help-4

[woeful oxide]

anyways moving on lets look at the formula for discriminant

[rare wharf]

oh they are the same? delta=discriminant okok just making sure (i had to look up spell aswell )

[humble chasm]

the discriminant being
$b^2 - 4ac$

[rocky lotus]

9k

[woeful oxide]

^

[rare wharf]

yes yes

[woeful oxide]

delta = change but in this context delta is the discrimanent

now

plug in your values

[humble chasm]

remember that the square root is not actually a part of the discriminant, but instead, what’s under the square root. some people mix that up.

[woeful oxide]

that you got from before

but dont solve yet

just plug them in and show me what you got

[humble chasm]

so when you think of this, think,

“taking the square root of the discriminant”

not

“the discriminant is the square root of b^2 - 4ac”

[rare wharf]

(4)^2-4(1)(-12)

[woeful oxide]

it should be -4

[rare wharf]

a=1 b=-4 c =-12

your right

[errant gyro]

i recommend putting the - in b as well because your b is actually negative

while, yes, it doesn't change the value, it ingrains in you the habit of remembering your signs

[woeful oxide]

lol i remember i was mid precalc final and i forgot about my signs halfway thru

[errant gyro]

because otherwise you get the habit of being super sloppy with them
and then when they actually matter, you get a wrong answer

[woeful oxide]

lost so many easy marks

couldve gotten a 100 ended up w an 87

[woeful oxide]

okay

so our discriminant is

$(-4)^2 -4(1)(-12)$

[rocky lotus]

haku

[rare wharf]

(4)2=16

[woeful oxide]

uhh

its not 16

well

4^2 is 16

[humble chasm]

ur thinking of 4 squared

[woeful oxide]

ohh

yeah

lets do it individually then

$(-4)^2 = 16$

[rocky lotus]

haku

[woeful oxide]

so $16 - 4(1)(-12)$

[rocky lotus]

haku

[woeful oxide]

now what

(just follow bedmas)

[rare wharf]

mb for the confusion i ment (4)^2 = 16 then -4 * 1 is -4 -4*-12 = 48

[woeful oxide]

okay hold on

lets take things

one at a time

wait no hes right

what is the final answer

[rare wharf]

then 16+48 =64

[woeful oxide]

yep

now we plug ∆ back into quad formula

[rare wharf]

wait

isnt it -

[errant gyro]

no

[woeful oxide]

no

[rare wharf]

16-58?

48

[humble chasm]

no

[woeful oxide]

negative times negative

is positive

[humble chasm]

this is -4*-12 which is 48

16 - 48
-32

[woeful oxide]

$16 -4(-12)$

[rocky lotus]

haku

[humble chasm]

oh

[woeful oxide]

$16 + 48$

[rocky lotus]

haku

[humble chasm]

wait maybe even i messed up HAHA

[woeful oxide]

lol happens to the best of us

[humble chasm]

no u gotta think of it like this

[woeful oxide]

or you could do the

VERY LONG WAY that i dont reccomend

[humble chasm]

$16-(-48)$

[rocky lotus]

9k

[humble chasm]

and then double slash

which means , slash through both negative signs

so $16 + (+48)$

[rocky lotus]

9k

[humble chasm]

aka, 18+48 🙂

16*

my apologies

[woeful oxide]

yeah

[humble chasm]

double slash is really useful

instead of writing like $-b$

[rocky lotus]

9k

[humble chasm]

i a prefer to write

[woeful oxide]

honestly i use my head for all of precalc

[humble chasm]

$-(b)$

[rocky lotus]

9k

[humble chasm]

it gives the sense of

$-1 * b$

[woeful oxide]

like i graph all of the functions in my head

[rocky lotus]

9k

[errant gyro]

let's refocus on the problem, folks

[woeful oxide]

it takes a lot of time doing this but like you still have to

okay yeah yeah

[rare wharf]

$4+-8/2

[woeful oxide]

uhh

very wrong format

but like ur pretty much right

that would give you

$4 +- 8/2$

[rocky lotus]

haku

[woeful oxide]

how do i use fractions again

is it n/

i forgot

[tidal swift]

$\frac{4 \pm 8}{2}$

[rocky lotus]

Seia

[rare wharf]

my first time trna use that comand haha

[woeful oxide]

yeah

now

simplify

[rare wharf]

no way she knows the +- to

[woeful oxide]

boom

ur done

$ \pm$

[tidal swift]

are you referring to me?

[woeful oxide]

oh wait i need to put this into a

yeah

$ 2 \pm 2$

nope wait

[tidal swift]

ah ok

[woeful oxide]

thats wrong

hold on

[rare wharf]

yes

[woeful oxide]

yk what uhh

[humble chasm]

refocus please

[woeful oxide]

yeah yeah exactly

[tidal swift]

let's not use OP's channel as a #latex-testing substitute

[woeful oxide]

@rare wharf

simplify

[humble chasm]

so we have this so far

[woeful oxide]

thats always your last step (in precalc atleast)

sometimes its ur first

[rare wharf]

combine like terms?

