precalculus

[torn oriole]

how dare you have a question

jk go for it

[trim fable]

so like

whats the difference between those

[viscid thistle]

Wait

[torn oriole]

nothing

[viscid thistle]

Why does i and ii say the same thing as iv and v?

[torn oriole]

its just highlighting who's who

[trim fable]

no clue tbh

oh so its the same thing

iii and vi?

so intervals of increase would be

(0.2,infinity)

decrease (0.2, - infinity)

oops

i meant

the opposite

so

[patent beacon]

There is a difference.

vi says to state where the function is positive/negative

iii says to state where the function is sloped upward/downward

[trim fable]

isn;t that the same

[patent beacon]

Nop. A function can be negative but sloped upward

[trim fable]

ohhhh

true

[torn oriole]

ohh

i misinterpreted the q

D:

[patent beacon]

You thought there was a '

[acoustic laurel]

since this is a precalculus channel, would it be feasible to use derivative rules to figure out the slopes?

[patent beacon]

The functions aren't too hard to graph

[acoustic laurel]

oh nvm then haha

[trim fable]

ok so

[patent beacon]

Getting positive slope and negative slope is easy here without calc tools

[trim fable]

intervals of increase: (-infinity to 0.2]
decrease: [0.2 to infinity)
f(x)>0 (-infinity to +infinity)
f(x)<0 never

right? @patent beacon

[patent beacon]

Why 0.2?

[trim fable]

thats the y intercept

so increases till there then

decreases after that point

[patent beacon]

The y-intercept is when y = 0.2

[trim fable]

ye

oh..

it would be 0

oopsy

[patent beacon]

The values over which you're giving intervals of increase/decrease are x values

[trim fable]

😛

soo then 0!!

[patent beacon]

Also, the y-intercept just happens to be the point where the function touches the Y-axis. It doesn't have any bearing on increase/decrease

[trim fable]

ye

i totally forgot that hehe

[patent beacon]

You will want to find the domain, as that totally matters here

[trim fable]

ye

can we do b together too

so i can see if im doing it right

ill tell u my thoughts and u tell me if im right?

is thats alright

xintercept none

y= -0.1111111111

lol

domain is uh

x element of the real

right?

0<y<=0.1111 (range)

@patent beacon

wait im so confused. .

[patent beacon]

There are x values you can't plug in

[trim fable]

the graphs not the same

idk how to graph these

like ones goes like

[patent beacon]

They are quite different yeah

[trim fable]

why tho

idk how to graph them

without desmos

[patent beacon]

What happens if you plug x = 3?

[trim fable]

oh

asymptote at 3

coz it will be equal to 0

so like it would be at +3 and -3?

why -3 tho?

oh nvm 😛

makes sense

and the horizontal asymptote is just

[patent beacon]

Yeah, you divide by 0 if x = ±3, so that causes an asymptote

[trim fable]

the ratio of the coefficients

so then it would be at 1

right?

[patent beacon]

Only if the degrees are the same

[trim fable]

oh really?

what if they aren't

[patent beacon]

You have a constant over a quadratic here

[trim fable]

ye

[viscid thistle]

(Vertical asymptote to clarify for x= +-3)

[acoustic laurel]

if bottom is bigger, horizontal asymptote is zero

[trim fable]

ye

[patent beacon]

The bottom gets larger, while the top stays the same. The horizontal asym is 0

[trim fable]

huh

[acoustic laurel]

bigger on top, do synthetic division to get a slant aymptote

[trim fable]

oh

hmm

this stuff is honestly so different

and confusing lol

[patent beacon]

The asymptotes are most of the work. With them, graphing should become much easier

[trim fable]

ye

its weird how

1 graph

becomes 3

or its just

discontinuous

how would u right the domain?

x cant equal +- 3?

[patent beacon]

Yep, that's pretty much it

[trim fable]

ohh

and for y

it would be y cant equal to 1?

[patent beacon]

The range is all reals except 0

[trim fable]

why 0?

isn't it 1?

OH RIGHT

[patent beacon]

The horizontal asym is 0

[trim fable]

coz

[patent beacon]

Lel

[trim fable]

😛

coz its 1 and a quadratic

and they have to be the same

to do the dividing

[patent beacon]

"bigger power wins" is a way to think about it

[trim fable]

ok

how would u do

the intervals of increase for something like this

[patent beacon]

You'll want a sketch to help you out

[trim fable]

ye i sketched it

[patent beacon]

The graph "turns around" in between the two asymptotes

[trim fable]

ye

[patent beacon]

So at x = 0 it switches direction

[trim fable]

the reflected quadratic

looks like that

this graphs so weird

i dont get it LOL

but ill accept it

so the

interval of increase is

(- infinity to -3) U (3 to positive infinity) ?

oh no but the second is decreasing

so (-3,0)

then decreases (0,3)

@patent beacon

so my final answers for this are

interval of increase: (-infinity, 3) U (-3,0)
decrease (3,infinity) U (0, 3 )

then f(x)>0 (- infinity, -3) U (3, infinity)
f(x) <0 (-3,3)

right?

or no?

ohhhh...

umm help ;-;

@patent beacon 😅

now the inverse ;c

idk how to find the asymptote of a

@torn oriole help lol

[acoustic laurel]

lol im here

just trying to refresh stuff

[trim fable]

oh

[acoustic laurel]

for a

[torn swift]

find f inverse for all of these?

