help-13

[cedar kiln]

@heavy crypt Has your question been resolved?

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[livid hound]

what have you tried?

[wraith dagger]

Cringelord

Cringelord

Cringelord

[livid hound]

try combining them into a single fraction

the term on the right can also be simplified

[wraith dagger]

Cringelord

[austere merlin]

put on the same denominator and factor cos

multiply cos by cos^2

to get cos^3

[wraith dagger]

Cringelord

[austere merlin]

yes

now factor a cos(x), you should get a familiar identity

[wraith dagger]

Cringelord

[austere merlin]

which simplifies to $\frac{cos(x)(cos^2(x)+sin^2(x))}{cos^2(x)}$

[wraith dagger]

Aram

[austere merlin]

do u know what cos^2 +sin^2 is

yes

so you can simplify that

yes

you factor cos(x) like you would factor x^2+x = x(x+1)

[cedar kiln]

@opaque mantle Has your question been resolved?

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[eternal hedge]

Hello guys! How can I show that the following limit exists:

[wraith dagger]

HeyMilkshake

[eternal hedge]

by using the formal Limit definition:

[wraith dagger]

HeyMilkshake

[eternal hedge]

By evaluating the limit across some directions, anmely y=x, x=y, y=mx, x=y^2 and y=x^2 I have the candidate L=2

so I plugged everything in the definition and ended up with:

[wraith dagger]

HeyMilkshake

[eternal hedge]

now, How can I construct inequalities such that I can prove that:

[wraith dagger]

HeyMilkshake

[austere merlin]

you might want to go into polar coordinates

[eternal hedge]

I have to use the definition for this one.

Not an option to use any other path

[hasty pelican]

I would look into rationalization here

[eternal hedge]

so, maybe multiply the left side by the conjugate to simplify?

maybe that leads to somethign easier to prove?

How didn't I remember that. lol

[hasty pelican]

$$\frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1}$$
$$ = \frac{(x^2 + y^2)(\sqrt{x^2 + y^2 + 1} + 1)}{x^2 + y^2 + 1 - 1}$$
$$= \sqrt{x^2 + y^2 + 1} + 1$$

[eternal hedge]

I saw it

genious!!

let me pick from here

😄

[wraith dagger]

King Dedede

[hasty pelican]

I'm making a lot of mistakes

But you get the point

[eternal hedge]

but I get it

[hasty pelican]

😄

[eternal hedge]

yea

**** latex

I feel like an idiot for not seeing this... 😦

🤣

[hasty pelican]

Lol it happens

Rationalizing just refreshes in my brain since I frequently use it

[eternal hedge]

Happy to provide you with a refresher. I think that I can handle it form here

many thanks!!!

Here's a 🍺 as appreciation

[hasty pelican]

No problem!

[hasty pelican]

I'm still in school 😂

I will have to wait a few years

[eternal hedge]

what grade r u in?

Im in my 1st university year

[hasty pelican]

11th

[eternal hedge]

keep it up, cuz your maths is top notch 😄

.close

[cedar kiln]

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[hasty pelican]

Thanks

[cedar kiln]

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[upper basalt]

Can I get some help?

[fallen solar]

just substitute i guess

[upper basalt]

wdym?

I am confused

[fallen solar]

hint: the starting value of f(x) (in this context) occurs when x = 0

[upper basalt]

still dont know

Im lost

[fallen solar]

do you know what (0, 2000) represents?

if the graph of a function f(x) passes through a point (a, b), then b = f(a)

[upper basalt]

No

[fallen solar]

damn

[upper basalt]

I got the line though

f(x) = 2000 * exp( (-ln100)*x/2)

inst that right?

[fallen solar]

wha-

why'd you write it like that though

and how did you even find that

[violet rapids]

well k is right

[upper basalt]

oh shit wrong

nvm

I did it wrong

[violet rapids]

so you should have $f(x) = 2000b^x$

[wraith dagger]

sills

[upper basalt]

but idk the base

[violet rapids]

you have another ordered pair

(2, 20)

[upper basalt]

20?

2?

what is the base

[fallen solar]

no it's right

but

you have written it in the wrong form

idk why did you even write it in that form

[upper basalt]

but k still is 2000

[upper basalt]

thats the way I was taught

[fallen solar]

do you know what it means for a function to pass through a point?

