precalculus

[glacial maple]

and keeps us informed

I am doubting myself too :P

[daring crest]

@neon vale 2 is extraneous because if you plug it in the original equation, you will notice that the answer to the square root expression will equal to a negative number, which is not true for real numbers.

[neon vale]

Thank you

[solar orchid]

Doing precalc online course

uhhh

Can someone explain Exponential Functions to me

Why do they always start at (0,1) it seems

[frigid dragon]

https://gyazo.com/7d9e6d8f6cdd94ef08e499384d5034e7

<@&286206848099549185> Can someone help explain what the w o u represents?

[sullen shoal]

$w\cdotu$

Fucking shit bot

$ w \cdot u $

[obsidian monolith]

Samtell:

[frigid dragon]

So its just multiplying 2 functions?

then i'd multiple 7 after?

Or would i just plug the 7 into x?

after multiplying the 2 functions orrrr

[sullen shoal]

Oh I was looking at it wrong, but yes, essentially you would multiply the two equations together and then solve for 7

Shit

Never mind

[frigid dragon]

Its

Basically

I put U into W

of the X

[sullen shoal]

$ (w\circ u)(x) = w(u(x)) $

[obsidian monolith]

Samtell:

[sullen shoal]

Sorry about that

[frigid dragon]

yeahhh

so i do put 7 into the equation

[sullen shoal]

Yeah

[frigid dragon]

ayye thanks man

[sullen shoal]

Yeah anytime, fucked up a little hehe

[frigid dragon]

https://gyazo.com/362ee8b5dffee1e950fdf6dddee50be6

Kind of confused how im suppose to apply the log rule into this

<@&286206848099549185>

@sullen shoal

[viscid thistle]

log(stuff)=2

base^(=2) =stuff

4^2 =-5x

@frigid dragon

16/-5=x

=pup -8+log base 4 (16)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@languid wind

[viscid thistle]

Proved

[rare zephyr]

Hey uhhh

In my guide book

About differentiations

It states that if y=ax^n

Then dy/dx=anx^n-1

What does it mean and why is that?

the n in the second equation are both the same value

[spring thunder]

What is buzzing you, the multiplication by a or the x^n->nx^(n-1)?

or everything

[rare zephyr]

Second one

Why the gradient is equal to anx^n-1. Implying both of the n's are the same value

e.g.

=pup plot y=4x^n

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@rare zephyr

[rare zephyr]

oops I meant

[spring thunder]

Do you the binomial formula ? (ie the expansion of (a+b)^n) my internet is being shit rn errr

[rare zephyr]

=pup plot y=4x^5

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@rare zephyr

[rare zephyr]

I dont know how a gradient in that graph equals to what Ive just stated

[spring thunder]

"Do you the binomial formula ? (ie the expansion of (a+b)^n)"

[rare zephyr]

I know the binomial formula

[spring thunder]

OK noice

$$(a+b)^n = \sum_{k=0}^n{\binom{n}{k}a^{n-k}b^{k}}$$

[obsidian monolith]

emeric75:

[spring thunder]

(If I didn't typo shit)

So now

Let f : x->x^n (n positive integer)

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

[obsidian monolith]

emeric75:

[spring thunder]

So now

$$f'(x)=\lim_{h\to0}\frac{(x+h)^n-x^n}{h}$$

[obsidian monolith]

emeric75:

[spring thunder]

So expand (x+h)^n using the binomial formula

And simplify as much dope as you can

[rare zephyr]

I understand now thanks

[spring thunder]

this only proves it for n positive integer it's also provable for non-integer powers but it's a bit less of a piece of cake

[serene heath]

you can just use quotient rule to prove for negative powers

given that you proved for nonnegative beforehand

[sharp pagoda]

Can anyone help me with number 8? The answer is 5 cm but I dont understand where you pull out the five from

[viscid thistle]

If no one else helps in 5m, feel free to mention or PM me

[viscid thistle]

Why is -5x^2+55x-50 have negative roots but positive roots when it = 0

[flat terrace]

How do we derive the value of e?

[viscid thistle]

one way to get it is through thinking about compound interest

you start by saying something like, I have 50 bucks and it doubles in a year

[flat terrace]

Continuously compound?

[viscid thistle]

eventually yes

so at first you double so that's multiplying by 2

or another way to think of it is adding 100% more

to what you already have 100% of

so you can write it as 50+50

now if you want you can think, what if I add 50% more twice instead of 100% once

50 * (1+1) now is becoming 50 * (1+.5)*(1+.5)

now what if you do 33.33% more 3 times

and so on

$\lim_{n \to \infty} (1+\frac{1}{n})^n = e$

[obsidian monolith]

Empty Axes:

[flat terrace]

OH okay, it that the only method?

[viscid thistle]

which is a little more than 2

no it's not

I think it's the intuitive answer though

there are other ways to go about it though

what function has its value at every value of x also equal to the slope at that value?

e^x

well that's one such answer

also you can use anything you'd find the natural log for as well

like area between a hyperbola and a line arranged in a specific way

$\log_e(x) = \int_1^x \frac{1}{t}dt$

[obsidian monolith]

Empty Axes:

[viscid thistle]

idk probably some other ways I'm not considering

you could say from looking at the hyperbolic trig functions and looking at the x and y components on the unit hyperbola x^2-y^2=1 that

$e^t = \cosh(t) + \sinh(t)$

[obsidian monolith]

Empty Axes:

[viscid thistle]

so there

kinda similar I suppose

arguable what comes first or not though at this stage

[vast trellis]

Hi

How do I find the direction of the resulting vector of two other vectors

Is it SW or is it pointing NE

[lapis nimbus]

How would I solve for x if: log x = 6

I can't find an example anywhere in my book

[solemn tiger]

A side note, a logarithm without a written base is assumed to be in base 10. The natural log is assumed to be in base e.

[lapis nimbus]

Oh ok, got it

Thanks

So x=10^6

[flat terrace]

=tex {\log{9}7} + {\log{9}11} + \frac{\log{9}2}{3} + \frac{\log{9}5}{3}

[granite stirrup]

Command disabled

The sever owner has disabled that command in this location.

[hexed ermine]

?

,$ \log_{9}(7)+\log_{9}(11)+\frac{\log_{9}(2)}{3}+\frac{\log_{9}(5)}{3}

[obsidian monolith]

PJS:

[flat terrace]

yeah how do you do that? as in condense? @hexed ermine

I got this

,$ \log{9}(77)+ \frac{\log{9}(10){3}}

[obsidian monolith]

SteelBlades135:
Compile Error! Click the reaction for details. (You may edit your message)

[flat terrace]

,$ \log{9}(77)+ \frac{\log{9}(10)}{3}

[obsidian monolith]

SteelBlades135:

[flat terrace]

idk what to do with the 3?