[trim fable]

its the same thing as #1 so

[acoustic laurel]

do you take the inverse then mark the asymps?

[torn swift]

one tactic, graph and then reflect everything against the line, y=x

[trim fable]

I have to do that whole thing for #2

the parts in #1

idk how to find the asymptotes

[torn swift]

but it's jut saying to negate f(x) and then redo your analysis?

[acoustic laurel]

it wouldn;t be the inverse then

[trim fable]

well oops

the graphs flips soo

[torn swift]

think about what a vertical asymptote is

what does it look like?

[trim fable]

ik what it is but the denominater of 1/x^2-2

wont be 0

or..

would it be

idk

[torn swift]

vertical asymptote is a vertical line

[trim fable]

ye

[torn swift]

vertical lines have what slope?

[trim fable]

undefined

[acoustic laurel]

get the denominator and set it equal to zero

since it is undefined, we'd have to do that

[torn swift]

and so if this function is to have a vertical asymptote, then we need to find where it is undefined

[trim fable]

so it would be a decimal..

[acoustic laurel]

and look for the x's where it would be undefined

[trim fable]

+- 2

square root of 2*

[acoustic laurel]

yeah

[trim fable]

oooof

ok

[torn swift]

ok, you killed my logical progression to the answer, but then follow his method and then you can see what I was building to

[trim fable]

uh

what

sorry

@torn swift

[acoustic laurel]

no, i think it was me who killed it

[trim fable]

oh

[acoustic laurel]

sorry for interfering

[trim fable]

hey guys thanks

both of u!!

i get it 😛

[torn swift]

was going to help you see why you need to set denominator to 0, though some people prefer to cut to the chase, which is fine, but it jumped the gun for me at least lol

[trim fable]

😛

[torn swift]

are there any horizontal asymptotes though?

[trim fable]

ye at 0

[acoustic laurel]

if the degree of the polynomial (highest exponent of x) in the denominator is greater than the one at the top, then horizontal asymptote would be zero

[trim fable]

ye

😛

thanks

hey question so

the interval of increase for this graph is

(0 to square root of 2) U (Square root of 2 to infinity)

right?

[acoustic laurel]

be sure to specify whether this is the positive or negative case of the square root of 2 lol

[trim fable]

ye

ok

[acoustic laurel]

it would be different for both

[trim fable]

but thats right

right?

[harsh cipher]

Hi guys,

sorry to interrupt

[acoustic laurel]

mhmm I think you'd need to say whether it is the positive or negative one, because the graph's behavior is different as the approach the positive or negative

[trim fable]

ye ik

[harsh cipher]

where did "a" come from?

[trim fable]

oh

a is just the coefficient value

so like

[harsh cipher]

why do we put it there in the first place?

[acoustic laurel]

oh wait lol u are right

[trim fable]

who me?

ye i recently learned that

[acoustic laurel]

if you meant the positive one

[harsh cipher]

I got to run to work. I'll be back later thanks!

[trim fable]

ye

[acoustic laurel]

haha i was thinking of both cases because it was in words

[trim fable]

ur welcome

well ye im good at math 😛

but like

I could be better

im not great

i used to be idk

😛

I love math

i just make such dumb mistakes on tests

and for this my teacher didn't teach it she made us do an investigation ourselves and went through it so quick yesterday which is why i was confused

but i get it now yay!!

thanks

i had to self teach myself lol

[acoustic laurel]

lol some people are taking a prereq course to mine, and I'd have to say that they usually know more than me in some instances despite me having already completed the prereq course lol

[trim fable]

huh

wait what grade are u in?

[acoustic laurel]

12

lol let's take this to #chill

[trim fable]

oh sameee

wait so

ur 16?

are u turning 17 soon

[acoustic laurel]

yea

[patent beacon]

Nah you guys are fine here, this is on topic

[acoustic laurel]

17 next year

[trim fable]

huh so then how are u in grade 12

oh wait

canada does grades by year like

[acoustic laurel]

I didn't take kindergarten.

[trim fable]

people born in the same year are in the same class

ohhhhh

cooool

[acoustic laurel]

im an immigrant to the us lmao

coming here was pretty weird

the math curriculum isn't so different though

[trim fable]

oh

[acoustic laurel]

progress in elementary school was pretty slow in my opinion

i used to live close to canada

somewhere in northern washington state

[trim fable]

ohh

[acoustic laurel]

I've been to vancouver a couple times now, but I haven't had much exposure to how the math curriculum is structured there

[trim fable]

ohh

[acoustic laurel]

precalc felt pretty weird to me since it essentially was like algebra ii

like a review of it

only thing new for us were some other trig identities

[trim fable]

oh

hey i have one more question

@acoustic laurel

[acoustic laurel]

yea sure

[trim fable]

this one

😛

[acoustic laurel]

lol we have been dealing with a special something

haha it starts with an a

[trim fable]

what do u mean

[acoustic laurel]

polynomials are continuous

hehe

[trim fable]

ohhhhhhhhh

OH RIGHT

and these aren't

[acoustic laurel]

and rational functions have those things that start with the letter a

[trim fable]

letter a?

asymptotes

LOL

[acoustic laurel]

asymptotes! lmao

[trim fable]

lololol

ohh

makes senseeee

omgggg

wowww

yeyey im done this 😛

I still have soo much left

lol

[acoustic laurel]

mathematical mathematics

[trim fable]

I have to do all this

LOL

is this stuff similar

or different

[acoustic laurel]

should be similar

[trim fable]

ye ok

good 😛

[acoustic laurel]

except you will have to factor some polynomials to cancel out holes

[trim fable]

ye

wait so

when u have a hole

u will have a circle

at that point?

