this
Try ratio test
yeah i think it is the ratio test but im not sure how to get it into that form
What do you think the coefficient is
x?
No. Review definition 10.1.1
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/10%3A_Power_Series/10.01%3A_Power_Series_and_Functions
so would it be the n^2 if so, when we seperate the x^(n+1) into x and x^n what do we do with that extra x
That's the incorrect coefficient
That extra x tells you the radius and part of the interval of convergence
See example 1 for a worked problem very similar to yours
https://tutorial.math.lamar.edu/classes/calcii/powerseries.aspx
In this section we will give the definition of the power series as well as the definition of the radius of convergence and interval of convergence for a power series. We will also illustrate how the Ratio Test and Root Test can be used to determine the radius and interval of convergence for a power series.
thank you this helped, im trying it now
I don’t get how to get rid of the n at the bottom
Would that be right?
i'm not sure if that is what you wrote but $\frac{n^2}{(n+1)^2}$ isn't equal to $\frac{1}{2n+1}$
Lin Xia
oh wait yeah oaky thank you
oaky thank you both I think i have it now
I really appreciate the help
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can anyone explain me spans as subspaces in linear algebra
do you know span
no
do you know vector space
idk the technical terms very much
but i have just started
i know matrices
subspace is the part of the bigger vector?
do you know what's R^3
subspace is a vectorspace
it's a subset
not a vector
R^3 means no. of 3 vector elements
again idk much technical terms...that's why i am here
space would be a better word here rather than no. of
R = set of real numbers
how do i write it
ok
yes
can vector space be on xyz plane?
i read that span is all the sets of linear combinations
ok
if you have 2 vectors in R^3 and look at their span u get a plane (if the vectors are not in the same direction)
what's the span of (1,0,0) and (0,1,0)?
gtg
personal reasons
i dont think that was what the why was directed at 🙏🏻
This is a good question to answer though. Feel free to guess if needed. What's the set of all linear combinations of (1,0,0) and (0,1,0)?
doesn't there need three vectors
no
Yes, that's true
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You are the wizard gng
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I dont know how to answer the last box
little stuck
the easiest way would be to calculate using the smallest h mentioned in the question here
can you see how you can get the value of this expression for h = 0.08 from here
isnt the smallest h -0.62
you found f(π+h)-f(h) = -0.62
thats not a value of h
you check for the smallest h in magnitude here for the best approximation
yea h =-0.67 is the largest followed by h=0.08 then -0.23
but $(\pi+h)-f(\pi)$ doesnt correlate with each other because h=-0.62 -> -0.62, h=0.23 -> 0.23, h=0.08 -> -0.08
ø
i might be overthinking
it's curved so the values you get for small h would still be a bit off
as the magnitude of h becomes smaller you get a better value for the slope asked in the last question
what is curved?
the curve you experimented with in the question
it's just very slightly off from a straight line
yeah I agree
the smallest h in magnitude discussed in the question is 0.08
only if i zoom in, it will look like a straight line
yeah
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would prefer using divisibility, instead of modular arithmetic or something like that, did arrive at a few conclusions like x+j | y-x, feel like this is important but cant make any use of it
maybe i gotta use the condition that if a|b and b is not equal to 0 then |a| <= |b|
but
dont see how i can use it either, help is appreciated
🩷 dont want the exact method to prove this either, any hints or leads would help
hint: construct such $x,y$
Annie Maqionde
though not sure if it will work
, use with discretion
construct as in ? like assume some values of x and y ?
i tried to express y as a multiple of j
and i succeeded in that too but
i couldn't use it further
okay i did express y in terms of some jf where f is an integer
expressing x in terms of a j
hmmm
there's a solution which i found; so it does work.
🐲 so i am on the right path it seems
it dosnt necessarily have to be a multiple of j, but it can be.
oh you meant express it in some sort of jq + r
nah it does end up being a multiple of j actually
well you see what works
i gave you a big enough hint
@sand night Has your question been resolved?
honestly actually i just noticed y can not be expressed as a multiple of j 😭 i made a small mistake
this just set me back for quite some time
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This seems really hard to integrate
(x^2-4)^3 is not x^6-64
Remember the Pascal’s triangle coefficients if you’re stuck, 1 3 3 1
You can simplify it to:
$$ \int_4^4 \big(1+\frac{(x^2-4)^3}{36}\big)^{(1/2)}$$
S0S4
is the upper bound 9 or 4?
