yes
#precalculus
burnt sonnet
fallow slate
oh wait i think i know why it goes up now
burnt sonnet
yes
burnt sonnet
yes
fallow slate
yeah
fallow slate
exactly!
latent siren
It might help you if you also graph JUST f(x) = (x-4) (x+3)
burnt sonnet
there is 1 more significant thing
latent siren
then look at the relation of that to the two you have
burnt sonnet
I am alrdy helping someone @latent siren
oh sry
I tohught u were askign a question
sry
haha
latent siren
no lol x)
burnt sonnet
Can f(x) be zero?
fallow slate
oh no
burnt sonnet
can f(x)*x?
burnt sonnet
what if x=0?
fallow slate
wait...
burnt sonnet
😄
fallow slate
well for the red graph, it seems that it touches the origin (0,0)
but for the blue graph, if x = 0 then f(x) is undefined
fallow slate
oh wait both graphs cross the x axi
burnt sonnet
f(0)*0=?
fallow slate
yes
burnt sonnet
no
fallow slate
0
burnt sonnet
both does not cross
fallow slate
oh wait youre right
burnt sonnet
f(x)=1/(x-4)(x+3) is never 0
latent siren
because that would be 1/-12
fallow slate
the y-intercept
burnt sonnet
f(x) never crosses the x-axis
yes that would be the y-intercept
those are the 3 three differences
like Koolaidwannabe said it can be intressting to look at (x-3)(x+4) to realize about the zeroes, but It seemd like you understood this
fallow slate
well i think i understand this much better now
burnt sonnet
Awesome
fallow slate
Thank you so much!
latent siren
in a function that never has a non zero numerator you can only approach 0 when x -> +- infinity
burnt sonnet
it could be shifted up and down maybe 😛
latent siren
(as there's no vertical translation xD)
burnt sonnet
;D
but I agree if it is on the form f(x)=g(x)/p(x) if g(x) is never 0 then f(x) is never 0. 😄
Time to sleep now peace out ppl
opaque lantern
SOMEONE HELP PLZ
uncut mulch
show work
terse ravine
me?
uncut mulch
yes
terse ravine
oops
terse ravine
no I think the green part is wrong
uncut mulch
yes
nova ravine
@terse ravine where did you get 4x from?
terse ravine
brb I'm working on a problem atm.
opaque idol
Can anyone help?
velvet granite
@opaque idol Wassup
velvet granite
For cubic functions the equation is y=a(x-h)^3+k
H is your left or right and K is your up and down
opaque idol
Which one is this for?
velvet granite
What you should do is sketch your original function or parent function which is x^3 then, make translations
3
Use this one y=a(x-h)^3+k
opaque idol
Will that make the graph get stretched vertically?
velvet granite
The a value i believe is the only one that stretches the graph
opaque idol
Wym by that?
For 3a I did (x-5)^3+7
That’s the equation
But for 3b
Idk if there’s a way to make it stretched
velvet granite
one second plz
opaque idol
Ofc
velvet granite
I think it just means the a value is 2
velvet granite
y=2(x)^3-2 ye
valid echo
Could someone possibly help me with this
velvet granite
@opaque idol Im not sure because its hard to read
velvet granite
Yeah im not sure about 4, so sorry
viscid thistle
can someone explain what a limit is? i'm very familiar with assymptopes. i don't understand the notations used and what it's asking for in limits however
velvet granite
@terse ravine did you find the derivatives using implicit
patent beacon
What did you do? Haha
viscid thistle
vanish use derivatives
velvet granite
You need to do the full derivative before you plug in numbers
The substitution part when you plug in the x and y values comes at the end @terse ravine
terse ravine
I thought I did in the second step
velvet granite
No you just plugged stuff in
terse ravine
second step in blue..
uncut mulch
sign error
terse ravine
darn where
uncut mulch
when differentiating the 4y^3
viscid heron
What's going on
terse ravine
Oh thats supposed to be -12y^2y'
viscid heron
Yep
velvet granite
dude implicit differentiation is the worst
viscid heron
Wait till you get to integration
terse ravine
No its fun because it teaches you to pay attention
velvet granite
Well online classes dont help
viscid heron
True... It's the training of a ninja
velvet granite
looked good to me
velvet granite
GN
crystal osprey
sleep tight
jade heron
@loud marsh
Displacement is the vector quantity counterpart for distance
So its distance/position ascribed with direction
loud marsh
also have not taken calculus, this came up in my advanced functions class
so idk what a vector is yet
jade heron
So a displacement-time graph tells u ur displacement at each time
Ah ok
Vector quantity is a quantity with a direvtion
Scalar is one without
For displacement it matters the direction you're going
For distance it doesnt
I'll draw u something
So the total distance travelled is 1+2+1+1 metres
But the displacement is 1 metres
Yes?