[hexed ermine]

Write 3 as log base 9

log_9(x)=3

Wait hold on

3log_9(77)/3+log_9(10)/3

(3log_9(77)+log_9(10))/3

log_9(77^3)+log_9(10))/3

log_9(77^3*10)/3

log_9(77^3*10)/log_9(9^3)

Reverse change of base formula

log_(729)(4565330) lol

[fickle moat]

6. Assume that the annual sales of a small manufacturer can be modeled by a linear function and that sales were $15,000 in 1988 and $47,000 in 1993. Let x = 0 represent 1988 and f(x) represent annual sales, and write a formula for f(x).

do i have to use A=Pe^rt?

[fickle moat]

<@&286206848099549185>

[serene heath]

no thats an exponential function

it says linear function

some thing like f(x)=ax+b

[fickle moat]

so where do i begin

1988 = $15000
1993 = $47000
let X = 0

[serene heath]

so f(x) is the anuual sales

and x is the number of years that have passed since 1988

if its linear, you have $f(x)=Ax+B$

[obsidian monolith]

lemon catto:

[fickle moat]

before i do that

do i have to find 1988's annual sale?

....

[serene heath]

youre already given the sales in 1988

which were 15000

so in 1988 the sales were 15000

[fickle moat]

what do i have to do with 47000

hello?

i dont get it

i just want to know how to write a formula

[serene heath]

sorry my internet was bein meg gae

anyways you have f(x)=Ax+B

and you know when x=0 f(x)=15000

plug that in to find B

[fickle moat]

so
15000x+47000?

[serene heath]

that isnt one of the answer options

[fickle moat]

i know that

[serene heath]

so it cant be righ

plug x=0 and f(x)=1500 into f(x)=Ax+B

and find B

[fickle moat]

so f(x)= 15000x+47000 is correct answer?

15000x0 +47000

[serene heath]

there is no point in me just telling you the answer

[fickle moat]

47000?

[serene heath]

you need to understand how to get the answer

[fickle moat]

and i came for help

[pulsar lynx]

can anyone help me with an assignment ?

[serene heath]

right so plug in x=0 and f(x)=15000 into f(x)=Ax+B

and tell me what you get

[fickle moat]

f(x) = root x+3 , g(x) = 8x-7 ; find (f o g)(x).
root 8x-7+3
root 8x-4
whats next step?

2 root 2x-4?

[spring thunder]

$\sqrt{8x-4} = \sqrt{4(2x-1)}$ right

[obsidian monolith]

emeric75:

[fickle moat]

yep

oh

🤦

[spring thunder]

[fickle moat]

i should write it

on paper

[quiet marlin]

I'm supposed to demonstrate why it's algebraically impossible to solve for r in this ordinary annuity equation. I think it's because it isn't possible to isolate r. Does that look/sound correct?

[rocky bison]

try it

See what happens

You'll run into problems fast 👀

maybe you'll even do some binomial expansion

[solemn tiger]

I see a binomial expansion to the 360th power

jfc

[quiet marlin]

yeah lol, no matter what it ends with r on both sides

[solemn tiger]

Not evaluating this, screw that

[rocky bison]

But then you've gotta solve some question for an equation of degree 360

[solemn tiger]

[rocky bison]

lmao

[quiet marlin]

don't need to solve for it, just explain why it's not possible

[rocky bison]

finding roots for quintics is hard

[quiet marlin]

is my assumption correct though?

(using purely algebraic operations)

[rocky bison]

I mean you can't solveit

What's your reasoning though

Why is it not possible to isolate r

You said you gotta demonstrate it

[quiet marlin]

Ah, right. From what I've tried the only outcomes are r being on both sides of the equation/cancelling itself out

[rocky bison]

nah

You can't cancel it

because it's not r/12 to a power

it's r/12+1

Which means

,tex \left(a+b\right)^n=\sum_{k=0}^n\begin{pmatrix}n\cr k\end{pmatrix}a^{n-k}b^k

[obsidian monolith]

Pseudo:

[rocky bison]

And when n=12*30

That's a nah

[quiet marlin]

okay lol. thank you

[clear coral]

Could someone plz help me with some unit circle thing

@rocky bison

How am I supposed to do this

@quiet marlin ?

I find unit circle stuff normally ez but this question format I have never seen b4

[rocky bison]

Be a smart ass

And find the value of theta such that sin(theta)=-5

Given

,tex \sin\left(x\right)=\frac{e^{ix}-e^{-ix}}{2i}

[obsidian monolith]

Pseudo:

[rocky bison]

👀

[fallen cloud]

lol

those were fun

[gilded heath]

Can someone help em

PLease

My equation is 8^2/3 times 4^1/2

Which ik im using that product proporties

so I need to add thoes fractions

which I cam up with 8/6

question is do I mutiply the 8 and 4

and do factor tree of 32

[solemn tiger]

Unless you meant this instead.

If you meant the latter, follow PEMDAS (or equivalent order of operations method)

[limpid plover]

Isn't it PEDMAS?

[solemn tiger]

"or equivalent order of operations method"

[limpid plover]

I am specifically talking about PEDMAS

Which is not equivalent to PEMDAS, technically

[fringe stream]

no

[limpid plover]

But equivalent for all practical purposes

[fringe stream]

its pemdas

[thick raptor]

BODMAS

[limpid plover]

BODMAS and PEMDAS

Hmmm

Only US has Division and Multiplication switched

As expected of them

Canada and New Zealand use BEDMAS, standing for Brackets, Exponents, Division/Multiplication, Addition/Subtraction.

Most common in the UK, India, Bangladesh and Australia[11] and some other English-speaking countries are BODMAS meaning Brackets, Order, Division/Multiplication, Addition/Subtraction. Nigeria and some other West African countries also use BODMAS. Similarly in the UK, BIDMAS is used, standing for Brackets, Indices, Division/Multiplication, Addition/Subtraction

[serene heath]

BODMAS master race

[limpid plover]

BODMAS master race

Ye

Fuck BIDMAS tho

Sounds weird as fuck

And BEDMAS too

[pine rampart]

no one gives a fuck

[slender river]

yep

[ebon thistle]

What the fuck is all of this PEMDAS shit I've been hearing of

Crazy as shit

[full token]

People who don't use pemdas scare me

[kind pewter]

[thick raptor]

[chilly badger]

ALL HAIL PEMDAS

[spice lodge]

trig is where its at xD

[slender river]

?