[woeful oxide]

what terms would you combine?

just

acc wait

i think i used the wrong saying

anyways

[woeful oxide]

how do you simplify this

[humble chasm]

ok

OP

when you have a plus or minus sign, think of it as two different problems you have to solve

so this

is actually

$4+8/2$

[woeful oxide]

he should probaby simplify first

[rocky lotus]

9k

[humble chasm]

and

$4-8/2$

[woeful oxide]

makes the equation a lot easier to visualize

[rocky lotus]

9k

[humble chasm]

so u gotta solve both of those

[tawny moon]

$\frac{4+8}{2}$ and $\frac{4-8}{2}$

[rocky lotus]

Hanako

[woeful oxide]

im quitting latex too many sweats

[rare wharf]

here is it like this?

[errant gyro]

correct!

[rare wharf]

Mb for handwriting

Nicee

Ok lemme put in it in to see if it’s correct

[humble chasm]

uhhh

[tidal swift]

signs!!!

[humble chasm]

that would turn into positive 48. not negative.

🙂

[rare wharf]

where am i missing?

[woeful oxide]

signs

[rare wharf]

aw ok

[humble chasm]

op, can i teaxh you a trick that will really help you with problems like this ?

[rare wharf]

yea where

[woeful oxide]

-4 times -12 will never be negative

a negative times a negative is ALWAYS positive

[tidal swift]

-4(1)(-12) right here

[humble chasm]

has any of your math teachers ever taught you about the double slash technique ?

[woeful oxide]

nope

[humble chasm]

it makes it really easy. i’d love to show you

[woeful oxide]

i look at it and im like hmm cool cool and i do it in my head and call it a day

[humble chasm]

so if i have something like

$-a(-b)$

[rocky lotus]

9k

[humble chasm]

basically i like to keep the first negative sign as is and treat a as if its positive

so if i do that, i get

$-(-ab)$

[rocky lotus]

9k

[humble chasm]

following so far?

now just kinda…”SLASH” through both the negative signs.

like literally slash through them

so you get

$+(+ab)$

[rocky lotus]

9k

[humble chasm]

and that’s just positive ab 🙂

[errant gyro]

somehow OP did 16+48 but wrote 16-48

[gloomy sorrel]

It's ✨magic ✨

[rare wharf]

6 and 2 is wrong on the websight

[errant gyro]

16-48 gives us problems under the square root anyway, so something would have gone very wrong there

[woeful oxide]

try 2, 6

[errant gyro]

6 and **-**2

the negative sign!!

[woeful oxide]

^

i wasnt reading

[rare wharf]

bruh i missed my -

okok

[woeful oxide]

dude work on ur signs 😭

[humble chasm]

i can’t wait to show you factoring by grouping 😭

[rare wharf]

yay its right

[woeful oxide]

dawg hes going to hate it

that is like ONLY signs

[humble chasm]

alright can i show u factoring by grouping now

[rare wharf]

okok 9k how would u do this problem a lot faster

[errant gyro]

print this out and paste this above your PC lol

[humble chasm]

alright

[rare wharf]

brah lmao

[humble chasm]

let’s look at ur original problem just so we can regroup

[rare wharf]

okok

[humble chasm]

$x^2-4x=12$

[rocky lotus]

9k

[humble chasm]

alright so just to cover the basics

here’s how i look at it

[woeful oxide]

wait a minute

[humble chasm]

factoring by grouping is a really easy way to solve this problem.

sometimes factoring by grouping doesn’t work

in the cases where it doesn’t work, we have the quadratic formula as a backup.

[woeful oxide]

isnt this just

[humble chasm]

that’s how i look at it.

[woeful oxide]

nope its not

nevermind ignore me

[humble chasm]

following so far ?

[rare wharf]

also rq define factoring im kinda confused when poeple ask me to do it

[woeful oxide]

breaking down numbers into different numbers

uhh basically like

[rare wharf]

do you know before hand when its not going to work

[humble chasm]

factoring is the process of breaking a big expression into a product of smaller pieces

[woeful oxide]

the whole point is exploiting equations

[humble chasm]

no, you kinda just discover it as you try it, at least that’s how i look at it

[woeful oxide]

you do more of that in gr 11

[humble chasm]

anyways let’s refocus ok

[rare wharf]

okok

[humble chasm]

so there’s a few things you need to understand for this to work

have you ever heard of the zero product property?

it’s very easy.

[rare wharf]

yes

[humble chasm]

if you haven’t, i’ll teach you it

[rare wharf]

i just leaned it

[humble chasm]

great!

[rare wharf]

but yk i have not used it alot

[humble chasm]

so we know when

$AB = 0$

[rare wharf]

but i have it in my book here

[rocky lotus]

9k

[humble chasm]

$A or B must equal 0$

[rocky lotus]

9k

[humble chasm]

whoops

anyways u get it

when AB=0, A or B must equal 0

following ?

[rare wharf]

yes

a * b must - 0

[humble chasm]

alright so let’s go back to ur original problem

$x^2-4x=12$

[rocky lotus]

9k

[rare wharf]

a being 1 and b being -4

[humble chasm]

first thing you’re gonna wanna do is set the entire equation equal to 0. you know how to do that right ?

[rare wharf]

yes

-12 from both sides

[humble chasm]

good job

$x^2-4x-12=0$

[rocky lotus]

9k

[rare wharf]

A=1 and B= -4

[humble chasm]

well hold on

take a look at this

[rare wharf]

okok

[humble chasm]

what does this look similar to?