[acoustic laurel]

to find removable discontinuities, which are fixable by filling in one point

[trim fable]

uh what

[acoustic laurel]

not a circle, the circle is just there to tell us where the hole is

[trim fable]

ye 😛

ik

lol

[acoustic laurel]

lmao

[trim fable]

thats what i meant

lol

to show where it is 😛

which means its discontinuous at that point

[acoustic laurel]

yup yup

[trim fable]

ok yay

well im gonna go finish this 😛

Thanks soo much

for the help

It means a lot!!

[acoustic laurel]

thou art welcom

[trim fable]

lol

[trim fable]

@acoustic laurel

sorry

I have 1 question

how would u find the x intercept?

nvm

😛

i got it

[acoustic laurel]

Sure lol

Oh crap msg wont send

Ayeaye

[trim fable]

uh

what would the domain be hehe

x cant equal to 2

[acoustic laurel]

@trim fable yeah

Sorry for the late reply

[trim fable]

its alright 😛

@acoustic laurel

is the v. asymptote for b 1.33 and horizonta is 0.33?

[serene heath]

How did u get those

[acoustic laurel]

For b its should be -3

For vert asympt

Horizontal is -2/1 which is -2

How did you get those?

[trim fable]

uh

oops

i did f by accident..

hehe

LOL

ooooops

coz i thought b was under a 😅

@acoustic laurel

[acoustic laurel]

Lol sorry for the delays

Im preparing for a school dance now

[trim fable]

ohh

niceeee

[acoustic laurel]

You are correct

[trim fable]

yay

question

for e

if u factor out the

[acoustic laurel]

Yeah

[trim fable]

• u can cancel

which would

make a hole

so then

no asymptote?

[acoustic laurel]

The graph can have a hole and an asymptote simultaneously

[trim fable]

oh

so for e

it will just have a hole?

[acoustic laurel]

It will also have a horz asymp and vert asymp

With the hole

Thanks for pointing this out. It has been so thought provoking lmao

I never would have factored out -1

[serene heath]

@trim fable -1.33

[acoustic laurel]

Oh shit

You’re right lemon

[trim fable]

oh @serene heath ye

i had -1.33 lol

for c

h.a=1.5

v.a=-0.5

right?

[acoustic laurel]

Yeah

[trim fable]

yeyeyey

for d

v.a =0.333

h.a=0.6666

@acoustic laurel

sorry i wanna make sure coz i just learned this all 😛

i still dont get

e

no holes?

[patent beacon]

Asymptote at x = 2

[trim fable]

wait so

I do e the same way as the rest?

so theres no hole?

coz if u factor the negative it can

cancel out

oh wait

u cant

then x will be negative

[slow roost]

So i got 0.164 but the key says -0.164

[slim lintel]

yeah it's negative

[slow roost]

what am I doing wrong?

[slim lintel]

subtract

take the log

[slow roost]

i did

[slim lintel]

multiply by negative 1

then divide by 12

[slow roost]

o.

can u not divide by -12?

[slim lintel]

you can

[slow roost]

i did that but it gave me a positive output

[slim lintel]

log(94) = 1.973

divided by -12

[slow roost]

idk if this is weird but i did 1/10^-12x=94

[slim lintel]

-0.16444

[slow roost]

then my log form was

log1/10*94=-12x

and then in my calculator i did

[slim lintel]

claire

idk if this is weird but i did 1/10^-12x=94

Oct 20, 2019, 12:00 AM

that's fine, but you need to get rid of the negative in -12x if you're going do it that way

[slow roost]

log1/10*94 all over -12

ohhh

i see

mb

[slim lintel]

no problem

[slow roost]

ty!

[brazen river]

I'm trying to read a paper and it has this notation of a function in it. Can someone clarify what it means? I don't know what this notation is called, so I can't do a web search.

[serene heath]

A function f(x,y)

Takes 2 inputs x and y and maps them into a real number in that interval

[glacial island]

i dont understand where they got the numbers to plug in and the limit??

[serene heath]

Wym

[acoustic laurel]

You plug in zero because the h approaches zero in the limit

If that’s what you mean

[glacial island]

is it always zero limit?

[acoustic laurel]

Not always

[glacial island]

and like plugging it in as in, where they found f(h +3) etc

[acoustic laurel]

The reason we get the limit towards zero is because we want to get the slope

For that, we treat h+3 as another term

We foil and combine like terms

And then we cancel out the common h’s

Then we plug in zero

We cannot plug in zero before cancellation because that would give us undefined due to dividing by zero

[glacial island]

why is it squared so that you foil it?