[viscid thistle]

what is a natural log???

[slender river]

log with a base of e

[viscid thistle]

how is it diff from naked log

[slender river]

[spring thunder]

nsfw log

[quiet marlin]

a normal log has a base of 10, unless otherwise specificed

ln(x) = log_e(x)

[fallen cloud]

naked log...

wait

what if the derivative was nsfw?

[kind pewter]

😳

[fallen cloud]

I'll derive your tan so secant'll be squared with me 😉

[slender river]

lewd and confusing

[spring thunder]

"Lewd analysis : a new approach to differentiation"

[slender river]

ono lol

Lewd-valued functions

[dense zealot]

u forgot sec(x)

[hexed ermine]

sec(xy)

[spice lodge]

sec (c)

[pine rampart]

succ(0)=1

[gritty blaze]

It's sec(x)=1/cos(x).

[vast trellis]

Hi

X^4+8=0. Find all complex solutions.

How do I find the polar form and it’s degrees? I know the OG rectangular form is a+bi but is the x the a and the 8 the bi?

[patent beacon]

x⁴ - 8i² = 0
Now it's a difference of squares and you can factor it pretty easy

[limpid plover]

Woah

I didn't realise that you could do that

[vast trellis]

Me neither...our teacher didn’t tell us to do that

[viscid thistle]

you can also subtract 8 from both sides and take the 4th root

sprinkle 4th roots of unity for flavoring

[patent beacon]

Otherwise
x⁴ = -8
x⁴ = -8[cos(2kπ) + isin(2kπ)]
x = (-8)^(1/4) • [cos(kπ/2) + isin(kπ/2)]

Which should yield four solutions for choices of k

[vast trellis]

Wait. Could you do u^2-8i instead and then square the results

Since u would represent the ^2 anyways

[patent beacon]

You mean u² - 8i²?

[viscid thistle]

they want you to determine polar form?

I wouldn't use sine or cosine personally, just keep it all exponential

[vast trellis]

They want polar form

[viscid thistle]

which is why you shouldn't use sine and cosine

[viscid thistle]

because that's polar form split up into rectangular form components

cis is what I mean by exponential

$cis(t) = e^{i t}$

[obsidian monolith]

Empty Axes:

[viscid thistle]

that is polar form

[vale pewter]

ew who uses t

[viscid thistle]

adults

now shoo child

[vale pewter]

-.-

[vast trellis]

Lol

[rocky bison]

wtf is cis

=pup cis

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@rocky bison

[vale pewter]

Cos + i*sin

[serene heath]

cis gendered euler

[vale pewter]

Ah yeah it actually means not trans

Disregard that cos and sin thing

[serene heath]

bigot

😡

[vale pewter]

wait r u user?

[serene heath]

no i am me

[viscid thistle]

Anyone have advice for things to study before starting a precalc class? (it's 4 months long) I'm already going through the art of problem solving - introduction to algebra

[hexed ermine]

You dont need to know much

Just make sure your algebra is good

[rocky bison]

@viscid thistle precalc would be good

👀

[viscid thistle]

@hexed ermine yeap, got that down with art of problem solving

[hexed ermine]

You're pretty much good

[viscid thistle]

Why do we factor out the .5?

What is this form we are trying to get the function in and why are we trying to get it there?

[serene heath]

hmm not sure what theyre doing here

doesnt even seem right

translation to the left by pi units is correct

to go from 4sin(0.5x) to 4sin(0.5x+pi)

[solemn tiger]

To get x on its own so we may have our equation of the form:
y = asin(b(x - h)) + k

Where:
|a| = amplitude
2pi/b = period
h = horizontal translation
k = vertical translation

[bleak dawn]

Can I have some help with 66, 75 and 81

[flat turret]

66: $$4\sin^2 x=4-4\cos^2 x$$

[obsidian monolith]

Gonzo17:

[flat turret]

then you got a quadratic equation

75: use the formula for sin(2x)

81: sin(x+y)

[bleak dawn]

Ty

[viscid thistle]

hey can someone help me w pre calc

[flat turret]

anything specific?

[viscid thistle]

yeah this one problem

can I pm it to u

[flat turret]

go ahead

[rare apex]

can someone help me with trig identities?

[sharp mulch]

I can try @rare apex I am studying trig identities thoroughly at the moment.

[rare apex]

1-tan^4

how does the answer become 2sec^2 x - sec^2 x

i have a test tmrw on this bs

[sharp mulch]

Can somebody help me with function transformations? Particularly horizontal transformations, I thought I got them but I am having a mental lapse.

@rare apex, sorry haven't gotten to that one yet 😦 I have studied the addition identities, e.g. sin(a+b) and negative angle identities, e.g. sin(-a) so far. I can explain clearly where those come from but not the one you mentioned

[flat turret]

@rare apex $$1-\tan^4 x=(1-\tan^2 t)(1+\tan^2 t)=(1-\frac{\sin^2 t}{\cos^2 t})(1+\frac{\sin^2}{\cos^2})=(1-\frac{\sin^2 t}{\cos^2 t})\frac{\sin^2 t+\cos^2 t}{\cos^2 t}$$

[obsidian monolith]

Gonzo17:

[viscid thistle]

$$\frac{x}{2\pi}r^2$$

[obsidian monolith]

--:

[sharp mulch]

Why exactly when you have f(x) and g(x) = f(x+2) then f(x+2) results in the graph of f(x) shifted by 2 units to the left? Essentially subtracting 2 units to every x-coordinate of f(x)?

[flat turret]

$$=(1-\frac{\sin^2 t}{\cos^2 t})\cdot\frac{1}{\cos^2 t}=sec^2 t-\frac{\sin^2 t}{\cos^4 t}$$

[obsidian monolith]

Gonzo17:

[flat turret]

@rare apex What was supposed to be the result?

[rare apex]

2sec^2 (x) - sec^2 (x)

[flat turret]

wut that's just sec^2 (x)

[fickle moat]

u should look at trig identies

All you need to know to do trig identies

[dry swan]

anyone around want to help me with my precalc

[slate niche]

@dry swan

whats up

im at the unit circle rn lmao

[fickle moat]

unit circle 😄

[slate niche]

yes

i had to memorize that

i mean i passed with a 100 so

its either you get a 0 or a 100

@fickle moat

and you had 5 minutes to fill it out ><

[fickle moat]

lol

i got 93/95

i messed up in product to sum

[rare sail]

I need help with this

I’m wondering if what I’m doing is correct lol

[hexed ermine]

Yes p=1

[rare sail]

okay thank you!