If u travel 6 miles to school then 6 miles back your distance travelled is 12mi
But your total displacement is 0mi
loud marsh
OHH
okay i get the concept of displacement now, but how would that be shown in a graph vs time
jade heron
Ok so
Since displacement is vector quantity it matters where its measured from
Right?
Where u start basically
So imagine at time 0, you start at 0 displacement meaning that's where u start
By 10 seconds you move 10metres away
Your displacement is then 10 metres
Suppose u then move back to where u began after 5 seconds
Then ur displacement is 0
So you could draw this as a graoh
loud marsh
for context these are the questions
dont worry not an exam just a practice assignment
no ban pls
jade heron
Great so
Qn 3
If you have no change in displacement as time goes on
How would u expect the graph to look
loud marsh
line is straight
jade heron
Straight where
Up down left rigjt diagonal?
(I get what you're trying to say you just need to word it appropriately)
jade heron
Yep, you should say no slope or flat or something like horizontal line
loud marsh
AHH OKOK
jade heron
Straight line could be at a slope too !
loud marsh
well the "no change in displacement as time increases" sounds like the previous question but the speed part throws me off
idk how to visualize that
jade heron
Well if there's no change in your displacement as time increases what does that mean if u gave a physical interpretation
jade heron
Uhhh put it in a real life context
Suppose there was a runner
What would it mean if their displacement did not change as time went on ?
loud marsh
if they stay still? a circle would have displacement at certain points so it cant be that
loud marsh
a circle would be like the first example right where u end up at 0 displacement
^^no speed
well 0
loud marsh
that makes full sense thanks
jade heron
Or in a context it could.mean its zero only at a specific point
Like if u throw a ball up in the air
There's only one point where the ball is stationary
loud marsh
yea like the instantaneous rate of change
jade heron
Yea!! Ugot it
loud marsh
W
willow bear
what's a?
bitter sandal
oh the index the one below the greek "E"
willow bear
sigma is not an E
i'm saying that you did not define a variable named a and neither has the problem
so until you write down the sigma notation it is impossible to tell whether or not it is the one you are after
it is possible to make one which looks like $\sum_{a=0}^{...} (...)$; it is also possible to make one which does not
obsidian monolithBOT
Ann:
bitter sandal
sorry im bad at asking.
for 1000 days of donating should my "n" or "a" start at 0?
willow bear
alright well if you wanna do this then you will need to start at n=0, so as to include the first day where the sum is 1000 pesos
and technically you will also need to end the sum at 999 rather than 1000
even though that actually doesn't make much of a difference here as the term you included but shouldn't have is actually just 0
bitter sandal
ahh right² thank you
rugged linden
Hello i am going through serge lang's calculus
i came across a few confusing concepts
i wrote some notes on them but unfortunately im not sure if my understanding is on point
apologies for the grammar
foggy condor
sick steppe
Binomial Theorem?
willow bear
what's giving you trouble here
grim sand
I don’t know what to do
willow bear
you have your formula for the number of infected students as a function of time: $$P(t) = \frac{506}{1 + 4.5e^{-0.1t}}$$ and you are asked to find the value of $t$ such that $P(t) = 260$
obsidian monolithBOT
Ann:
willow bear
in effect you are asked to solve the equation $\frac{506}{1 + 4.5e^{-0.1t}} = 260$ for $t$.
obsidian monolithBOT
Ann:
willow bear
are you able to do that? Y/N
grim sand
How do I solve the bottom part
willow bear
what do you mean by "solve the bottom part"?