[acoustic laurel]

Yeah

Just foil it

[glacial island]

mm okay thank you

[summer sierra]

Suppose that a scientist has 100 mmilligrams of a radioactive substance that decays exponentially. After 35 hours 50 mg of the substance remains.

A. Based on the information above, write a function A(t) that models the amount of the radioactive substance remaining after t hours

[acoustic laurel]

you start with 100 mg

[summer sierra]

yea i kinda figured

[acoustic laurel]

A(t) = 100(0.5)^(1/35)t

[summer sierra]

thank you

[dull ginkgo]

I'm not sure if this is a valid counterexample for 53, but it says the number L.

If your L is infinity, that's not a number

@rugged gate

[hoary valley]

Can I classify this function as a trigonometric function?

[proud sparrow]

@rugged gate

I suspect $f(x)=x\sin\left(\frac{1}{x}\right)$ as $x\rightarrow 0$

[obsidian monolith]

Element118:

[proud sparrow]

it doesn't keep getting closer

it gets closer and further and closer and further

@hoary valley well, typically we don't

the only functions we can classify are sin, cos, tan, csc, sec, cot, and MAYBE their inverses

[acoustic laurel]

Why further?

[willow bear]

classification for the sake of classification is pointless

[proud sparrow]

@acoustic laurel well, graph it out and it oscillates around a lot

it really depends what you mean by "f(x) gets closer to L"

[native sequoia]

x^2 gets closer to -1 as x goes to 0

[proud sparrow]

ahhaha that too

[acoustic laurel]

My head hurts

[hoary valley]

It's a question in my book..

[willow bear]

what's your book's definition of "trigonometric function" then

[willow bear]

no answer...

[hoary valley]

@willow bear

[dull ginkgo]

Seems like a trig function then

[dull ginkgo]

Yeah it’s a trig function

[willow bear]

...

yeah no that isn't remotely what i was asking for but ok

[harsh cipher]

Hi

video explanation said he can foil this out by using pascals triangle

I've googled it didn't understand it. What is he doing there?

the value 1 3 3 1. how did it happen?

[willow bear]

aight so

do you know what Pascal's triangle is per se?

@harsh cipher

no answer...

[nova dew]

$\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$

[obsidian monolith]

CoolShot:

[nova dew]

when you don't know how to solve limits so you literally just put multiple values closer and closer to 5 and manually try to see what it's approaching

[willow bear]

no

[nova dew]

what would be a proper way to do it though

[willow bear]

that's not the way to do that

you can simplify the fraction

factor the numerator

[nova dew]

Oh wow I'm actually braindead

$= \lim_{x \to 5} x+5$

[obsidian monolith]

CoolShot:

[nova dew]

And that's = 10

Welp at least I got the answer

I'm literally back from an exam where I manually calculated for 4, 4.5, 4.9 and decided 10

wow ok thanks

[hoary valley]

What is this question? Can someone explain it to me.

[willow bear]

there is no way to explain it without just giving away the exact solution the book wants

[hoary valley]

Is it asking me "what's the absolute value function looks like" ?

[willow bear]

no

[hoary valley]

Tell me the answer Ann

[willow bear]

[flint stirrup]

@hoary valley

what does the given f look like

[hoary valley]

@flint stirrup Like a V shape

How to solve this problem ?

[hybrid charm]

by observation

[hoary valley]

@hybrid charm I don't get it.

I know how to composite..

[hybrid charm]

you see there sec(t)tan(t) and t²
you can choose f and g

[willow bear]

tbh

$v = v \circ \mathrm{id}$

[obsidian monolith]

Ann:

[willow bear]

[hoary valley]

Ok so f(t)=sec(t)tan(t) and g(t)= t^2 ?

What is " id" ?

[willow bear]

identity function

this is just to make a point

the point is that f and g aren't uniquely determined

[hoary valley]

How to write the square root using my keyboard ?

[hybrid charm]

√

[willow bear]

sqrt()

[hoary valley]

sqrt(4)

[obsidian monolith]

Ann:

[hoary valley]

It doesn't have a specific symbol for it? like for example we use ^ for exponents

[willow bear]

i mean you can write x^(1/2) for sqrt(x) if you want

but no it doesn't have its own plaintext symbol

[hoary valley]

Ah ok

[trail badge]

Smartphones have symbol √

But there isn't no symbol like that in pc

[hybrid charm]

smart pc have that symbol

[jaunty mason]

i found the equation but how to find the values of the roots?

[willow bear]

rational root theorem probably

[jaunty mason]

the what

[willow bear]

https://en.wikipedia.org/wiki/Rational_root_theorem

[dim jungle]

you can easily use vietas to bash

(a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ac) 🙂

[heady jewel]

ya

and you can find ab+bc+ca from (a+b+c)²-a²-b²-c²

[dense oar]

If you’re given a specific root like 4 and 3i, then you can use the sum and product method. If you’re trying to find a function of the lowest degree it would be cubic because 3i and -3i are a pair. (X-4) would be a root and then the sum and product method is (3i)(-3i) = 9 and the sum is -3i + 3i = 0. You can do x^2 - sum + product and multiply by (x-4) to find the function. For 3i and -3i, it would be x^2 - 0x + 9 (x^2 + 9) and then multiplied by (x-4) to find the function

[viscid thistle]

What's the difference between vertical stretch and horizontal stretch?