Is this correct?

[hexed ermine]

Close

Wait no that's correct

[rare sail]

Oh lol

Okay thank you again!!!

[viscid thistle]

SUBSCRIBE TO PEWDIEPIE

[keen gyro]

If a = b
Then ab = b^2
Then ab - a^2 = b^2 - a^2
Then a(b-a) = (b+a) (b-a)
Therefore, a = b + a and because, a = b
Then, a = 2a
Therefore 1=2
Therefor i have proven maths wrong and this server should be deleted

[hexed ermine]

Nice dividing by 0 now I see

[keen gyro]

@hexed ermine well thanks for ruining it!

[hexed ermine]

:(

Is it time for me to drink you

[keen gyro]

No, but i think it is time for me to eat you

[hexed ermine]

Good

But she died a couple years back:///

[serene heath]

what the fuck

[rare sail]

I was wondering if this was correct

[serene heath]

you forgot about the 3

in 3(x+1)

actually nvm

[rare sail]

Oh okay lol

[daring patrol]

3(x+1) = y-2 → 3x+3 = y-2 → 3x - y = -5 → y - 3x = 5

they're the same equations

[turbid mauve]

󠂪󠂯

[vast trellis]

Anyone on for a quick check on if I got an answer right?

[spring thunder]

[vast trellis]

Wait

The cos should be 315 as well

[serene heath]

you sure bout your modulus?

[vast trellis]

Thanks

[spring thunder]

hmm 🤷

[vast trellis]

Is the polar form 8(cos360+isin360)? If so how would I explain it on a test?

I mean, I know it’s a 360 circle with a radius of 8 (because 8^2 )

[serene heath]

can you take a closer pic?

[vast trellis]

Sure but they will be in two parts

Ugh I hate mylabsplus because if you type in the equation I a different form (even though it’s technically right ) they’ll mark it as wrong

[serene heath]

so you wanna convert that into polar right

[vast trellis]

Yes. And an equation

[serene heath]

do you know the relations between caretesian and polar form?

[vast trellis]

No

I do know the CIS thing and turning rectangular equations into polar equations

Matheay gave me this but there were times where it was completely wrong, which is why I don’t rely on it

I check my answers for it mostly

[serene heath]

$x=r\cos(\theta)$ and $y=r\sin(\theta)$

[obsidian monolith]

lemon catto:

[serene heath]

meaning $r^{2}=x^{2}+y^{2}$

[obsidian monolith]

lemon catto:

[vast trellis]

So would I need to find the cos and sin of 8

[serene heath]

why?

you have $x^{2}+y^{2}=64$

[obsidian monolith]

lemon catto:

[serene heath]

meaning $r^{2}=64$

[obsidian monolith]

lemon catto:

[serene heath]

so $r=8$

[obsidian monolith]

lemon catto:

[serene heath]

thats your polar form

[vast trellis]

Since ok. Thanks

I thought polar form had to have sin and cos that’s why

[serene heath]

it does

rip

there we go

that ones better

can you see how x=rcostheta

and y=rsintheta

[elfin night]

What is the Q?

Integral..?

This kind of deal?

opps thats wrong

$ \sin\left(4x\right)\sin\left(2x\right) = \ 2\sin\left(2x\right)^2\cos\left(2x\right) = \ 8\sin\left(x\right)^2\cos\left(x\right)^2\left(\cos\left(x\right)^2-\sin\left(x\right)^2\right) $

[obsidian monolith]

Colen:

[elfin night]

I mean like

Expand out I guess

"Express each product as a sum" is a weird question

Usually you are trying to make it more compact not less...

[elfin night]

smh where u go

$ \left(\frac{1}{2i}\right)^2\left(e^{4xi}-e^{-4xi}\right)\left(e^{2xi}-e^{-2xi}\right) \ \left(\frac{1}{2i}\right)^2\left(e^{6xi}-e^{2xi}-e^{-2xi}+e^{-6xi}\right) \ -\frac{1}{4}\left(\left(e^{6xi}+e^{-6xi}\right)-\left(e^{2xi}+e^{-2xi}\right)\right) \ \frac{1}{2}\left(\frac{1}{2}\left(e^{2xi}+e^{-2xi}\right)-\frac{1}{2}\left(e^{6xi}+e^{-6xi}\right)\right) \ \therefore \frac{1}{2}\left(\cos\left(2x\right)-\cos\left(6x\right)\right)\ =\sin\left(4x\right)\sin\left(2x\right) $

[obsidian monolith]

Colen:

[elfin night]

Idk where you went but if you're still there thats a way to do it

[slate niche]

https://media.discordapp.net/attachments/468168945900191748/519347150174289930/image0.jpg?width=507&height=676

I might need help on how to do these, I have my sign test sheet but the answers from the key, don't match it. If someone could help me with this.

[floral tide]

does polar bears use polar coordinates tho?

or how do they navigate

[slender river]

[viscid thistle]

We start with the equation in the form asin(bx+c)+d and so I left the c value as 1.25 like its expressed on the graph. The problem wants me to convert it to a value involving pie, I guess thats the value it would have had before the transformations were applied to this function. My question is, why do I factor out and multiply 2pie/5 by it when the b term does not multiply the c term in the standard form of this function listed above
What I'm trying to udnerstand is why in step 3/4 in the second image I have to factor out 2pie/5 and multiply the x and c variables when the c variable was never being multiplied by the b term in the standard form of the function asin(bx+c)+d

@elfin night

[elfin night]

y ping me tho

[viscid thistle]

Didn't want to ping all the helpers at once and saw you online

its np if your busy

<@&286206848099549185>

The step I'm wrestling with is the scond part of step 3/4 where it says 7sin(2pie/5(x))= 7sin(2pie/5(x+1.25) . Why are we multiplying the c term by the b term when that was not the case in the original standard form of the equation asin(bx+c)+d?

[dry swan]

what's up guys

anyone around to help me study for my quiz tomorrow? :X

[tiny verge]

@viscid thistle your original equation was just asin(bx), there wasn't any c.. it was just 2pi/5 x. Notice?
Now the question states the amount by which x changes, so the new x = old x + change.
So the equation now is gonna be asin(b(new x)) or asin(b(old x + change))

Makes sense?