[rare wharf]

umm

[humble chasm]

focus on the left side

forget about the =0 for now. we’ll come back to that later

this is a quadratic equation in its general form

$ax^2+bx+c$

[rocky lotus]

9k

9k

[humble chasm]

see?

[woeful oxide]

ps you need to learn how to factor well

are you in grade 10

[rare wharf]

i remember something the teacher said about we have to change add something to make it a aquare number?

[woeful oxide]

thats for an entirly different equation

[humble chasm]

yeah here just focus on this

[humble chasm]

.

[rare wharf]

dayum ok

[errant gyro]

you're possibly talking about completing the square, which is not happening here

[humble chasm]

.

[rare wharf]

im with you

[humble chasm]

do u see

ok

now remember what i said about factoring.

it’s.

taking a bigger equation -> breaking it down into smaller products

[rare wharf]

i see how this equation is a quadratic is its genral form

[humble chasm]

what is a, b, and c in your equation here

[rare wharf]

a=1 b=-4 c=-12

[humble chasm]

close

you’re just missing c

nice

ok

now in order to factor by grouping you’re gonna start by multiplying ac

what is ac here?

[rare wharf]

-4

1x-4

[humble chasm]

no

[rare wharf]

1*-4

[humble chasm]

that’s ab

[rare wharf]

oh

ac ok

its -12

[humble chasm]

nice!

now

we are looking for two numbers which:

multiply to ac (-12)
and add up to b (-4)

[rare wharf]

i see now

[humble chasm]

so what two numbers add up to b and multiply to ac?

[rare wharf]

-6 and -2

[humble chasm]

close

remember that b is -4 and ac is -12

[rare wharf]

oh yea -x- is +

so -2 and 6

i can see how that is a faster way to solve

[humble chasm]

close again, that adds up to 4.

[rare wharf]

oh adds

[humble chasm]

you’re looking for something that adds to -4

and multiplies to -12

you’re so close

just oneeee small change

[rare wharf]

hm

[humble chasm]

🤔

cmon you’re essentially at the finish line brother u got this

[rare wharf]

i wana say -6 and 2

[humble chasm]

what’s making you question yourself here?

-6 and 2:
multiply together to -12
and add up to -4

no?

[rare wharf]

didnt i say that?

[humble chasm]

no u said -6 and 2

[humble chasm]

here

or

other way around

u said -2 and 6

[rare wharf]

ohhhh

bruh

ok

[humble chasm]

yea

it’s ok

ok so we have -6 and 2

nice

[rare wharf]

i see

[humble chasm]

now let’s circle back to your original equation

$x^2-4x-12$

[rocky lotus]

9k

[humble chasm]

isn’t that the same as

$x^2 -6x + 2x - 12$

[rocky lotus]

9k

[humble chasm]

because when you combine -6x and 2x you just get -4x, which is the same thing as the original, right?

[rare wharf]

are those the same?

[humble chasm]

yeah, look at -6x and 2x

what happens when you combine them?

just brings u right back to -4x, right?

here let me put this in forms of PEMDAS so maybe it’ll make it easier to see

isn’t that the same as

$x^2 (-6x + 2x) - 12$

[rocky lotus]

9k

[humble chasm]

whoops

isn’t that the same as

$x^2 + (-6x + 2x) - 12$

[rocky lotus]

9k

[humble chasm]

follow pemdas. parenthesis first

see how it just brings u back to -4x?

which is what your original equation was
$x^2-4x-12$

[rocky lotus]

9k

[rare wharf]

i see

[humble chasm]

nice

so let’s work with this

$x^2 -6x + 2x - 12$

[rare wharf]

can we try my next problem on here?

[rocky lotus]

9k

[rare wharf]

solve with factoring?

[humble chasm]

yeah

[humble chasm]

so imagine this

right

[rare wharf]

yes

[humble chasm]

but think of the + sign in the middle as sort of like

a wall that divides the first half of that equation from the second

well then we have

$x^2 -6$

[rocky lotus]

9k

[humble chasm]

and

$2x-12$

[rocky lotus]

9k

[humble chasm]

following so far ?

[rare wharf]

yea

[humble chasm]

ok so

[humble chasm]

let’s focus on this one

btw

that’s supposed to be 6x not 6

my apologies

$x^2-6x$

[rocky lotus]

9k

[humble chasm]

well this is the same as
$x(x-6)$

[rocky lotus]

9k

[humble chasm]

see how i’ve just factored x out of it?

i took a bigger equation and broke it down into 2 smaller products

[rare wharf]

yea

[humble chasm]

i want you to do the same for this one

factor 2 out of it

[rare wharf]

2(x-12)?

[humble chasm]

close!

[rare wharf]

or x(2-12)

[humble chasm]

that would break your equation because 2*-12 is 24.

-24*

you can’t factor x out of it

because -12 doesn’t have an x

[humble chasm]

you were on the right track here

but one small mistake

[rare wharf]

hm i dont know how else i would do it

[humble chasm]

$2x-12$

[rocky lotus]

9k

[humble chasm]

you’re looking for:

“how can i break this into something where AB = (2x-12)”

[humble chasm]

remember what i did here

i had $x^2-6x$

[rocky lotus]

9k

[humble chasm]

i broke it down to
$x(x-6)$

[rocky lotus]

9k

[humble chasm]

the reason i was able to take x out of that is because x^2 is just x times x.