[acoustic laurel]

Vertical stretch happens when the entire function is multiplied by a number, while horizontal stretch would mean that the x value of the function got divided by a number

[viscid thistle]

Ooooh

[acoustic laurel]

try it with desmos

https://www.desmos.com/calculator/4zuhhzzvxq

@viscid thistle

[acoustic laurel]

stuff done on the horizontal is usually done on the x values, while stuff on vertical is with the entire function

[viscid thistle]

I can see why is it called a vertical stretch, but not why is it call an horizontal stretch...

[acoustic laurel]

let's take x^2 for instance

(6x)^2 is different from 6x^2

[patent beacon]

That example is iffy, because for √x, every horizontal stretch can be seen as a vertical compression instead

[acoustic laurel]

yeah

[patent beacon]

Cause √[4x] = 2√x

[acoustic laurel]

what I'm arguing is that it wouldn't be same scaling factor

but there is some degree of change to the horizontal realm as a vertical stretch is done

[patent beacon]

Yes. The horizonal compression by 4 is the same thing as a vertical stretch by 2

[patent beacon]

Rofl

[acoustic laurel]

both lmao

LOL

[viscid thistle]

;_;

Well... It makes sense...

But why does both function work...

[patent beacon]

Because √[4x] = 2√x

Remember that √[ab] = √a √b

[acoustic laurel]

$sqrt{4x} = 2sqrt{x}$ $2sqrt{x} = 2sqrt{x}$

LOL

[viscid thistle]

Can that happen with other numbers? Like ✓[25x] = 5✓x ???

[patent beacon]

Close, it's \sqrt

Yes, it can!

[viscid thistle]

:00000

[acoustic laurel]

Think of it as being part of one term

[viscid thistle]

Sorry guys, I didn't get to study algebra 2

[native timber]

rooting is implicitly exponentiation and exponentiation is distributive over multiplication

[patent beacon]

Why sorry? That's what we're here for

[acoustic laurel]

rooting is exponentiation to a fraction

[viscid thistle]

I think of it as, 5 is the square root of 25

That's why the have an equality (?

[native timber]

$\sqrt{25x} = \left(25x\right)^{\frac{1}{2}} = 25^{\frac{1}{2}} \cdot x^{\frac{1}{2}}$

[obsidian monolith]

Jiramide:

[viscid thistle]

[patent beacon]

$\sqrt{25x} = \sqrt{25}\sqrt{x} = 5\sqrt{x}$

[obsidian monolith]

Kaynex:

[viscid thistle]

Oh I see it!

[patent beacon]

And, by definition, $\sqrt{x} = x^{1/2}$

[obsidian monolith]

Kaynex:

[viscid thistle]

That's more difficult...

[patent beacon]

It's the same thing! Just a difference in notation

[jaunty mason]

How to do questions like these, where you have to construct an equation for polynomials with modified roots

[viscid thistle]

I haven't see fractions exponents in my life

[native timber]

you can generalize it to $\sqrt[n]{x} = x^\frac{1}{n}$

[obsidian monolith]

Jiramide:

[native timber]

if you want to see why it's defined as such, take $x^\frac{1}{2} \cdot x^\frac{1}{2}$ and determine what it's equal to using power laws

[obsidian monolith]

Jiramide:

[patent beacon]

That is what a fractional exponent is

[viscid thistle]

That up the would add to x^1

[native timber]

right, and that's just x

[viscid thistle]

Right

[native timber]

what other number has the property where if it's multiplied to itself gives back x?

[viscid thistle]

... Square root

[patent beacon]

@jaunty mason
Vieta's formulas come to mind

[native timber]

hence you arrive at the conclusion that $\left(x^\frac{1}{2}\right)^2 = \sqrt{x}^2 \rightarrow x^\frac{1}{2} = \sqrt{x}$

[obsidian monolith]

Jiramide:

[native timber]

wrong arrow

what's implication arrow

[jaunty mason]

what is a vieta

oh that

yeah ik that, but how do i relate that here?

sum alpha = -b/a

[viscid thistle]

Oh my my... I know exponents and square roots "cancel" each other in cross operations but...

[jaunty mason]

but how do i find sum alpha for the "new" roots

[acoustic laurel]

Also, to solidify: the reason why an exponent of zero on a number leads to one is because: $x^\frac{1}{2} \cdot x^\frac{-1}{2} = \frac{x^\frac{1}{2}}{x^\frac{1}{2}} = 1$

[jaunty mason]

@patent beacon

[viscid thistle]

Where did the square root sign came from

[native timber]

ok im kinda tired usin latex ill just type it out

[obsidian monolith]

Chairman Rovic:

[native timber]

you yourself said that (x^(1/2))^2 = \sqrt{x}^2 when i asked you "what other number has t he property where if it's multiplied to itself gives back x"

[viscid thistle]

Yeah, a square root...

[native timber]

right so we have the equation (x^(1/2))^2 = \sqrt{x}^2, we can then get rid of the square by square rooting both sides giving us x^(1/2) = \sqrt{x}

[viscid thistle]

Yeah... I guess, it's hard to picture but I think I can agree with that

[jaunty mason]

@patent beacon Any clue?

he just went offline, yeet

This got kind of buried, can anyone help me with my question: https://discordapp.com/channels/268882317391429632/363224154469826562/635574486703538216 ?