[viscid thistle]

So could I write the equation as f(x)= asin(b(x+c)) +d instead of asin(bx+c)+d?

@tiny verge

Ultimately the same thing, but the first form is a little cleaner

So its the translation from the first form to the second form that is still stumping me. I feel really comfortable with the rest of the transformations but how we get from (bx+c) to (b(x+c)) is the sticking point for me.

@viscid thistle

Well, bx+c isn't equal to b(x+c) -- you use different numbers-- it's just another of writing bx + c

Ok that makes sense, but what is the intution behind when we use one or the other? For example, in the practice problems listed a little ways above they have me start out with the form (bx+c) and have me end with the form (b(x+c)) which is causing me confusion.

Why translate to the second form? What is the intuition here?

[tiny verge]

okay, so you need to know, in your question there was NO c term.

it was just sin(bx)+d

and hence got transformed into sin(b(x+e))+d where e is any variable

if it was sin(bx+c)+d, and the question was the same, that is if it said the x was changing

then the equation would be sin(b(x+e)+c)+d

@viscid thistle

and in case it's still unclear how or why I replaced x by x+e, I mentioned the intuition behind that in my last message

[viscid thistle]

b(x+c) does the phase shift first and then changes the period, bx+c changes the period and then does the phase shift

[viscid thistle]

Can someone tell me how to do this?

#precalculus

[serene heath]

do you know the formula for the arc length, given the radius and the angle substended at the centre?

[fallen cloud]

r * theta

where theta is the angle in radians

[viscid thistle]

I honestly don’t know how to do any of this was sick for 2 weeks and they gave me some notes but idk what am looking at.

[limber bone]

ok its jusut a formula

radius * angle in radians = the length of the arc subtended by the angle

so u can convert the angle to radians

then plug in the formula

[viscid thistle]

Well thanks anyway I don’t know what am doing lol

[solemn tiger]

You have arc length (s), and you have theta (in degrees)

If you convert your degrees to radians and plug it straight into the above formula, it's easily evaluated.

[viscid thistle]

so something like 3pi = r(105)?

[solemn tiger]

"theta must be in radian measure"

[hasty wolf]

See I never bother using the "arc length formula"

I always think of it in terms of proportions.

θ / 360° (or 2π or τ rad) = arc length / circumference = sector area / circle area

I'd say that makes it easier to remember because it's not just a formula --- it's the whole concept of "part over whole" in three different ways. 😃

[earnest finch]

Anyone here on graphing tan and cot and sec and csc

I am like screwed for that

[fallen cloud]

=pup graph tan

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@fallen cloud

[fallen cloud]

=pup graph cot

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@fallen cloud

[earnest finch]

No but like

[fallen cloud]

=pup graph sec

[earnest finch]

With transformations

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@fallen cloud

[earnest finch]

And application problems

[fallen cloud]

o

[full token]

=pup graph -tan

[granite stirrup]

Do more with Wolfram|Alpha pro

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@full token

[full token]

@earnest finch yea I can help with transformations

If it's not too late D:

[viscid thistle]

Why do we divide x by the .5 instead of multiply? The notation makes it look like multiplication.

[hasty wolf]

@viscid thistle : You're right, that is a confusing thing about transformations. The thing to remember is that anything inside the function seems to be applied "backwards". Same as how y = (x - 2)² translates y = x² right by two instead of left.

As for why, what's really going on behind the scenes is that you're moving the coordinate axes, so the graph moves opposite to what you do. y = (x - 2)² moves the coordinate axes two to the left, which has the effect of moving the graph two to the right. Same with cos(x/2) -- the x-axis shrinks in half, which has the effect of blowing up the graph.

Think of it like this -- if you have an alarm that's set for 8:00, and you want to move it an hour earlier but the "set alarm" button is broken, you could set your clock an hour later.

[viscid thistle]

So that makes sense, but the notation still seems wrong. look at the formula at the top of this image f(x)=asin(bx+c)+d is how it is expressed, yet I'm supposed to dived x by b not multiply as the notation suggests

[hasty wolf]

It "seems wrong" but that's how it works. The inside stuff is backwards, but if you remember that you're basically good.

y = a sin(bx + c) + d is the simplest-LOOKING algebraically

[obsidian monolith]

DMAshura:

[viscid thistle]

Yes!

that makes so much sense

[hasty wolf]

That way c is how much it goes to the right, and b is how much you multiply the x by

It's still backwards but it's at least easier to see what's going on

So if you have

$y = -3\cos\left(\frac{x-3}{2}\right)+1$

[obsidian monolith]

DMAshura:

[hasty wolf]

Then it's moved 3 to the right and stretched horizontally by 2

(If you've ever learned about conics that should actually look sort of familiar seeing things in that format)

[viscid thistle]

Right because dividing the x input means it takes longer for us to get through a period of the function. Ok that is intuitive.

[hasty wolf]

Eyyyyy nice 😄

[viscid thistle]

Could we do a call and I screenshare a problem with you ?

[hasty wolf]

I can't unfortunately

But understand that by the same token

$y = \sin(2x) = \sin\left(\frac{x}{\tfrac{1}{2}}\right)$

[obsidian monolith]

DMAshura:

[hasty wolf]

That shrinks the period.

If you see why, you're gold.

[viscid thistle]

Because we cycle around the unit circle twice as fast?

[hasty wolf]

Yup!

Yeah if you think of things in terms of the unit circle that helps a lot.

[viscid thistle]

Ahh, yeah that makes sense. It's more the notation that is tripping me up.

[hasty wolf]

You'll get used to it. 😃

[viscid thistle]

Yeah, practice practice practice 😃

Like sometimes I see the expression represented this way

And sometimes its f(x)= asin(bx+c)+d

Looks similar to the form you presented but with the b multiplying rather than dividing

[hasty wolf]

Yeah, it's equivalent though

THe dividing just makes it a bit easier to see the scale factor

But you definitely want to factor the b out either way

Otherwise you won't have the right horizontal shift

[viscid thistle]

It's just in this form "f(x)= asin(bx+c)+d" the b is not multiplying the c term

so its not immediately clear to me how we can factor it out

[hasty wolf]

Well, think about just bx + c

Can you write it as b(x + _ ) ?

What would the _ need to be so that when you redistribute it you get bx + c?

[viscid thistle]

c/b?