-6x is just x times -6

[humble chasm]

so i’ve essentially turned this

into

$AB = x(x-6), A = x, B = (x-6)$

[rocky lotus]

9k

[humble chasm]

does that make sense?

[rare wharf]

ab = (2+x) -12?

[humble chasm]

no 🙁

[rare wharf]

sorry im a little lost on the logic here

[woeful oxide]

yall its been like

2 hours

[humble chasm]

you were on the right track here!

it would be $2(x-6)$

[rocky lotus]

9k

[humble chasm]

do you see how that is the same as $2x-12$

[rocky lotus]

9k

[humble chasm]

because if you were to multiply out 2(x-6), you’d end up with 2x-12

go do it on a piece of paper if you’re curious

[rare wharf]

im on question 4 so at my current pace il be done in 62 hours

[humble chasm]

do u see it :p

[rare wharf]

i am this method does seam faster then doing a quadratic formula

[humble chasm]

nice

so basically

[humble chasm]

we have this

[humble chasm]

and we have this

[humble chasm]

remember this ? @rare wharf

[rare wharf]

yes

[humble chasm]

$x(x-6) + 2(x-6)$

[rocky lotus]

9k

[humble chasm]

do you see what i did

[humble chasm]

when u remember what we did when we factored

$x^2 - 6x = x(x-6), 2x-12= 2(x-6)$

[rocky lotus]

9k

[rare wharf]

yea i see

[humble chasm]

yeah, i kinda just swapped the original with their factored versions

$x(x-6) + 2(x-6)$

[rocky lotus]

9k

[humble chasm]

so now all u gotta do is take this

take $x-6$ out of it

[rocky lotus]

9k

[humble chasm]

and that leaves u with $x+2$

[rocky lotus]

9k

[humble chasm]

see it?

[rare wharf]

ywa

yes

[humble chasm]

on a piece of paper

i want u to go multiply this out

$(x-6)(x+2)$

[rocky lotus]

9k

[humble chasm]

because in order to understand it, i want u to actually see that when u multiply that out, u will end up with your original equation, which was this

so go multiply those out and lmk when you’re done

so u can actually see it in action

cuz i feel like that’s the best way to grasp what ive just showed u

ping me when you’re finished multiplying it out

[rare wharf]

ok

[humble chasm]

if you’re having trouble just remember FOIL

firsts
outers
inners
lasts

[rare wharf]

is foil a different thing then pendmas

[humble chasm]

it’s just a trick u can remember when multiplying stuff like this

[rare wharf]

ok

[humble chasm]

these are called binomials

you’re multiplying two binomials together

[rare wharf]

i really apricate your help but i have to go, when i come back i will try it and see if i can get the same awsner

thanks for sticking with me 😭 @humble chasm

and @woeful oxide

[humble chasm]

you’re good man, good luck!

[rare wharf]

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• Show your work, and if possible, explain where you are stuck.
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[nova lance]

For question 19, I don't understand why the answer is A, x-7 should be a horizontal translation 7 units to the right.

[stark wedge]

your marked answer of B is correct

by the looks of it

[nova lance]

Oh

This is the textbooks justification

Am I missing something?

[stark wedge]

no they just fucked it up

most likely they meant for the question to say 9(x+7)^2

[nova lance]

Oh lmao

Thank you so much

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[vale dock]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[severe lodge]

I need help with finding Vo and Io

[severe lodge]

for Io i thought maybe i need to find the voltages after R1 and R2 and use that to find the current but that doesnt make sense

I have Kcl Kvl Voltage Divider and Current Divider that I can use

Kcl all current in = All current out
Kvl all voltage in = All voltage out
Votlage Divider Vo = Vin * (R2/ R1 + R2)
Voltage Divider VR1 = Vin * (R1/ R1 + R1)
Current Divider = IR1 = Iin (R2/ R1 + R1)

Use voltage divider with a voltage source feeding series resistors.

Use current divider with a current source feeding parallel resistors.

Don’t use either if there’s a bridge/extra connection between the divider nodes (that breaks pure series/parallel).

[jovial edge]

VR1=VR2=V(R1&R2) since R1 and R2 are parallel

V(R1&2)=I*(R1&2)

VS=V(R1&2)+Vo

[severe lodge]

Does this work @jovial edge ?

wait nvm

its wont work for series

[jovial edge]

Is Vt=V(R1&2)?

What's Vt

[severe lodge]

Vt should be Vs

i just called it Vt

[jovial edge]

Yeah that's wrong, Vs is the voltage R4 and R3 or R4+R3 since they're parallel

[severe lodge]

oh wait so i was kinda close right just wrong voltage

[jovial edge]

Yeah that Vt should the th voltage R1, R2 and R1&R2

Oh wait

[severe lodge]

its over R1+R2 isnt it

[jovial edge]

Okay, so should know how to combine two parallel res into one resistance

[severe lodge]

yes (1/R1 +1/R2)^-1

but if you do that doesnt it remove the connection that is I0?