[viscid thistle]

The best thing is to go to another channel an wait for help there

[acoustic laurel]

@jaunty mason hey

we could prolly do this

first, get the factors of the last term

then do synthetic division with plus/minus each factor until you get a remainder of zero

then get the opposite of the factor and add or subtract that to x

[jaunty mason]

factors of -10 ?

[acoustic laurel]

then get the polynomial from synthetic division

yeah

[jaunty mason]

I'm kind of new to this

is synthetic division same as long division or are they separate things?

[dense oar]

They are separate

[jaunty mason]

fuck

[dense oar]

Synthetic division is much easier if you have a factor that’s like (x-5) but if you have something like (6x+7) it’s probably harder

[rigid sun]

Let’s do

[acoustic laurel]

do long division for that instead

[rigid sun]

Long division

[acoustic laurel]

why long division

not for this one, we're trying to get roots

and it would be simpler with synth div anyway

[jaunty mason]

patrickJMT's vid says synthetic division is a "shortcut" for long

[acoustic laurel]

yeah, it is

but synth div has its limitation

like the variable having a coefficient

mhmm wait nvm it could be done

we could do long division, but you will have to deal with fractions lol

[jaunty mason]

then do synthetic division with plus/minus each factor until you get a remainder of zero
then get the opposite of the factor and add or subtract that to x

[acoustic laurel]

but just do synth division

[jaunty mason]

Can you elaborate a bit on this?

[acoustic laurel]

for this one, if you're doing for example 5 as the divisor

and you got zero for remain

[native timber]

dividing seems unnecessary if you can just use factor theorem

[acoustic laurel]

make sure it is written as (x-5) for the term

[jaunty mason]

what are we even dividing ?

[acoustic laurel]

the factors of -10 to get a remainder of zero

[jaunty mason]

ah

[acoustic laurel]

just so we decompose the polynomial

and get zeroes

the polynomial we get from synth div can be further factored too

especially if it is a quadratic

which would be in this case

Once you do synth division, you would get a binomial and a trinomial (which can be further factored), that would give you: three binomials that have roots at either alpha beta and gamma

[rigid sun]

How is that going to help

we already knew that the roots were alpha beta and gamma

[acoustic laurel]

omg dude im just giving a guide

just so he or she gets the actual constants alpha beta and gamma

[rigid sun]

Or just

Do less force*distance

divide the whole thing synthetically and you’ll prove it

[acoustic laurel]

yes

that's what I just said

you will have to factor the other trinomial from the synth div though to get two more roots

then proving u + 2*alpha = 1 would be plug and play

and once you get the three constants alpha beta and gamma: just add them up in some fashion according to what the problem says for each binomial for them to have the root

[rigid sun]

a+b+y=1

[jaunty mason]

i'm still stuck at the division part 😆

[acoustic laurel]

what difficulties are you experiencing?

[jaunty mason]

yes, so i have the factors of -10, how do i divide them

what is the numerator, what's the denominator

[acoustic laurel]

https://www.youtube.com/watch?v=1byR9UEQJN0

lol i feel lazy but this deserves some instruction with some actual audio for retention

[jaunty mason]

no, i understand that, i simply don't know what expressions i'm supposed to be dividing together

[acoustic laurel]

mhmm

you get a factor of -10

[jaunty mason]

a single pair?

[acoustic laurel]

and then you get the coefficients of each term of the polynomial

not a pair

[jaunty mason]

arbitrarily pick a factor of -10?

[acoustic laurel]

yeah

you need something that has zero as a remainder

because that you allow you to break the polynomial to a binomial times a trinomial

[jaunty mason]

so trial-and-error with all factors?

[acoustic laurel]

no remainders mean that they break up to that

yeah

not all

[jaunty mason]

not 1 and -10 ofc

[acoustic laurel]

since you might eventually get lucky or get it on first try

[jaunty mason]

so, first i should try diving the original polynomial with 5?

[acoustic laurel]

yeah, try

[jaunty mason]

i'm getting -95, is that correct?

i know it's not zero but did i do the division correctly?

[rigid sun]

Just divide using alpha beta and gamma

[acoustic laurel]

mhmm but why

[jaunty mason]

divide what with alpha beta gamma

[acoustic laurel]

that would leave us with a symbolic answer

[rigid sun]

F(x)

[jaunty mason]

the original polynmial with the product of alpha beta gamma?

[rigid sun]

yes

x^3-x^2

Then some other garbage

I forgot

[jaunty mason]

synthetic?

[rigid sun]

Mhm

[acoustic laurel]

we will need the roots to write the actual cubic function with the roots anyway

just continue on with what you're doing

im afraid i don't have pen and paper

[rigid sun]

Wdym

I think he has to do both

[jaunty mason]

1 , -9, -93, -940

something tells me that's not right

[acoustic laurel]

how are you doing it?

https://cdn.discordapp.com/attachments/363224154469826562/635574483716931594/unknown.png

[limber bone]

u+2a is the sum of roots ig

[jaunty mason]

.
10 | 1 -1 -3 -10
10 90 870
1 9 87 -860

[acoustic laurel]

yeah

[jaunty mason]

i did that but

doing it wrong?