[hasty wolf]

Yup

So if you have sin(2x + π/3)

You'd write it as sin( 2(x + π/6) )

[viscid thistle]

ahhhhh

[hasty wolf]

And you can always check if you did that right by real quick re-distributing in your head

I call factoring "un-distributing" for that reason 😛

[viscid thistle]

That example is really helpful

[serene heath]

[hasty wolf]

Awesome!

bbiab

[viscid thistle]

Thanks for the help, see you around!

[hasty wolf]

Back.

[viscid thistle]

Something I have been thinking about is how the direction of shift of a periodic function sort of becomes meaningless since we can shift it far enough to just back to where we started

I sure there is a practical application to physics or something, but mathamatically it seems like shifting a sine function +2pi or -2pi is basically the same thing

[hasty wolf]

That's true.

[viscid thistle]

I'm gonna go sleep now, I should be on a lot this weekend though

[viscid thistle]

How do I write this in simplest form? Like what do I need to transform

Yknow

[spring thunder]

like get rid of the nested fractions

[potent onyx]

^

[spring thunder]

basically rewrite 1/R1 + 1/R2 + 1/R3 as a single fraction

[viscid thistle]

Kinda like rationalize the denominator

[spring thunder]

then inverse it

[viscid thistle]

?

Can I just multiply them out

[spring thunder]

they're added to each other not multiplied :/

[viscid thistle]

Lel

[potent onyx]

If you'd do it that way you'd have to: R(1/R1 + 1/R2 + 1/R3) = 1

[viscid thistle]

1/(1/R1R2R3)

?

[potent onyx]

You multiply the denom out

[viscid thistle]

(1/R1R2R3) ^2?

Or * R

[spring thunder]

$$\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} = \frac{R_2R_3}{R_1R_2R_3} + \frac{R_1R_3}{R_1R_2R_3}+\frac{R_1R_2}{R_1R_2R_3}$$ $$\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} = \frac{R_2R_3+R_1R_3+R_1R_2}{R_1R_2R_3}$$

[obsidian monolith]

emeric75:

[spring thunder]

rewrite them with the same denominator

add them

them inverse the resulting sum to get R

[pseudo nexus]

Hey ideally need help on this test

[viscid thistle]

Olkay thanks

[spring thunder]

[pseudo nexus]

You understand any of that 😭

[viscid thistle]

Lmao trig is pretty much know it or you don’t

[pseudo nexus]

This is precal

Wait is trig precal🤔

[viscid thistle]

Trigonometry is the study of angles

In real calculus

I’m pretty sure it’s used to find values for graphs idk how tho

I’m only p11

[pseudo nexus]

What’s p11

[viscid thistle]

Grade 11

Precalc

[pseudo nexus]

I’m I. High school and Idk any of this test

Ima sophomore tho

Was hopping this Discord could help

[hasty wolf]

@pseudo nexus From #❓how-to-get-help :

"6. Do not ask for help on tests. Any violation of this will lead to appropriate action being taken at mod discretion."

[serene heath]

👏

[solemn tiger]

[viscid thistle]

Rate = x+3 and x-3upstream

But what is x?

[grim bridge]

Anyone here good at input-output in regards to a matrix

[solemn tiger]

[grim bridge]

It already has the answers I would just appreciate an explanation

[muted mason]

any good sources of studying pre-calculus?

[white ferry]

Compound interest is getting me rn

I need to find the total return of an investment of 1000 with 5 percent returns for 65 days per 4 years

But this formula just doesnt work for some reason, ik I'm doing something wrong

Can someone please help

[viscid thistle]

You're getting help in another channel, right?

[viscid thistle]

how do find the focus of this: y=-(x+3)^2-1

and directrix

[hasty wolf]

@muted mason purplemath.com has good articles

[muted mason]

Ty

[viscid thistle]

Hello

[potent onyx]

@muted mason I recommend:
Trig: Michael Corral's - http://mecmath.net/trig/trigbook.pdf
Algebra/Misc: Precalculus Mathematics in a Nutshell (PDF available online). Other various sources would be good for this as well.
Basic function knowledge: Gelfand's Functions and Graphs (PDF online)

[vale pewter]

How do you evaluate $$\sum_{x=0}^{10^{10} - 2} \frac{3x^2 +9x + 7}{(x+3x+2)^3}$$

[obsidian monolith]

Plum:

[hexed ermine]

Brute force

[vale pewter]

xd

@thick raptor

[cyan karma]

The trajectory of the stone can be modelled by d =37+7t- 4.9t^2, where t is the time in seconds and d is the vertical distance in meters.
Can someone please enlighten me as to how to find the value of the derivative of this equation after 2 seconds. I can graph it on desmos, but from there I get stuck.

[hexed ermine]

Differentiate and plug 2 in for t

Thisll give you the instantaneous rate of change in velocity at t=2

[obsidian monolith]

Yue:

[cyan karma]

So my equation to start would then become d=37+72-4.92^2 once I have those answered then plug that into that equation format?

I'm learning derivatives and they threw this curveball which stumped me.

[solemn tiger]

I think it'd be a lot faster to just learn how to take derivatives of terms with powers

[obsidian monolith]

Yue:

[solemn tiger]

The derivative of a constant is 0.

That should be enough to do that problem, I think. 🤔

[cyan karma]

Thanks I'll give it a go, currently google is my guide as tutor can't be bothered tutoring. (all done through correspondence)

[cyan karma]

So just to double check on my answer at t=2 would be 15.7? From my course it said to simplify again from 31.4 but 31.4 is where it sits at 2 seconds.

[solemn tiger]

Mmm, I think it should be negative.

[obsidian monolith]

Yue:

[solemn tiger]

Using derivative power rule on each term:

[obsidian monolith]

Yue:

[fresh pike]

d = 37+7t-4,9t^2
d' = 0+7.1-4,9.2.t
d' = 7-4,9.2t
7-4,9.2.2 = -12,6

[cyan karma]

But if t=2 does that not mean we change all the (t)s in the equation to two?

Going by the original formula I mean.

[fresh pike]

You use the t=2 after you derive

[cyan karma]

With your derivative and anti-derivative I am still getting lost. I understand the basics of reversing the equation but I don't get where you get 0+7.1-4,9.2.t from.

[fresh pike]

When you apply dy/dx to a constant you get 0
dy/dx 37 = 0
when you use it on a variable you get 1
dy/dx t = 1
and when you use it on x^n you get n.x^(n-1)
dy/dx t^2=2t^(2-1)=2t^1=2t

if that's what you're asking

if you need the basics for dy/dx, then it's a bit too complex for me to explain it right now. :/

[cyan karma]

That's it, that's where I was going wrong.
No I have a video that is going over that part at the moment with number replacement and the next video is going to cover dy/dx so I think I need to watch these before I try to tackle the question and then explain it.