[jovial edge]

Yeah so we combine R1 and R2 into one

[severe lodge]

k

[jovial edge]

We'll get to that later

[severe lodge]

k

[jovial edge]

Do the same for R3&4

[severe lodge]

k

[jovial edge]

Then you will have a series of 2 res

[severe lodge]

yep

[jovial edge]

Could you find the voltages of both res?

[severe lodge]

yea give me a second

R12 = 157.9

R34 = 35.0244

[jovial edge]

VR12 is also VR1 and VR2

VR34 is also VR3 and VR4 or Vo

Now you have all the voltages you can start to find the current

And using that node rule which I forget its name

To do the last part

Sorry have to be quick since Imma have a test in an hour

[severe lodge]

when you say this VR12 = VR1 + VR2 ?

[severe lodge]

all good thank you so much

[jovial edge]

VR12=VR1=VR2

[severe lodge]

oh

cause series

[jovial edge]

Cause they're parallel

R1 and R2 are parallel

[severe lodge]

thats wha i meant 💀

parallel same voltage

[jovial edge]

Bro you said series

[severe lodge]

misinput all good

[jovial edge]

Anyway you should be able to do the problem

Imma take a nap b4 the test

[severe lodge]

good luck

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[flat locust]

consider total distance D

ajay starts from A and vijay starts from B

in the first meeting They meet 500 m away from A

and time is same

so velocity of Ajay / velocity of vijay = 500 / D - 500

do the same for second meeting

you can equate the ratios because speed is constant and find out the distance b/w them

first do this

[unborn locust]

bet

yea yea now I got the distance

ty

i can do the rest on my own

[unborn locust]

thanks a lot

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[swift ermine]

yo

[swift ermine]

can someone help me

i got this question on my test

[hot tide]

split it into two sums

[swift ermine]

i did that

it didnt work

[hot tide]

why not

[swift ermine]

it gave me a different value

[hot tide]

send your work

[swift ermine]

aight

[hot tide]

$\sum_{r=1}^{15} 2^r + 12 \sum_{r=1}^{15} r$

[rocky lotus]

artemetra

[reef adder]

Lmfao

[hot tide]

what?

[dire cloud]

no thats right idk what shes laughing for

[hot tide]

xd

[swift ermine]

see

[hot tide]

huh

[swift ermine]

see

im confused asf

[dire cloud]

yeah no the question seems wrong

[reef adder]

it is

LOL

[swift ermine]

no it the exact question i got in the test

[hot tide]

you get that answer if you have -12r

[swift ermine]

i remember

[hot tide]

instead

[reef adder]

the statement is wrong

[steady charm]

,w sum 2^r + 12r from 1 to 15

[rocky lotus]

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[fossil mulch]

Maybe you didn't

[hot tide]

,w sum 2^r - 12r from 1 to 15

[rocky lotus]

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Requested by arte_m_etra

[hot tide]

yes exactly

it's supposed to be -

it's a typo

[swift ermine]

oh maybe i misread the question

idk

hmmmm

[hot tide]

possibly

[swift ermine]

is there some other way i could solve it

[hot tide]

no this is the right way

[swift ermine]

without splitting it

[reef adder]

no just split the sums

[swift ermine]

ok

[hot tide]

cuz again then you get 65534−1440

by splitting

[swift ermine]

i see

yeah so either i made a typo

or the question is wrong

idk

[reef adder]

you did

[swift ermine]

aight

thanks

.close

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[ripe sapphire]

I understand the idea of concavity is that the graph curves up (could "hold water") but I don't see how to pick the points for that on the graph. Like would it concave up between (-2,0) and (2,-1) or how does that work?

[pine prairie]

(2,1) is not on the graph

Did you mean (2, -1)

[ripe sapphire]

sorry, meant (2,-1)

Yes

mb

[pine prairie]

Yes

The way I think about it is like this

If someone asks you why at (-2, 0) the first derivative is negative

How would you explain that?

[ripe sapphire]

I'd be confused

Aren't derivatives taught in calc? I'm just starting precalc.

[pine prairie]

…

Wtf

Concavity is defined by the second derivative

😭

Here I’ll give you a visual intuition for it

[ripe sapphire]

Oops, my teacher didn't explain it beyond how it looks and holding water lol.

[pliant nova]

it asks what INTERVALS not points

[ripe sapphire]

So hypothetically i.e. [-2,2] ?

[pliant nova]

i think theyd be open so (-2, 2)

[pine prairie]

Consider the purple function

Look at the red dot and the red line

If you zoom in a lot to the red dot, the purple function looks like the red line

Would you agree?

[ripe sapphire]

Yes

[pliant nova]

dont really need to teach them derivatives right now they can just learn that properly in a calc class

[pine prairie]

We call the red line the tangent (straight) line at the red point

Now if we try to approximate our function instead of a linear straight line

Maybe we try, what parabola is the “closest” to the purple line at the red point

(Picture coming)

(Yes I moved the point for clarity)

At that point do you agree that this red parabola looks very similar to the purple function?

[ripe sapphire]

yep

[pine prairie]

Okay this doesn’t look like the purple function

[ripe sapphire]

it does not..

[pine prairie]

As you move around the red point the red function is so far away from the purple function!

[pine prairie]

This is better!