[fallen solar]

also can you explain how you got this

[upper basalt]

wait

f(x) = C * exp(kX)

[violet rapids]

post your work ^^

[upper basalt]

f(0) = 2000 ---> C = 2000

thats what I am doing

So now f(x) = 2000*exp(kx)

f(2) = 20

20 = 2000*exp(2k)

1/100 = exp(2k)

ln(1/100) = 2k

ln1 - ln 100 = 2k

k = (ln 1 - ln 100)/2

= (-ln 100)/2

f(x) = 2000 * exp( (-ln100)*x/2)

@violet rapids

@fallen solar

I need help finding the base

[fallen solar]

now just do the same thing, except instead of using exp, just consider an arbitrary base b

[upper basalt]

still need help though

idk the ase

[fallen solar]

exp() function is just another way of representing e^x
and instead of that, do b^x for some base b

[fallen solar]

you have to find that...

[upper basalt]

and I need help

you have any idea how hard it was just to find the first answer?

[fallen solar]

uhhhhhhh

wat

[upper basalt]

exactly, uhhhh

dude

[fallen solar]

do you understand why f(0) = 2000 and f(2) = 20?

[upper basalt]

20 is the base?

[violet rapids]

no

[upper basalt]

not really

[fallen solar]

how did you use that in your solution then

[fallen solar]

.

[upper basalt]

20 = kb^2
20 = 2000b^2

@fallen solar

b=0.1?

[fallen solar]

yes

[upper basalt]

a better solution is 2000(1/10)^x

really

b is 0.1?

[violet rapids]

bro stop guessing

and try to think

[upper basalt]

Im not guessing

20 = 2000b^2 is 0.1, -0.1

@violet rapids

1/10

[cedar kiln]

@upper basalt Has your question been resolved?

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[cedar kiln]

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[grizzled rivet]

<@&286206848099549185>

u should help lol

[south dove]

I can help, but I thought this was a math discord?

[cedar kiln]

@grizzled rivet Has your question been resolved?

[grizzled rivet]

whoops

i forgto about tht ol

[eternal comet]

I've seen physics problems here, engineering problems

But not biology before

damn

Also, it has to do with the definition of species

not even that. Thought it was one of those questions

[south dove]

the question asks the utility of genetics for classification, tricky

[eternal comet]

I mean here it is a reductio ad absurdum

[south dove]

I could only eliminate the 1st answer really, the rest could be correct

I'd be mad if I got a question like this

[grizzled rivet]

@south dove

i need help with this one

now

[south dove]

not sure

[shy oriole]

<@&268886789983436800>

[grizzled rivet]

bruh

[trim sentinel]

This is not math

[grizzled rivet]

why u have to be a snitch

..

[south dove]

mathematically if you count there are 5 mutations

[grizzled rivet]

exactly

and u have to COUNT the differences

so it has math

add me in case i get banned lol

[south dove]

that should help, but the rest is theory, can't really help with that

[gleaming pasture]

@grizzled rivet Please go to any number of biology or general homework help servers instead of asking it here

You're more likely to receive help and you wont be cluttering up the help channels that way

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[lunar cave]

A point on the terminal side of Angle0 has the coordinates..... Where is angle0? is it center or is that the coordinate?

[ocean holly]

Can u guys help me

🙏🏼😭

[violet rapids]

#❓how-to-get-help

[violet rapids]

could you post the full problem?

[lunar cave]

yes I can

I just want to verify that I am answering the questions based off the coordinate given as Angle0

[cedar kiln]

@lunar cave Has your question been resolved?

[cedar kiln]

@lunar cave Has your question been resolved?

[lunar cave]

I figured it out

[cedar kiln]

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[frosty oxide]

How did that happen?

Doesn't arcsin(x) have a domain restriction?

[modern compass]

yeah, i think they meant the limit to go to 0

someone probably copy/pasted and forgot to fix it

[deft forum]

Yea

[frosty oxide]

alright

I've found a few mistakes in this book before

That makes sense

thanks a lot!

.close

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[cedar kiln]

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[warm vector]

I need help with 26 e).( it’s not a test, it’s a resource from Nelson.net)

[rigid ridge]

if you don't mind me asking, what have you yellowed out here

[warm vector]

A bunch of numbers for confusion

For the question

So any idea of how I would do the last part e, I’ve done the rest

?

[cedar kiln]

@warm vector Has your question been resolved?

[warm vector]

<@&286206848099549185>

Anyone?