grim sand
The 1+4.5e
willow bear
no, what do you mean by "solve the bottom part"? you're solving the entire equation, not just part of it.
willow bear
there are many possible first steps
i would prefer if you did most of the work on your own
so let's try something simpler. forget your problem for a moment.
here's another equation, completely unrelated to the first: $$\frac{50}{4+3x} = 2$$ are you able to solve this equation for $x$? Y/N
obsidian monolithBOT
Ann:
willow bear
uh what
grim sand
Do i get x by itself first
willow bear
how ELSE would you solve an equation if not by getting x by itself??
where did you get $2 \cdot 50 - \frac{4}{3}$ from anyway? not only is that not the answer, it doesn't even equal 32 !!
obsidian monolithBOT
Ann:
willow bear
you realize this makes absolutely no sense right
"change the / to *"
nonsense, absolute bullshit
willow bear
alright so first off
grim sand
?
willow bear
just to nip this thing in the bud which i've seen for the 1000th time today
when you write \verb|a/b+c| in plaintext like this, it reads as $\frac{a}{b} + c$, NOT as $\frac{a}{b+c}$ !!!
obsidian monolithBOT
Ann:
willow bear
if you want to type a fraction in plaintext but its num or denom are a sum or a difference, you need parentheses, like so: 50/(4 + 3x)
grim sand
Oh ok
grim sand
it auto capitalize
willow bear
so you meant to write x = 2(4+3x)?
grim sand
Yes
willow bear
then why did the left-hand side become x?
why x?
are you asserting x = 50? for absolutely no reason?
grim sand
the x is supposed to be 50
willow bear
you're doing something that makes no sense \
if its supposed to be 50 then why the FUCK do you choose not to write 50
grim sand
Cuz I was gonna add it in last
willow bear
you do realize you can't just replace random shit with x willy-nilly like that
you arlready have an x
you don't know if it's equal to 50
(and spoiler alert it fucking ISN'T)
you can't just use x again for a different thing
grim sand
50= 2(4+3x)
willow bear
THERE we go
grim sand
50=8+6x
willow bear
would it have killed you to write that right at the beginning
grim sand
No I think
willow bear
this simple matter of multiplying both sides by (4+3x) is apparently too arcane
it's okay from that point at least... but jeez
grim sand
Yes
willow bear
and now it's time to get back to that scary complicated, impossible-to-solve monster of an equation: $$\frac{506}{1 + 4.5e^{-0.1t}} = 260$$
obsidian monolithBOT
Ann:
willow bear
i am purposefully exaggerating the difficulty with these epithets
but the arcane idea of multiplying both sides by (1 + 4.5 e^(-0.1t) ), inaccessible as it may be to mere mortals, still applies here, and one may then get the following, simpler equation:
$260(1 + 4.5e^{-0.1t}) = 506$
obsidian monolithBOT
Ann:
willow bear
yeah cause it is easy
youre just stuck with a mental block that i'm trying to help you overcome
which i'm probably not very successful at
willow bear
did the flowery language obscure the idea too much?
grim sand
e or t
willow bear
$$\frac{506}{1 + 4.5e^{-0.1t}} = 260$$ multiply both sides by $(1+ 4.5e^{-0.1t})$:
$$260(1 + 4.5e^{-0.1t}) = 506$$
obsidian monolithBOT
Ann:
willow bear
does this make sense to you Y/N
willow bear
yeah so are you able to continue from here or do you need further assistance
willow bear
get e^(-0.1t) by itself first so that you have an equation that looks like e^(-0.1t) = (a number)
then remember the relationship between exponentials and logarithms
honest radish
What is the difference between gradient and derivative?
Is a derivative or gradient of x^2 equal to 2x?
grim sand
e^(-0.1t) = 0.21
viscid thistle
@honest radish this channel is occupied, please proceed with a free one.
patent beacon
@honest radish
The word gradient is reserved for spaces other than just functions on R. I can take a gradient on a function over R³.