[ripe sapphire]

so is the red dot an inflection point (for the purple function)?

[pine prairie]

Consider the “best” parabolic approximation to the purple function at the red point

The concavity of this parabolic approximation is the concavity of the red point on the purple function

[pine prairie]

Inflection points are where both concave up and concave down parabolas approximate the function equally well

Let me give you an example

[silver birch]

if you want to know in broad terms without going into derivatives for concavity, it's concave up if the rate at which the function changes is going up, and it's concave down if the rate at which the function changes is going down

basically an "acceleration" of the function

[pine prairie]

Okay the red point got drawn over but you can guess where it is

[ripe sapphire]

so in the graph, it's concave up from [-0.5,4] and [6,inf]?

[pine prairie]

Here both the yellow and blue parabolic approximations sort look just as good as each other

That’s an inflection point

Kinda

(To be more specific, on the 2 sides of a point, the concavity of the best parabolic approximation needs to have different signs)

So for example, in f(x) = 0

At 0 (and every other point as well), both concave up and down parabolas approximate f(x) equally good

But it’s not an inflection point because the concavity doesn’t change signs (ie the best parabolic approximation doesn’t change concavity on either side of 0)

[silver birch]

actually from (-2, 4)

you don't want to know if the function is decreasing or increasing, moreso the rate at which the function is changing values

[pine prairie]

Tbh geometrically my explanation is as correct as the algebraic ones, slope and concavity are defined by the “best linear and quadratic approximations” which are precisely the straight and parabolic lines ive been drawing

[silver birch]

but the parabolic approximation is a good enough method for knowing if a function is concave up or down

[pine prairie]

So I’m not cheating the explanation, it’s a very well argued and correct interpretation of concavity

[silver birch]

it is indeed

[analog herald]

if f''(x) < 0 its concave downward; if f''(x) > 0 is concave upward

[pine prairie]

Yes but they haven’t learned calculus yet

[analog herald]

oops

if it concave up, the slope is increasing as you move left to right

(im not sure if its called monotonicity)

for concave down, the slope is decreasing as you move left to right

,w graph y = x^2

[rocky lotus]

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[analog herald]

this is concave up

[pine prairie]

Yes but then you need to know what the slope is

Which if you haven’t done calculus you probably don’t know either

[analog herald]

i didnt do calculus but like cuz i do geom so i know lol

like its a tangent line to slope

[pine prairie]

Yes but you’re not really helping the OP

[analog herald]

oops

my ba

*bad

[ripe sapphire]

I understand this definition in theory. In practice, does that mean that it includes the part of a graph from the left that initially slopes downward?

Like in the given graph from my assignment question, would we define [-2,2] as concave up because it forms the upward-facing parabola or just [-,05,2] since that's the range where the slope is actually increasing?

[pine prairie]

First one

Think of the parabolas

It is precisely what concavity is so you won’t be fooled

I’ve secretly swept the calculus theory under the rug of “best approximation”

But you can visually interpret and see what best approximation should look like so I don’t feel a need to expand on that

[ripe sapphire]

ah

so concave up at [-2,2], and would the second one be concave up [5,8] or [5,inf]

and concave down at [2,5] and the first one would be [inf,-2] or [-6,-2]?

[pine prairie]

(5, inf)

Don’t include the end points

The end points are inflection points, their concavity is 0

0 is neither positive nor negative

So a concavity of 0 is neither up nor down

And (-inf, -2) for the concave down

It might look crazy but it’s still slightly curving up/down when it goes high/low

So the best approximations are still on the same side as the intervals you’ve otherwise suggested

[ripe sapphire]

Got it

Thamk you very much

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[bleak grail]

i'm trying to understand the discrete fourier transform. when you take your samples and transform them, i understand that the numbers that pop out represent the amplitude and phase shift of a particular frequency, but what frequencies are they exactly?

if i have a vector x of samples, and then i get a vector X thats been transformed, what do the numbers in X represent

[flat locust]

claim another channel

[tame osprey]

Idk which one to go to

[flat locust]

#help-45

[tame osprey]

Okay thank you

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[midnight pier]

how do i approach this?

[molten trellis]

Put x=2

You will get E

Then differentiate

And put x=2 again

You will get D

[midnight pier]

can i do this without caculous pls

[molten trellis]

Or you can use binomial theorem

[midnight pier]

its pre calc question

havent done that either yet i forgot

[molten trellis]

Hmm

Without binomial and calc

Let's see

[distant galleon]

||Divide the left hand side by (x-2) to get E. Then, (LHS-E)/(x-2) leaves a remainder of D when divided by (x-2). etc.||

[molten trellis]

Idk man you gotta just expand it ig

[midnight pier]

ok ill try this

[midnight pier]

how do u know 2 is a root of the function

oh nvm its given

i get remainder 524/x-2

is that correct for E?

[distant galleon]

,w 2x^4 - 9x^3 +145 x^2 - 9x + 2, x=2

[rocky lotus]

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[distant galleon]

$E=524$, not $\frac{524}{x-2}$.

[rocky lotus]

Civil Service Pigeon

[midnight pier]

um

how so?

I did this

[distant galleon]

You divided both sides by $x-2$, remember?