[acoustic laurel]

-1 and 10 is 9

[jaunty mason]

yeah

so 860 for the remainder?

[acoustic laurel]

make sure to do both positive and negative cases

of each factor

[jaunty mason]

so i got 860, what next @rigid sun ?

[rigid sun]

Woah

Tf

Ur remainder should always be 0 if it’s a factor of the polynomial

[jaunty mason]

[acoustic laurel]

put that aside, 10 doesn't work. do another factor

[rigid sun]

^ it means 10 is the wrong guess

[jaunty mason]

let's say i got 0, then what?

the resulting polynomial is my answer?

[acoustic laurel]

then you get the thing you divided with, put it in (x-(stuff you got)) form times the (trinomial you just got)

[trim fable]

heyy

[rigid sun]

In that case if u got 0, it would mean (x-10) works

[trim fable]

will u help me in the

probability channel

😛

[jaunty mason]

wait a second, where did we modify the roots?

we're supposed to find the eq for (- alpha), beta and gamma

[acoustic laurel]

we will get three roots eventually

[jaunty mason]

also

define stuff you got

is it the trinomial i just got or the factor?

[acoustic laurel]

both

them multiplied to each other

[jaunty mason]

wot

[acoustic laurel]

that's why you are synth dividing

[jaunty mason]

put it in (x-(stuff you got))

[acoustic laurel]

to break them apart

[jaunty mason]

let's say the factor was 5

[acoustic laurel]

x - (the divisor you had a zero remainder with)

as one factor of the polynomial

[jaunty mason]

so it'd be (x - 5)(resulting polynomial) ?

[acoustic laurel]

yeah

[jaunty mason]

gotcha

but we just divided the original eq with a factor of the last term

how does this help us find the eq for (- alpha), beta and gamma?

[acoustic laurel]

because the last term is the goi we need to satisfy to have a zero remainder

[trim fable]

@acoustic laurel after u help taha will u help me in probability

[acoustic laurel]

a zero remainder means the number we are dividing with is a factor of the polynomial

@trim fable i could review probability, but im afraid im not too good haha

we'll see

[trim fable]

ok lol

[jaunty mason]

ok sure, how is the factor significant?

[acoustic laurel]

we focus on last term because it gives us problems with remainders

[jaunty mason]

uh-huh

what i mean to say is

[acoustic laurel]

if it doesn't end up zero, then we know the term doesn't go all the way in

[jaunty mason]

we got (x - 5)(resulting polynomial)

now what?

[acoustic laurel]

break up the resulting polynomial

[jaunty mason]

factorise it?

[acoustic laurel]

it would be a quadratic

yeah

then after that you will get three binomials multiplied to each other

[jaunty mason]

(x - 5)(x + a)(x + b)

now what

[acoustic laurel]

which will give you roots

[jaunty mason]

roots of what?

how do we know which is alpha, beta and gamma?

[acoustic laurel]

roots are when the function equals zero

we will get there

[jaunty mason]

right, so (x - 5)(x + a)(x + b)

[acoustic laurel]

u + 2 alpha is just alpha + beta + gamma if you substitute -alph + beta + gam from u

https://cdn.discordapp.com/attachments/363224154469826562/635574483716931594/unknown.png

we don't need to get alpha beta and gamma sepcifically for this problem

[jaunty mason]

so what do we do?

[acoustic laurel]

you plug in the roots to alpha + beta + gam = 1 to prove that u + 2 * alpha is indeed one

[jaunty mason]

excuse me wtf

i already proved u + 2 alpha = 1

was all this to prove u + 2 alpha = 1?

[acoustic laurel]

How?

[jaunty mason]

REEEEEE

[acoustic laurel]

no

[jaunty mason]

oh

[acoustic laurel]

you could use the roots for the cubic equation

[jaunty mason]

• b / a = sum alpha

that

[acoustic laurel]

what?

[jaunty mason]

-b / a = alpha + beta + gamma; for any cubic polynomial

[acoustic laurel]

but how would you know which one is the b and a value from the way you have factored it?

[jaunty mason]

you dont understand

b and a from ax^3 + bx^2 + cx + d

here - b / a = - (-1) / 1 = 1

[acoustic laurel]

alright, then no need for roots then. but use the roots for the cubic equation

[jaunty mason]

yes

[acoustic laurel]

I never learnt that

and also I was thinking of the (x+a) and (x+b) you mentioned earlier

with the (x-5) factor

[jaunty mason]

yeah sry

so um can you help me with finding the eq with the roots (-alpha), beta, gamma?

[acoustic laurel]

yeah sure but lol this has come a long way

[jaunty mason]

haha yeah

i should've been more clear i had already done the first part

[acoustic laurel]

for that, you could assign the label alpha beta and gamma to each of the roots

https://cdn.discordapp.com/attachments/363224154469826562/635574483716931594/unknown.png

[jaunty mason]

and expand?

(x - alpha)(x - beta)(x - gamma) = 0, right?