Derivative is usually just an R → R deal, but there's special types of derivatives too
When it's just a real variable function, gradient and derivative mean the same thing.
viscid thistle
@grim sand you still stuck?
lament fiber
Find the range of f(x)= 2cosec2x + secx+ cosecx
can someone ping me when they answer I might lose internet now and have to check tomorrow instead of now
I tried to change it all into sins and cosines and using the multiple and submultiple things
I got
[cos(x/2) + sin(x/2)]/[sin(x/2) * cos(x)]
where do I go from here
glass umbra
Hey, I know this is pretty simple pre calc but I am still a bit confused.
i need to find the domain of this function.
unborn hemlock
Hello, I require assistance with this question.
I broke down the right side, and rewrote it as:
$2[\cos(\frac{c}{2}+\frac{d}{2})\cos(\frac{c}{2}-\frac{d}{2})]$
obsidian monolithBOT
Ainsley:
viscid thistle
oh
unborn hemlock
Anyways, I used the compound formula for cosine and ended up with this.
$2\cos^2(\frac{c}{2})\cos^2(\frac{d}{2})-2\sin^2(\frac{c}{2})\sin^2(\frac{d}{2})$
obsidian monolithBOT
Ainsley:
viscid thistle
is it asking you to derive it
unborn hemlock
In a way, yes.
I know one of the double angle identities could be used, just don't know how exactly.
Ping me for a response so I don't lose track.
viscid thistle
hmm
well in that identity, at 2cos(c+d / 2) they seem to average out the angles, and lengthen the hypotnuse
that'd give you a pretty large adjacent side
cos(c-d / 2) is under 1 so it prob makes 2cos(c + d / 2) smaller
meaning that 2cos(c + d) is 1/(c-d / 2) times bigger than cosc + cosd
violet knoll
How do I type in the next step? I am pretty sure its sin (pi/2) cosA(x) + cos(pi/2) sin(x)
is this wrong?
patent lance
I have no clue how to find this
i got as far as setting it up
but i dont know how to find y2
patent lance
I figured it out sorry shoulda put that in
astral sedge
all good
astral sedge
no problem
:)
I have covered remainder and factor theorem in class but I don’t know how to do this question.
astral sedge
I’m having trouble doing this question
I figured out the equation in terms of how many blue marbles there were but when it comes to subbing it in I get a negative fraction and that is not possible
distant nova
I have covered remainder and factor theorem in class but I don’t know how to do this question.
@astral sedge the remainder when P(x) is divided by (x-a) will be P(a), when divided by (x+a) will be P(-a) and when divided by (ax+b) will be P(-b/a)
so P(p)=p^3
Maybe that'd help?
obsidian monolithBOT
Tiessie:
jolly creek
Does anyone know the answer to I^I where I = infinity
Is it just infinity or is it just undefined
And can anyone help me understand what is l'hôspital's rule
bold meadow
Can anyone send me some stuffs(like curriculum or book) of precalc?
viscid thistle
@jolly creek In calculus inf^inf -> inf (use a^b = e^(b * ln (a) ) and simplify).
honest depot
hi all. i have a question that i need help on. based on my teachers key i am wrong but ive done all the steps
the question is simplifying
sin(x-pi)
using the sum and difference identities
i get : sinxcos(pi) - cosxsin(pi)
which is
-sinx - (cosx(0))
so the answer should be -sinx right?
but my teacher's key says: tanx
is she wrong?
viscid thistle
-sin(x) is correct.
arctic kestrel
velvet granite
@arctic kestrel Are you allowed to use a calculator?
arctic kestrel
yeah
velvet granite
alright well find what they make for 400 rooms at 50 dollars minus the cost of service for each room
arctic kestrel
yeah
but it's weird
i got that but it's just
it doesn't make sense to me
for some reason
arctic kestrel
Okay
lament fiber
can someone please answer my question from yesterday
I even pinged the Helper role
and mentioned that the q didn't get answered
lament fiber
<@&286206848099549185>
humble pagoda
yo can someone help him
knotty echo
is anyone able to help with a question on cofunctions?