[rocky lotus]

Civil Service Pigeon

[midnight pier]

oh yea

correct

i forgot

[distant galleon]

Can you do the rest yourself now?

[midnight pier]

ill try

[midnight pier]

i got D = 261

its not correct tho right?

[distant galleon]

!show

[vale dock]

Show your work, and if possible, explain where you are stuck.

[distant galleon]

Divide the left hand side by (x-2) to get E. Then, (LHS-E)/(x-2) leaves a remainder of D when divided by (x-2) etc.

you only found (LHS-E)/(x-2)

you never actually found the remainder (D) when it's divided by (x-2)

[midnight pier]

so I do it again?

[distant galleon]

$f(x)=2x^4-9x^3+145x^2-9x+2$. Then $$f(x)=A(x-2)^4+B(x-2)^3+C(x-2)^2+D(x-2)+E$$ $$f(x)-E=A(x-2)^4+B(x-2)^3+C(x-2)^2+D(x-2)$$ $$\frac{f(x)-E}{x-2}=A(x-2)^3+B(x-2)^2+C(x-2)+D$$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

actually

yeah just divide it by $\frac{f(x)-E}{x-2}$ again to find $D$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

cause if you use the remainder theorem, you still have to do the division again after anyway

[midnight pier]

alr ty

i Got 5/x-2

[distant galleon]

!show

[vale dock]

Show your work, and if possible, explain where you are stuck.

[distant galleon]

I'm confused

where did 5 come from

[midnight pier]

it needs +5 to become 261

[distant galleon]

-266 + 5 is -261, which is not 261

[midnight pier]

how so

[distant galleon]

edited

brain works faster than fingers

[midnight pier]

ohh i forgot the negative

yea ur right

yea its 527

which is correct

[distant galleon]

mhm

Imma go now, but as a check, your values of $A$, $B$, $C$, $D$, and $E$ should add to $1199$.

[rocky lotus]

Civil Service Pigeon

[midnight pier]

.close

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[urban raft]

Any tips on how I should simplify this? I think I need to change the second term so that there is a constant factor inside the square root.

[stark wedge]

well you can (and probably should) use $\sqrt{a^2} = |a|$, but can you also give some more context for this?

[rocky lotus]

Ann

[stark wedge]

i am suspicious of the way this is presented and would like to know if there's any other info or instructions here

[urban raft]

Well I am told to simplify it when x = pi / 4, but it is said that it needs to be simplified earlier and it should be simplfied to ()+()-()

[stark wedge]

ok right x=pi/4

this means x is in the first quadrant and so some of the headache is gone-ish

although

[stark wedge]

send me a screenshot please.

i would like to see it directly instead of retold.

[urban raft]

Well it's not in English

[stark wedge]

ok can you give me a literal translation of these bits

i dont speak finnish unfortunately

[urban raft]

Let's simplify the expression in a common form before making the substitution. First and third term are already in their simplified form, so we can only look into the second term. When the term is changed so that only the constant factor is inside the square root will the expression be simplified to ()+()-()

[stark wedge]

ok, so it says to keep the first and last terms as-is.

and it also tells you to keep the constant factor (the eight) under the root.

so the middle term needs to be rewritten

and because $x$ is in the first quadrant, $\sin(x)$ and $\cos(x)$ are both positive, and so you dont need the modulus bars when cancelling out a square inside a square root.

[rocky lotus]

Ann

[urban raft]

Yeah

[stark wedge]

so do it

[urban raft]

So then the middle term would be sqrt(8) * 2sin(x) * cos(x) * sin(x)?

[stark wedge]

indeed

[urban raft]

Oh wait it's correct xd it said that it doesn't allow sin or cos before, so I was a bit confused

I think I can finish this on my own hopefully

Thankies for the help!

.close

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[lament hazel]

i think i understand a

u can take out 3 from the second vector, so u get:

[ 1
2
3]
so the same vector, which would be a line

for b and c, im a bit confused though, mainly deciding if its a plane or if it takes up all of R^3

[drifting hornet]

for (b), there are only 2 vectors

[lament hazel]

yeah

[drifting hornet]

https://www.youtube.com/watch?v=k7RM-ot2NWY
Skip to 5:50

I dont think i can explain it better than this

[lament hazel]

ok thank you

[drifting hornet]

(in general, if you have n vectors, they span at most n-dimensional space, spanning exactly n-dimensional space iff they're linearly independent)

[lament hazel]

ok so b is a plane and c covers all of r^3

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[soft cape]

does anyone know what #latex-testing even is

[cinder creek]

A channel for you to test/write out your latex code.

[soft cape]

whats latex code?

srry im still in middle school but I love math

[keen tundra]

Code that allows to write mathematical symbols

[soft cape]

oh

but i thought you can do that with a chrome extension

[keen tundra]

More complicated things

And it looks nice

[soft cape]

ohh

is it like python?

[keen tundra]

No

Example:
$this is math \mathbb{R}$

[rocky lotus]

ExpertSqueeSQUEE

[soft cape]

wha-

[spring jackal]

https://www.latex-project.org/ for ur reference :)

[soft cape]

thx

[spring jackal]

$\Q = \left{ \frac ab : a,b, \in \mathbb{Z}, ; \gcd(a,b) = 1 \right}$

[rocky lotus]

haseeb

[spring jackal]

sorry bout that, but there's an example of how you'd write latex and what it would look like

[keen tundra]

I had no good idea on what to write

[spring jackal]

i always like the epsilon-delta definition of the limit

looks pretty

[keen tundra]

I would write some things I studied recently but I am on my phone

[feral sapphire]

It's a markup language for documentation that is inclusive of generating scientific and mathematical symbols. Here it s used for generating math images.