sorry, this doesn't lead to nothing easy😅
your reasoning was correct but the expansion of the (x^2-4)^3 is bad as krish said
you should get to $$ \int_4^9 \sqrt{1+ \big( \frac{(x^2-4)^{3/2}}{6}\big)^2}$$
S0S4
mmm
It's a bit unclear to me because I don't understand your caligraphy that much, but I don't think that's the easier way to go
following from here it's straight forward
@unborn ridge Has your question been resolved?
pi_day
just simplify this first before differentiating
Oh
$\sqrt{x^2-4}$ does not simplify further
pi_day
Oh
you should test your calculations with some x like x=4 to see if it's even right
i don't know what this last step is
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Hey guys I just need help getting a quick explanation. Maybe I’m looking at Series wrong but for example, a series that’s (1/n)^3 how come the series converges to 1/2? Wouldn’t you be adding up an infinite amount of fractions and it would be divergent?
I’ve been struggling fully grasping the concept, I can go through the motions and do the different tests but I really want to satisfy this itch of actually understanding what’s going on
The associated sequence (1/n)^3 converges fast enough for the series to not diverge.
There are "tests" for this condition
Imagine you have a chcolate bar and cut it in half and then cut the half into 2 new halves and continue
So the series is basically adding up to 1/2 because the fractions get super small fast enough that it’s not really increasing anymore?
Thats the same as the sum 1/2 + 1/4 + 1/8 + ...
Yea, basically. Youd have to be a little more refined with your statement, since 1/n could theorically still satisfy said condition, but it doesnt.
Yeah I figured lol but thank you that’s really helpful
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Am I on the right track for 6?
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hi
Hi, do you need help with something in particular?
Stop spamming in every channel
yo no ablos ingles
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Hi there, im just doing problems on finding out if a vector field is conservative on a non-simply closed region and want to see if my work is correct
F(x,y) in top left, D is the region
circular with hole in the middle
r(t) is a parameterization that i chose enclosing the hole
does this look right?
i got not conservative
also don’t mind the random thetas, they should be t’s im just silly
yes but x² + y² should be 36
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thx for catching that
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oh i also wanted to remark that you can add and subtract vector fields v such that \int v \cdot dr = 0, so in particular the constant vector field (-10, 1)
.reopen
here so it’s easier
ohh so you can see that it’s off center in a way
about the origin
well not in the sense that you're saying it but in another sense yes it is off-center
but the point is that the computation is much simplified
hmm okay
so here it would suffice to instead consider the vector field (-7y/(x² + y²), 7x/(x² + y²))
so it’s fine to not include the extra constants in the problem ?
so long as you can justify why to your instructor
Okay that makes sense
in fact if you don't care about the actual number and only care about detecting if it's zero or nonzero you can do even more things
like you could also ignore the 7
but this is maybe a bit too higher-maths-core for now
yeah all i’m at rn is Linear Algebra and Calc3
i might take some more math courses in the future bc they’re pretty fun
but can’t stray too far off degree plan either so
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Sorry bro I js wanted help
you've already opened it. please send your question.
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Does this look…like it’s going in the right direction? (Top part and bottom part are different things)
,rccw
The top part determined holes x-3 and x+2 but no VA or x int
The bottom part is an attempt to get the slant asymptote
what is the ques
14
I believe you don't need to polynomial division this
you need to simplyfy or what?
The degree on top is higher than the degree on the bottom
After you factorized from the top and bottom there's a $x^2 - x - 6$ common on numerator and denominator
1 divided by 0 equals Infinity
The degree on top is 6 while the agree on bottom is only 3….
Yes
So what are you implying here
I have x/-3 remaining
Im kinda confused here, the polynomial you wrote on your work and in this picture is really different
So what problem did I write…
You wrote problem 8 💀
Fuck
Into problem 14
Let me write 14 then
Also are you supposed to simplify $f(x)$ or smth?
1 divided by 0 equals Infinity
Does this look any better?
I have to find….all of this
Horizontal or slant asymptote, vertical asymptote, holes, y and x intercepts and rnd behavior
🫡
It’s not horrible once I get into a groove
Right..
So to get the bottom to equal zero….
It’s x=3 and x=-1
Which of them works for a VA?
But here you can see the denominator's degree is greater
So it’s 0?
Veritcal asymptote happens when the deno is 0
So both of them works here
Are you sure..?
Yea
I also graphed the function out here
Oh ok
X intercept happens when the graph intersects the x axis
Or at $y = 0$
1 divided by 0 equals Infinity
So basically substitute $f(x) = 0$ and solve for $x$
1 divided by 0 equals Infinity
Also what do I do with the remaining top portion from the first part
The HA and SA?