I'm really hard stuck <@&286206848099549185>
i've tried substituting cosA with sin(pi -A)
using cofunction identity
but I'm getting nowhere
jade heron
A+B=pi/2 right?
knotty echo
yeah
wind igloo
@lament fiber Keep simplifying using half-angle formulas until everything is in sin(x) and cos(x).
knotty echo
i dont know what half angle formula is
jade heron
So u have sqrt(1-sin²B)
wind igloo
I wasn't talking to you, chene12.
jade heron
U understand why friend?
jade heron
Um
knotty echo
so just clarifying
jade heron
You should've got CosA=SinB
@jade heron this should solve it for u
knotty echo
and they you have pythagorean identitties
jade heron
Unnecessary work
Just plug in cosa =sinb into ur expression
Then ur one step from done
knotty echo
no i meant since cosA = sin(pi/2 -A) and pi/2 -A =B the cosA = sinB
just verifying cuz I need to show work
jade heron
Ye
jade heron
tan(pi/2-x)= cot x and vice versa
So we have
tanx+cotx=2
So tan²(x)+1=2tan(x)
so (tan(x)-1)²=0
So tan(x)=1
@knotty echo
knotty echo
@jade heron is this right?
jade heron
Um no dont do that
Once u have 1+tan²=2tan
Then u should solve for tan
As a quadratic
jade heron
Ok
If u had a quadratic
1+t²=2t
And you're being asked to solve for t
Can you do this?
knotty echo
so if we know tan we use arctan to find angle x?
jade heron
we trying to find angle x here I'm guessing
@knotty echo read the question
so if we know tan we use arctan to find angle x?
@knotty echo whay does the question want
knotty echo
I thought we had to find angle x then find the tangent of that
knotty echo
so we can skip past that step and just find tanx
knotty echo
that was my idea at first
jade heron
Seems kinda pointless
knotty echo
so I can just skip past my nonsense and straight up solve for tanx
using quad equation
jade heron
Um yeah 24% charge left on my phone i am yours to.use
jade heron
Ok so we have to realise here that
Cos²(x)+cos²(90-x)=1
Because cos²+sin²=1
Right?
knotty echo
yes
jade heron
So u can pair 1° with 89°
And 2° with 88°
So on so forth
How many pairs do we have?
jade heron
we have 1 2 3,...,44
Then we have : cos²45 and cos²90 that are unpaired
Luckily we know what cos45 and cos90 are
1/2 and 0 respectively
So we have 44 pairs that add to 1
Plus a half
obsidian monolithBOT
chene12:
knotty echo
oh sorry I read wrong
@jade heron can you explain why cos^245 is 1/2
I still dont get it
wait nvm
@jade heron thanks for your help today I'm done now
worn anvil
Can someone solve this? Teacher slapped it down without teaching us and I’ve got no clue how to fill this out
worn anvil
Thank you
lament fiber
thank you @wind igloo
lean tusk
how do I use a calculator for this?
solid lark
how do I use a calculator for this?
@lean tusk I’m pretty sure u put y=5cosX and y=-4 and then find the intercepts
lean tusk
oh ok ty
rare zephyr
blazing parrot
what is (- 2/7)^2
solid lark
what is (- 2/7)^2
@blazing parrot 4/49
fiery wren
4/49
blazing parrot
yes 4/49 not -4/49
fiery wren
jesus christ
you're right
this whole time i was trying to figure out
what i was doing wrong
thank you lol
do i use the same formula for
cos(theta + 2pi)
blazing parrot
what’s the period of the trig functions
bold meadow
2\pi
blazing parrot
what kind of equation do you think produces a shape like that
steel swan
yo so uh
should I skip pre calc
im rn a freshman taking algebra 2 h but I wanna take calc bc next year
willow bear
what is $O_3$?
obsidian monolithBOT
Ann:
drowsy helm
I got no idea
willow bear
does your book not define it anywhere?
drowsy helm
I don't think so
willow bear
can you look more closely in the section where matrices are first introduced
cause right now all i can think of is that $O_3$ might represent the 3-by-3 zero matrix, but this would make $A - \sqrt{2} O_3$ equal to just $A$
obsidian monolithBOT
Ann:
willow bear
ok great so $O_n$ \textbf{does} refer to the zero matrix of size $n$.
obsidian monolithBOT
Ann:
drowsy helm
yeah
willow bear
so?
drowsy helm
how do I solve it
drowsy helm
idk how to scale them but I can add them
willow bear
you've never had to multiply a matrix by a number?
so if i told you to, say, calculate 10A you would not be able to do it?
drowsy helm
oh wait so I gotta multiply sqrt 2 by all the 0s in the matrix
I can calculate that> so if i told you to, say, calculate 10A you would not be able to do it?