[soft cape]

this looks like hieroglyphics to me for some reason

even though i can understand it

[feral sapphire]

If you've ever opened any type of math or science textbook, they likely used LaTeX to create the text and images.

[soft cape]

oh nice

[vale dock]

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[stoic portal]

how would i go about starting this? i found the direction vector and thought about using the negative reciprocal to get a perpendicular direction vector but idk if that's fully right for this problem

[distant galleon]

yes that is relevant

[stoic portal]

ok slay

[distant galleon]

From that, you can also determine the fact that <a,b> and <-b,a> are perpendicular

[stoic portal]

what would that be useful for? just in general for future reference?

[distant galleon]

I mean

here for one?

[stoic portal]

true

the direction vector for the given line is <3,3>, so i would then use <-3,3>

[distant galleon]

mhm

which scales down to <-1,1> for conveninence

[stoic portal]

true true

now then i should find another point besides the given one right?

[distant galleon]

yes

[stoic portal]

i originally used the point <-5,3> in my previous attempts, does that work here still? (looking at when t=0)

[distant galleon]

read the question again

[stoic portal]

what am i missing besides (-2,-5)?

[stoic portal]

yeah i'm still not seeing things i feel lost

[distant galleon]

Yes that’s it

You need to use that point

Not (-5,3)

[stoic portal]

ok that's what was confusing me bc you said "yeah" when i asked if i should find another point besides the given point (-2,-5). sorry if that wasn't clear lol

ok i got it now tho

thanks!

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[torpid knot]

how to solve question 4 i came to 7.85cm idk if its right tho

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[errant gyro]

for future helpers: could you please translate the problem roughly into English?

[torpid knot]

yup

uno momento

[torpid knot]

From a rectangular piece of cardboard (length l = 60 , \text{cm}; width b = 40 , \text{cm}), a simple rectangular box is to be folded.
For this purpose, a square piece (side length x = 7 , \text{cm}) is cut out of each corner, and the side parts are folded up along the dashed line.

⸻

1.) Determine the dimensions a, b, c of the folded box.

2.) Calculate the volume of the box.

V = a \cdot b \cdot c

3.) Set up a formula for the volume V of the box so that the volume can be calculated for a given x.

V = (60 - 2x) \cdot (40 - 2x) \cdot x

4.) For different values of x, different-sized boxes are obtained.
Determine the box with the largest volume.

i need 4.)

[errant gyro]

you're not allowed to use calculus, i suppose?

[torpid knot]

we are

[errant gyro]

in that case, you can use the first derivative of V to find the turning point(s) of V. if one is obtained, that should be your max

if you get two, use the second derivative test to find which one the maximum is

this should be your attack plan, though would need checking from other helpers

[vale dock]

@torpid knot Has your question been resolved?

[vale dock]

@torpid knot Has your question been resolved?

[wintry oxide]

You don’t like the derivative answer ?

[errant gyro]

hi OP, if you are still around and would like a second opinion, you may ping helpers

[vale dock]

@torpid knot Has your question been resolved?

[vale dock]

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[toxic narwhal]

how to solve this in desmos? i dont even know where to start... i tried regression but didnt give me the correct answer

[toxic narwhal]

idk what to do

[vale dock]

@toxic narwhal Has your question been resolved?

[proper pike]

lemme take a look

Still here @toxic narwhal

[toxic narwhal]

yes

[proper pike]

I used to struggle with questions like these so I feel your pain lmao

But let me start with the fact I do not think you need desmos to solve this

Do you know the answer already

[toxic narwhal]

if i remember correctly i think its 33

[proper pike]

Yeah that’s what I got

Alright then

The question tells you this is a line

What forms of line equations do you know

[toxic narwhal]

linear quadratic and exponential

[proper pike]

If it specifies it’s a line, it’s linear

Quadratic and exponential isn’t a line as lines are, well, straight lmao

[toxic narwhal]

ah ok didnt catch that

[proper pike]

So you know this is a linear function now

What form should you probably use based on the question

[toxic narwhal]

point slope?

13+15=m(k-k+7)

[proper pike]

You’re definitely given two points, but this question specifically gives you the intercept

You should use intercept form as the last term is what you want to solve for

But no matter what form you use

You don’t know the slope

[toxic narwhal]

i found the slope

i think

its 4 no?

[proper pike]

What is it

Almost, double check your signs

[toxic narwhal]

-4

[proper pike]

That’s the slope. So if we use intercept form

Y=-4x+b

[toxic narwhal]

b is -5

[proper pike]

how did you get -5 ?

[toxic narwhal]

desmos

[proper pike]

B is supposed to be the answer to the question, and I think we both got 33

[toxic narwhal]

oh

[proper pike]

Remember b is the y intercept

[toxic narwhal]

yes

[proper pike]

Okay, let’s stick to point slope if you used that instead then