In this case the HA exists so i think that you should fill in the HA and probably state that SA does not exist in the SA spot
Degree of numerator < denominator
So it always exists at y = 0
Ohhh
For x and y intercepts I got 4-
For both-
Are there any holes?
<@&286206848099549185>
A hole happens when the same number makes the numerator and denominator equal 0
So none
Yep. Take smth like…
(x² - 25)/(x + 5)
End behaviour is basically what happens when x -> ±♾️
Yeah ik
And so, you figure that out by finding the horizontal asymptote
I have that
HA = 0 right?
Yes
So, what the means is the end behaviour is y -> 0
I thought you used the vertical asymptote to find it…
That’s what my paper is implying at least
Since I only have one HA and 2 VA’s
VA’s tell you when the y value shoots off to infinity or negative infinity around a certain x value.
HA’s tell you what the graph does as the x value approaches infinity or negative infinity
That’s exactly what those are. None of those have anything to do with end behaviour though
Aside from the fact that the numbers are in the wrong place, it’s right, lol
The numbers go underneath the sign.
3⁺ as an example
But that’s a TINY nitpick. Your answers are right
Do you know how to tell what the HA is?
It depends on the degree of the top and bottom
Just out of curiosity
If they are equal?
If the denominator has a higher degree it’s 0 if the numerator has a higher degree you have to do the weird dividy thing…
I don’t know- and my next problem has tha
(Context the next problem is 15)
Lucky for you, I would say it’s probably one of the easiest cases to find the HA
If the degree of the numbers for is equal to the degree of the denominator, you just divide the coefficients of the highest degree term
Oh
So for 15, what would you do?
Well idk if I can factor that….
So just…set the denominator equal to zero?
Cus I don’t think anything cancels either
Nope. So, what would your VA be then?
-2?
Yessir!!
That’s correct!!!! 😊
No holes
Since nothing cancels
x, and y-int’s are pretty easy too
Yessir. Both your x, and y intercepts are located at (0, 0)
Ok ok
Are you told to do rough sketches of graphs? Or just to find these values
I was told I can use the fancy calculator but I don’t have one on me
Hold on one sec
,w graph (2x)/(x+2)
Huzzah
So I think it’s…. Positive infinity
For both
Cus you can’t have positive and negative zero
For what?
X-> +/- 0?
Huh?
Limits as x approaches what?
0
Oh siad
Wait
Wrong thing
Mb I was looking at the intercept
-2?
Or ig 2 and -2
So, the limit as x approaches -2 from the left is +♾️. And the limit as x approaches -2 from the right is -♾️
So…. X -> -2^+ is inf and x-> -2^- is -inf?
Other way around
If you actually wanna calculate which ones which, if you’re allowed calculators, plug really really close values to -2 into your calculator. From the left, plug in smth like -2.0001. And from the right plug in -1.9999
I’m allowed to for that and that alone
But…I’m just telling her I put it in the calculator
RIGHT NEW PROBLEM
I factored it
Alr. So, you’ve factored properly!! Now just find your VA, and HA
I could go more though…
You’re right. I would’ve stoped there because I could figure out my values from there
But you’re right
Ok
Not the numerator and denominator can be factored even more
Ok I got 2 for VA
Is 2 the only one
I think it might help you to be able to factor a difference of squares
Wut
Do you know how to do that?
I don’t see how you could get another answer besides that
I can show my work
But the square root of 4 could be 2 or -2
But square rooting smth gives you both a positive and negative answer
That’s what I said
If you have :
a² - b² -> (a - b)(a + b)
^
Nvm. I just saw this!
Teehee
Now, is there any HA for this one?
And the degree on top is larger than the one on the bottom-
So you have to do the weird dividy thing right?
Yep. So you know that there isn’t an HA. But there might be an SA
Yep. Just make sure to account for every term when you do it
What I mean by that
Is if you have a missing term, add it with a 0 coefficient
x³ + 0x² - 16x + 0
lol!
Your way is…confusing me
So I went back to what I new
Knew*
And now I’m also confused
How do you do it then?
I just don’t put in the extra terms
Do you leave spaces to account for them?
No…
Then you’ll end up with a wrong answer
I’m sure your teacher wouldn’t show you an incorrect method
She would.
As long as the answer is correct, is there really a “correct” method?
Remainder doesn’t matter for the slant asymptote, no
So it’s just… -1/2x?
Exactly
Wait do you not have to fix that to make it x= 1/2?
No
The asymptote itself is the line y = -0.5x
In this case, no
I got x as 16…idk if that’s right
Find x and y intercepts…