Khan academy and youtube are paving way for my late education.

[thick raptor]

@vale pewter

[kind pewter]

Heya SA

[thick raptor]

$$\frac{3x^2+9x+7}{(x^2+3x+2)^2}=\frac1{x+1}-\frac1{x+2}+\frac1{(x+1)^2}+\frac1{(x+2)^2}$$

[obsidian monolith]

Simple_Art:

[thick raptor]

@kind pewter hay

[kind pewter]

Is for horses

[thick raptor]

$$\sum_{k=0}^n\frac1{k+1}-\frac1{k+2}+\frac1{(k+1)^2}+\frac1{(k+2)^2}=2H_{n+1}^{(2)}-\frac1{n+2}+\frac1{(n+2)^2}$$

Where H are the generalized harmonic numbres

[obsidian monolith]

Simple_Art:

[vale pewter]

[thick raptor]

[vale pewter]

there's a much better way

[thick raptor]

🤷

[vale pewter]

👀

[viscid thistle]

Ay you all dumb like me to :D

Is it chill if i post my problem in the chat?

[hexed ermine]

Sure

[stray obsidian]

I know where all the numbers go and the formula but how do u make The principal amount to the otherside to get the answer everytime i try it i get clunky numbers

[viscid thistle]

U need to ln both sides

So u take the natural log bc the formula has e in it

[stray obsidian]

i know

but how do i move the P

[viscid thistle]

Use log properties

If ln(a) = ln(b) - ln(c)

Then a = b/c

Bc ln(a) = ln(b/c)

So in this case u have xe^101.05= 22000

10 * 1.05

Not 101.5

So u ln the left and ride side

lnx + ln10.05 = ln22000

Are u good from there?

[stray obsidian]

im confused on where u getting 10.05

or is it *

[viscid thistle]

Bc it’s increasing at 5 percent

So 1.05 is the rate

Bc it’s increasing

[stray obsidian]

where does the 1 come in

thats where im confused

[viscid thistle]

Bc it’s 105 percent

Every year

If it’s increasing u need to add the rate to 1

[stray obsidian]

oh ok

[viscid thistle]

Decreasing u subtract

From 1

[stray obsidian]

oh thank you so much

[viscid thistle]

np

[pale zealot]

need help with this (Not sure if pre cal or not)

[patent beacon]

@pale zealot
Still looking for it?

[pale zealot]

yea

[patent beacon]

The right should be easy, it becomes
1/2 [u + 1/u] - 1

For the left:
u = x + y
Differentiate in terms of x:
du/dx = 1 + dy/dx

So in total we have:
du/dx - 1 = 1/2 [u + 1/u] - 1

[earnest finch]

wtf

Not precalc

if it was I would fail

lol

[rocky bison]

@earnest finch

Did you just want deriv of x^2

or did you want the proof for why

[earnest finch]

The deriv

Isn't it the power rule

[rocky bison]

Yes

[earnest finch]

And that's all

ok

I'm trying to study for calc BC

When I'm in precalc

[rocky bison]

,tex \frac{d}{dx}\left(x^{n}\right)=nx^{n-1}

[earnest finch]

So I'll be ready

[obsidian monolith]

Pseudo:

[rocky bison]

That's a good idea

[earnest finch]

I just started like a week ago

[rocky bison]

Best of luck!

[pale zealot]

@patent beacon ah that makes it easier although I'm still not really clear about how du/dx became 1+dy/dx (Sorry if I put it in the wrong area)

[patent beacon]

Just the derivative of u = x + y, in terms of x

[tawny stag]

https://gyazo.com/db8f4c3ddd17e9b29ba7b5987f20c88b

can someone help me out

[solemn tiger]

Consider factoring the numerator.

[tawny stag]

(x-2)(x-7)/(x-2)

then the x-2 crossout right

[viscid thistle]

Yeah, because the top is being multiplied by (x-2) and the bottom is being divided by it they cancel out

@tawny stag

[tawny stag]

i put x-7 as the answer but its wrong somehow

[viscid thistle]

That seems correct to me

can you take a picture of the problem so I can see if you are making an error?

[tawny stag]

yeah

https://gyazo.com/ee71d12350adbf5d758495d4fb398823

[solemn tiger]

"use y = mx + b format"

[viscid thistle]

Hmm

[solemn tiger]

Have you considered putting "y ="

[viscid thistle]

are you comfortable with y=mx+b?

[tawny stag]

its put y= for you

[solemn tiger]

Has it now

[tawny stag]

nvm youre right

[pale zealot]

@patent beacon ah thanks much appreciated

[viscid thistle]

Why is (pie/2(x-3)) a translation of 3 to the right? Wouldn't the fact that (x-3) is being multiplied by pie/2 change how far that translation shifted us to the right. So it would be a translation of 3/pie/2 to the right?

@narrow hawk

@lost pawn

[narrow hawk]

f(x) => f(x+a) means that the curve has moved a units to the right

pi is simply a multiple which makes sin give integer values

[viscid thistle]

Why does the period of the green function being multiplied by two mean that our horizontal translation only takes us half as far to the left?

<@&286206848099549185>

[flat turret]

the period doesn't get multiplied by two

If sin(x+1)=f(x) then red is f(x) and green is f(2x)

and so the value that appeared on some point a now appears in a/2

$green(\frac{x}{2})=f(2\frac{x}{2})=f(x)=red(x)$

[obsidian monolith]

Gonzo17:

[viscid thistle]

I think I'm missing the intution algebraically

I don't understand why the horizontal shift would be affected by the shortenting of the period of the function. Shouldn't it shift all points 1 to the left regardless of the functions period?

@flat turret

[flat turret]

shifting all points to the right is f(x-1)

f(ax) means the value f(x) is now achieved "a times earlier"

[viscid thistle]

Could you suggest a function to graph by hand so i can see it in action?

[flat turret]

you could just go with a straight line, like y=x+1

[viscid thistle]

"shifting all points to the right is f(x-1)
f(ax) means the value f(x) is now achieved "a times earlier" Can you clarify what you mean by the value of f(x), do you mean the value where it intercepts the y axis and x equals zero. Looking at this graph that does make more sense to me now. So if we have y=4x-1 the y intercept will be .25. That feels pretty comfortable but I'm having a little trouble transfering this intution to the graph of periodic function.