[acoustic laurel]

then add them as you see fit and make three binomials that would look like this (x-(some way alpha beta and gamma are added)) * (x-(another way alpha beta and gamma are added))

times (x-(another way alpha beta and gamma are added))

because it says there are three roots

one where it's neg alph plus bet plus gam

another where alpha minus beta plus gamma

and another where alpha plus beta minus gamma

then foil 'em

and set them equal to zero because they're roots

[viscid thistle]

Guys guys, am beyond confused

[acoustic laurel]

ey

[viscid thistle]

So to vertically stretch you multiply the function, and to horizontal stretch it you divided by a any number right???

[acoustic laurel]

that would be vertical compression hehe

[viscid thistle]

What's the difference

[acoustic laurel]

anything done with the independent variable has something to do with the horizontal realm

but like we explained

a vertical stretch of some value is a horizontal compression of some value

[viscid thistle]

[acoustic laurel]

like for two times square root of x versus square root of the product 4 times x

$2\sqrt{x} = \sqrt{4x}$

[obsidian monolith]

Chairman Rovic:

[viscid thistle]

So when stretch the function graph shirinks

[acoustic laurel]

wdym shrinks

[viscid thistle]

Like closes

[acoustic laurel]

I guess it goes inward

yeah

[viscid thistle]

Yeah yeah

And compress woul be outward

[acoustic laurel]

because it gets higher values for every value of x now if it is a vertical stretch

[viscid thistle]

Would*

[acoustic laurel]

vertical compress would be outward

be clear on whether you mean vertical or horizontal

[viscid thistle]

;_; okay okay

Let me draw it...

[acoustic laurel]

because a horizontal stretch would make graph thicc

like make it wider

vertical stretch would make the graph thinner

you can see the difference between big chungus and small chungus

[viscid thistle]

Okay, used this :)

Use*

[acoustic laurel]

for x^2

[viscid thistle]

Green one is vertical compress?

[acoustic laurel]

that looks more like horizontal stretch

[viscid thistle]

This is frustrating :'(

[acoustic laurel]

keep in mind that the function vertically compressed does not equal the function horizontally stretched

[viscid thistle]

That makes sense, vertically and horizontally are different stuff

[rigid sun]

It can be the same

But don’t assume it is

[acoustic laurel]

yup, in some cases

[viscid thistle]

But the word compress and stretch confuse me

[acoustic laurel]

sorry, if I implied that they can never be the same

they can be

[viscid thistle]

:(

Is there no "rule" I can just learn?

Like

Vertical stretches look this way

Vertical compression look this way

Horizontal stretch look this way

Horizontal compression looks this way

And then, what the frickiry frick is a compression and what's a stretch

And the difference between them

[acoustic laurel]

They do have to look a certain way, but since we do know that a vertical stretch is a horizontal compression to some degree: there is some ambiguity as to which one we should call them.

[viscid thistle]

Awesome.

[acoustic laurel]

we did go over the: you do stuff with the entire function vs only the independent variable

[viscid thistle]

Okay okay

Continue

When we do something just with the x ...

[acoustic laurel]

then it's a horizontal stretch or compression

[viscid thistle]

Am copying that

[rigid sun]

here’s a clue

Vertical manipulation doesn’t affect x intercepts

Horizontal manipulation doesn’t affect y intercept s

[viscid thistle]

What's a x intercept?

[rigid sun]

Where the graph intersects the x axis

[viscid thistle]

Uhmmmm

[acoustic laurel]

and y is zero

[rigid sun]

Y axis is the same thing, except x=0

[viscid thistle]

can you send an example

Am bad at visualizing math

--- when we affect the whole function is a vertical stretch?

[acoustic laurel]

or compression

[viscid thistle]

Are stretch and compression synonyms?

[acoustic laurel]

no no no

lol

[viscid thistle]

Oh my days, so what's the difference?

Stretch inward compression outward?

[acoustic laurel]

if vertical

[viscid thistle]

Uhmmm

[acoustic laurel]

I challenge thee to never say compression or stretch without saying either horizontal or vertical

that should clear things up

[viscid thistle]

Okay okay

Test me

[acoustic laurel]

I have f(x) =69x^2 -420

then I have g(x) = 1337(69x^2-420)

what happened from f to g

[viscid thistle]

F is multiplied by 1337

[acoustic laurel]

lol describe it with the terms we have been using

[viscid thistle]

oopsi

Uhmmm

Compression

[acoustic laurel]

it will be : vertical stretch. Dividing 1337 would be vertical compression

[viscid thistle]

I thought you didn't want me to say vertical.or horizontal...

[acoustic laurel]

never say compression or stretch without saying vertical or horizontal

[viscid thistle]

Ooh okay okay

Another one

[acoustic laurel]

LOL

ok

I have f(x) =x^2+1

and g(x) = 2749823742837928472934724(x^2+1)

[viscid thistle]

That's difficult :V

[acoustic laurel]

not really lol

[viscid thistle]

Okay

So

G is a vertical stretch because f is being multiplied

[acoustic laurel]

yup

multiplying the whole function is a vertical stretch. dividing the whole function would be a vertical compression. multiplying the variable is horizontal compression. dividing the variable would be a horizontal stretch.

[viscid thistle]

Give me horizontal questions

[acoustic laurel]

aye aye

f(x) = sqrt(x/472398793723) +69*420