@willow bear
willow bear
yeah again what's stopping you from calculating sqrt(2)*O_3 then
it's just going to be O_3 again
drowsy helm
oh aight so it's basically equal to A
willow bear
not "basically"
drowsy helm
gotcha, thanks
viscid thistle
any hints to what I got wrong?
willow bear
everything lmao there is not a single correct answer there
the domains of f+g, f-g and fg are all the same - the INTERSECTION of the domains of f and of g
the domain of f/g is that minus the points at which g(x) = 0
can you tell me what you think the domain of f is @viscid thistle
viscid thistle
oh I see, it cannot be -11 or lower
I forgot that I cannot root a negative number
willow bear
not all numbers are integers, frost
viscid thistle
Im gonna try something
willow bear
there are numbers which are higher than -11 but cannot be the input of f
in particular the number -10.5 would like to have a word with you
viscid thistle
domain of f is [-10,infintiy), right?
sick steppe
$10+x \geq 0 \implies x \geq -10$
obsidian monolithBOT
moshill1:
blissful ridge
Check domain of g(x) again
viscid thistle
(-inf,10]
But basically, to find the domain of F + G I just need to find the intersection though, right?
blissful ridge
Yes, intersection of domain of f(x) and g(x)
viscid thistle
ok, great, Now i know how to do this.
blissful ridge
👍
viscid thistle
my last class my teacher had touched on intersection but she didn't really explain it much so I didn't understand the importance of it
but now Ik how its utilized
rugged linden
hello i've been studying calculus over again and in the book I've been following we just introduced the least upper bound axiom
It got me thinking about this other thing i had heard about earlier
Basically my question is, would this be a proper way to complete the reals?
This is from my notes so if it makes no sense i'll be happy to clarify
also sorry for the typos
here's the least upper bound axiom for anyone who's interested in helping
i think this'd just be a fairly inefficient way of just doing dedekind cuts, right?
drowsy helm
hi can someone help me calculate the limit with the squeeze theorem?
toxic coral
Hii can someone walk me through at least one of these? Thanks
serene heath
@drowsy helm take the ln of it and use the fact that lnx<x for x>0
jagged glade
Hello there@toxic coral
Do you recall how to find the domain and range of functions?
drowsy helm
alright thanks
jagged glade
I see, so domain in this context is all real numbers that you can input into the function f such that f(x) is a valid number.
Like for question 1a, what kind of number you cannot put into the function?(so that f(x) will not be a valid real number)
toxic coral
hm I see
jagged glade
That number would need to be opt out from the domain
toxic coral
how do I "calculate" for that number
jagged glade
Lemme check if you know which number cannot be put into the x of f(x) for question 1a first before we continue :)
jagged glade
Yep
Now for the reasoning
You can say since x-1 = 0 when x=1, and anything divided by 0 is undefined, so the domain will be blah blah blah
toxic coral
oh so there's no way to actually have a "solution" for the domain?
knotty echo
Brief overview on domain restrictions https://courses.lumenlearning.com/waymakerintermediatealgebra/chapter/restricting-the-domain/#:~:text=That is%2C only real numbers,not be a real number.
Cuz we can't really go anywhere if you havent gotten these down
The domain of a function is ALWAYS and interval
therefore your solution should either be in setbuilder notation or interval notation
knotty echo
the range of the function is the DOMAIN OF THE INVERSE
To get the range of a function
find its inverse
and then GET the domain of the inverse
that would be the range of the original function
rugged linden
Hello i'm having some trouble knowing when to apply the identity $\sqrt{x^2} = \abs{x}$ and not
obsidian monolithBOT
rugged linden
sometimes, for example, we have something like $x^2 = 25$ and then taking the square root and applying the identity gives us $ \abs{x} = 5 \implies x = \pm 5$
obsidian monolithBOT
rugged linden
but other times we do this and it comes out to be nonsense
my question is when do i apply this and when do i not
What i mean to say it is clear sometimes we need to apply the identity, and other times we don't, and we can't have both at the same times so there must be some condition behind this that im not aware of
obsidian monolithBOT
chene12:
knotty echo
that should be clear
You take the absolute value of x when you are trying to solve for the principal square root
The principal square root is the positive number square root. Unless otherwise stated, "the square root" of a number refers ONLY to the principal square root. The square root of n^2 is the absolute value of n.