@flat turret

[flat turret]

the graph of f(ax) is a times narrower than the graph of f(x)

and $f(a \frac{x}{a})=f(x)$

[obsidian monolith]

Gonzo17:

[viscid thistle]

I'm comfortable with that so far

[flat turret]

so by the value of f(x) I meant f(x) for any x

you could take a little more complicated function, like $\frac{1}{x^2+1}$

[obsidian monolith]

Gonzo17:

[flat turret]

=pup graph y=1/(x^2+1)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@flat turret

[flat turret]

=pup compare y=1/(x^2+1) to y= 1/((10x)^2+1)

[granite stirrup]

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Query made by

@flat turret

[viscid thistle]

Ahh, so you mean we reach our output a times sooner

yeah that is intuitive

[flat turret]

wait let me present it with wolframalpha

[viscid thistle]

ok np

[flat turret]

=pup plot 1/(x^2+1), 1/(((4x)^2)+1)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@flat turret

[flat turret]

so the red is just 4 times closer to the y-axis than blue

[granite stirrup]

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@flat turret

[flat turret]

k now it's 4 times closer

[viscid thistle]

Ok so that makes sense because we reach all of our y values 4 times sooner which means our function is squished 4 times closer together on the x axis

What is really confusing me is the relationship between this and the vertical shift

*Horinzontal shift sorry

So for instance in the functions sin(x+1) and sin(2x+1) it seems to me that the function sin(2x+1) would be compressed by a factor of two but the horizontal shift of 1 should still move its points equally far

That is incorrect I know but I'm having trouble getting the intuition why

[flat turret]

in your case the horizontal shift happened first

and then the x to 2x

[viscid thistle]

I thought the order of transformatons demanded that the stretch/compression happend first. Maybe I need to write the function like this then? sin(2(x+1/2))

[flat turret]

=pup plot e^x, e^{2x}

[granite stirrup]

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Query made by

@flat turret

[flat turret]

whenever one of the functions can be represented as f(x) and the other as f(ax) you make the graph of f(x) a times narrower to get f(ax)

=pup plot sin x, sin(2x), sin (2(x+1))

[granite stirrup]

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Query made by

@flat turret

[flat turret]

so here you have blue being the original function f(x)

then red is f(2x) so two times closer to y-axis

and then if we set f(2x)=g(x), you have green=g(x+1)

so it's moved 1 to the left entirely

=pup plot sin x, sin(x+1), sin (2x+1)

[granite stirrup]

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Query made by

@flat turret

[flat turret]

here however you have blue=sin x=f(x)

then red is f(x+1) so it's 1 to the left from blue

and then if g(x)=f(x+1)=red, you have green=g(2x)

so it's 2 times closer to y-axis than red

so if you're trying to graph a function of a form f(ax) take f(x) and make it a times closer to y-axis

like when you have sin(2x+1)=f(2x) and then f(x)=sin(x+1) which you can already graph

idk if that makes sense to you

[viscid thistle]

I'm not sure how sin(2x+1)=f(2x). Where did our +1 value go? I think my issue was I have not been following the order of operations properly. I was applying the stretch/compression and then the vertical shift rather than the other way around.

[flat turret]

if sin(x+1)=f(x)

then how would f(2x) look like?

[viscid thistle]

Okay, got it

That part makes sense

[frigid raven]

Hello, could someone please help me with this problem

I don't understand the third part of the problem

I think I have all the formulas written down on my notes but none of them seem to work with this

[patent beacon]

If you wanted to know the 55th term, you'd need to double 54 times.

If you wanted to know the nth term, you'd need to double ___ times

[frigid raven]

n-1 times?

Wow

Thank you for clarifying that question

[patent beacon]

The wording was kinda everywhere. Glad it makes sense now

[frigid raven]

Is there anything wrong with this? I used the sum formula for geometric sequence. It worked on other problems except this

formula

[forest canopy]

You put n-1 and not n

In the box

[frigid raven]

Well I did that. 18 is the stopping point so I replaced n with 18 following the n-1 rule

I tried using 17 as well but didn't worked, it must be an issue with the site itself

[patent beacon]

Why not compute that? Then put it in

== 6*(1 - 4^(17)) / (-3)

[granite stirrup]

34‚359‚738‚366

[scenic charm]

Hi I need help trying to solve this question

7*5^(x+4)=125 *7^(x+2)

[frozen needle]

125 = 5³

you can divide by 125 and by 7 to get something more interesting

[spice canopy]

How do i find the inverse of y = -x^2 + 2x?

[frozen needle]

you mean, the inverse or a local reciprocal?

[spice canopy]

inverse

inverse function

[frozen needle]

1/y is the inverse of y

you're looking for an almost reciprocal function of y

[limber bone]

replace x with y

and solve for y XD

[spice canopy]

yeah thats what i tried but i dont know what to do after replacing them

thats how we learned it too

i got -y^2 + 2y = x

and i have to solve for y

[frozen needle]

is it usual in your country to identify cartesian equation and function?

[spice canopy]

i have no idea what those are

[frozen needle]

when you write y=-x²+2x, this notation is usually used to describe some curve while when you write y(x)=-x²+2x, this describes the image of an input x through a function y

And clearly, when you're swapping x and y, you're taking the "equation of a curve" point of view

while your ambition is to study a function

And I find that a bit weird

[spice canopy]

oh ok

we dont worry about the words that much lol

but yeah basically what i asking is to solve -y^2 + 2y = x for y

[scenic charm]

complete the square to solve for y

[spice canopy]

hmm lemme try that

wait how does that work

@scenic charm

i dont know what to add on the left side

sqrt(2)?

[scenic charm]

actually I think you can factor out the y's and make x = 0

[spice canopy]

ok

but then how do u calculate what y equals

[scenic charm]

making if 0 = -y(y+2) you can individually solve for each y

[spice canopy]

yeah it equals 0 or -2 but i dont want the zeros

i want to find the inverse function

[scenic charm]

oh then -y^2+2y was basically the answer

[spice canopy]

but i need to make it in y= format

[scenic charm]

wait you are supposed to solve for y

[spice canopy]

yeah so basically solving -y^2 + 2y = x for y

[scenic charm]

correct

[spice canopy]

and idk how

[scenic charm]

lemme try to solve it

[spice canopy]

k thank you

[scenic charm]

does it tell you to solve for the inverse or just y?

[spice canopy]

basically just solve that equation for y^