@rugged linden
rugged linden
wait wdym
So whenever I'm solving for a variable is when I do this
otherwise I can cancel the exponents of $(2)\left(\frac{1}{2}\right)$,
obsidian monolithBOT
rugged linden
Ah
knotty echo
i dont see exponents in your expression
rugged linden
I mean that whenever im solving for a variable I have to apply the identity, otherwise if i get $\sqrt{a^2}$ where a is a constant I can cancel the square root and the exponent of 2
obsidian monolithBOT
rugged linden
Yeah what i said was unclear, I'm tired mb
knotty echo
unless specified $\sqrt{a^2} = | a |$
obsidian monolithBOT
chene12:
knotty echo
for normal problems always take the positive solutions
paste the link into address
it will clarify
rugged linden
I'll actually like to ask for a specific piece of help. I remember I was trying to derive the quadratic formula
knotty echo
in case of the quadratic formula, you can take both negative and positive answers
rugged linden
And I had this $\sqrt{\frac{s}{4a^2}}$. the s term is a few terms im too lazy to remember
obsidian monolithBOT
knotty echo
The bottom is $\sqrt{4a^2} = 2a$
obsidian monolithBOT
chene12:
rugged linden
So here the answer would be $\frac{\sqrt{s}}{2a}$, I should take the positive root here, right?
obsidian monolithBOT
knotty echo
yes take positive square root
rugged linden
since a is a variable but not really a variable
it doesn't really vary but it's an abstraction
would it be correct to call that a dummy variable?
obsidian monolithBOT
chene12:
rugged linden
yes
knotty echo
try not to overthink in derivations
rugged linden
but we will still take the positive square root of 4a^2 since that is still positive- oh wait
knotty echo
because values can change when numbers are plugged
yes positive
default to positive square root when unsure
rugged linden
that implies $\sqrt{4 (-4)^2} = 2(-4) \implies \sqrt{64} = -8$
obsidian monolithBOT
rugged linden
since sqrt denotes the principal square root this is kinda funky
this square root and exponent crap always confused the heck out of me
knotty echo
$\sqrt{4 (-4)^2} = 2(4) \implies \sqrt{64} = 8$
obsidian monolithBOT
chene12:
knotty echo
since $(-4)^2=4^2$
obsidian monolithBOT
chene12:
rugged linden
no but weren't we saying that $\sqrt{4a^2} = 2a$?
obsidian monolithBOT
knotty echo
there are two possible solutions for this equation
mine is the second possibility
yours in the default one
rugged linden
aaa
knotty echo
none of them is wrong
rugged linden
So here this is equivalent to taking the absolute value thing
Yeah i know it's not wrong
What im trying to get my head around is when to use this identity and when not to
sometimes it's very useful
knotty echo
its just you need to know which one to consider over the other
rugged linden
like for example in the derivation for the quadratic formula the plus minus you get in front of the sqrt is very helpful bcs you applied the identity
knotty echo
and your solution needs to make sense in accordance to the problem
yes
that plus or minus tells you the value of the root from the vertex axis
rugged linden
but other times applying the identity tries to give you the plus minus thing to 2a in the denominator
knotty echo
yes 2a is the denominator
rugged linden
yea but if you apply the identity you get $\pm 2a$, not just $+2a$
obsidian monolithBOT
rugged linden
here what i'm saying is that in the formula you get $\pm 2a$, when it's supposed to be $2a$
obsidian monolithBOT
rugged linden
And which will give you four different roots for a quadratic, which is total nonsense
So what's confusing to me is is there a hard and fast rule that tells you when to apply it and when not to
knotty echo
$\pm2a = 2(\pm a)$
obsidian monolithBOT
chene12:
knotty echo
this says the a can be either positive or negative
-a simply negates the value of a
therefore you use 2a
rugged linden
wait what
You can get negative and positive answers without the rearrangement, no?
knotty echo
remember how applying a negative to a number changes the sign
-(-2) =2
you dont want that in a quadratic
but this scenario arises if you -2a
rugged linden
yeah i know what im confused about is why you used commutativity to rearrange there
knotty echo
the two is equivalent
sick steppe
You just put the plus/minus first typically
knotty echo
the idea is to note that 2 is constant
rugged linden
Ah that's what you meant
knotty echo
the value of a changes accordingly
rugged linden
regardless the plus minus doesn't change much does it
knotty echo
to embody all real values of a, you use a
rugged linden
$\pm (-a) = \pm(a)$
obsidian monolithBOT
rugged linden
or maybe $\mp a$
obsidian monolithBOT
rugged linden
though that only means something in the context of other $\pm$ signs
obsidian monolithBOT
knotty echo
a plus or minus on a variable shouldn't really matter
rugged linden
i see
knotty echo
since a variable can be either positive or negative
rugged linden
ahhhhhhh i see what you mean
knotty echo
a can be -4, or it can be 5, or anything in the real domain
knotty echo
-a is essentially -1 * a
rugged linden
Yea
knotty echo
it will negate the sign of a
rugged linden
Yeah
rugged linden
While that is true, when we have $\pm(-3)$ that doesn't mean that we have to pick the minus, we can pick the plus, so $\pm(-3) = \pm(3)$
obsidian monolithBOT
knotty echo
a $\pm$ on a number is different from a $\pm$ on a variable
obsidian monolithBOT
chene12:
rugged linden
yes i see
okay i get it now
So like, what you're saying is, if we have a variable x then the values it can take on are, say, all the real numbers
$\pm x$ says it can take on all of the real numbers too, so there's no real difference
obsidian monolithBOT
rugged linden
In fact, even if we have $-x$ on it's own, the variable is unchanged, it still takes on all the real numbers
obsidian monolithBOT
knotty echo
$\pm x$ is equivalent to purely saying x
obsidian monolithBOT
chene12:
knotty echo
since x can be positive or negative
obsidian monolithBOT
rugged linden
thank u for taking the time for explaining this to me btw
knotty echo
when a variable is negative
we imply that we switch the sign of x
if x is negative, -x is positive
and vice versa
lets say x= 3
then -x would be -3
@rugged linden
rugged linden
But when we say $\pm x$ we say that $x = 3$ and $x = -3$
obsidian monolithBOT
knotty echo
$\pm x$ is synonymous to just saying x
obsidian monolithBOT
chene12:
rugged linden
hrnnngng
knotty echo
except $\pm x$ is more explicit
obsidian monolithBOT
chene12:
rugged linden
are u sure about this
im sorry im tired af
i might just have to ask my teacher or something
tho he is a bit condescending oof
thanks for taking the time with me though, i appreciate it :)
knotty echo
i think you are overthinking
a variable is just ANY number
and $\pm$ is almost always used with the square root
obsidian monolithBOT
chene12:
topaz obsidian
gritty stratus
Hello does anyone know how to do this? It’s rotating conics but I have no idea what to do
dim ermine
does anyone know how to determine where a polynomial function is less than zero?
sick steppe
$p(x) < 0$ ?
obsidian monolithBOT
moshill1:
opaque idol
Can someone help me?
burnt sonnet
yes you do 🙂
opaque idol
It doesn’t give me an equation
burnt sonnet
but google quadratic equation
opaque idol
Y= or f(x)=x^2?
burnt sonnet
a general quadratic function can be on the form $y=ax^2+bx+c$
obsidian monolithBOT
egocarpo:
burnt sonnet
and now we have to decide a,b,c so that it has matches the description
burnt sonnet
we want a max point at (-3,-6)
opaque idol
X^2+3x-6?
burnt sonnet
okay try and see if x=-3 gives the function value -6
because what (-3,-6) means is if f(x) is your quadratic function
then f(-3)=-6
opaque idol
Ohh ok
burnt sonnet
y and f(x) are the same in this case
burnt sonnet
ok
opaque idol
-3=-6
burnt sonnet
you said that -3=-6 which is not true
burnt sonnet
$f(-3)=(-3)^2+(-3)*3-6=9-9-6=-6$