.close

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[plucky dirge]

Confused

[bold vine]

Basically:

1. Plug in x = -2 into h(x) = ax^2 + 1

2. Replace h(x) = h(-2) with 21.

You can then solve algebraically for a

[glad kestrel]

why'd you replace x with 2 and slap a negative in front of a

[bold vine]

Remember:

Example: If t = -4: t^2 = (-4)^2 = (-4)(-4) = 16 ≠ -t^2 = -(-4)^2 = -(-4)(-4) = -16

[cedar kiln]

@plucky dirge Has your question been resolved?

[cedar kiln]

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[gentle ore]

how do u do this

[gentle ore]

<@&286206848099549185>

[upper abyss]

What's the slope of x + 2y = 0?

[gentle ore]

1/-2

[upper abyss]

Very nice. Now for the tougher question, when does that curve have slope -1/2?

That is, for what values of x does dy/dx = -1/2

[gentle ore]

uhhhhh

im not sure i understand

[upper abyss]

I'll back up a bit. Can you calculate dy/dx of that curve?

[gentle ore]

ye gimme one sec

ok so dy/dx = -1/2y

[upper abyss]

Very nice! So note that you care about when dy/dx = -1/2

As that would be the same slope as the line

So you care about -1/2 = -1/2y

[gentle ore]

ahh ok i got to this step when i was going over the question and then didnt know what to do after

cuz i isolated for y and got y = -1/4 but i have no clue what to do w the value

[upper abyss]

So I thought you meant dy/dx = -1/(2y) as that would be correct

Then the solution to -1/2 = -1/(2y) is y = 1

So the point that we are looking for has a y-value of 1! We are just missing the x-value

[gentle ore]

OHHH

yk what i did

i thought the question was asking for perpendicular so i subbed in 2 instead of -1/2

and then got y=1/-4

[upper abyss]

Kek yeah that makes sense

[gentle ore]

lol ok so w the y-value, do u just sub it into the original curve to get x?

o ok i have figured out the answer

tysm!

[cedar kiln]

@gentle ore Has your question been resolved?

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[granite gate]

How was the integral of sqrt(1-y^2) with respect to x from 0 to sqrt(1-y^2) evaluated here? When I found the indefinite integral, I got this, which is not nice at all:

[wraith dagger]

fuguehoppy

[granite gate]

^ The result of my indefinite integral

Oh wait, I see, it's completely constant with respect to x

Oops lol

.close

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[patent flicker]

Help: When a number is increased by 20% the result is the same when it is decreased by 10% plus 12. What's the number?

[cedar kiln]

Please don't occupy multiple help channels.

[dire geode]

Call the number x and find an equation

x increased by 20% means ?

[patent flicker]

thats whats confusing me is it 20% of x or x+0.2

.close

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[frigid canyon]

i can't read

try sending a ss

[glad kestrel]

test?

[dire geode]

Looks like hw

[glad kestrel]

<@&268886789983436800> :(

[dire geode]

Wait no it is a test

Read the rules

.close

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[glad halo]

hello

[glad halo]

this is a simple one

how do you denote an integrated function

as in derived funciton is f'(x)

[cedar kiln]

@glad halo Has your question been resolved?

[sweet vapor]

What?

Do you mean how to denote an integrated function of a derivative function?

@glad halo

[dire geode]

$\int f$ is fine

[wraith dagger]

riemann

[bold hinge]

No its not

You forgot dx

[dire geode]

F

$F(x) = \int f(t) dt$

[wraith dagger]

riemann

[cedar kiln]

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[warped wave]

Qno 4

How do u prove it i am confused😕

[sweet vapor]

I’d assume it’s vectors

[warped wave]

The lesson is related to forces

[warped wave]

Yes it is

[sweet vapor]

Unfortunately, I’m doing vectors next chapter. So I can’t help

Sorry bro

There are smart people here tho, so don’t lose hope, someone will help

[warped wave]

Well i hope so

I just need the answer rn😉😉

It wasn't in online website

[sweet vapor]

We help with the problems but not give answers, that’s the policy

[warped wave]

😑

[sweet vapor]

What have u tried so far?

[warped wave]

Traingle law of vectors

Ok i will try that

[sweet vapor]

Actually, let me try doing it

U can also use the website cause I’m prob gonna be wrong

Depends on the country

We study vectors in yr 11 advanced math or year 11 physics

Idk about the USA, I don’t live there

Yea

No

I’m in 10th as of now

But I’m doing yr 11 content

[warped wave]

Why did i get bunch of weirdos

.clsoe

.close

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[bitter nymph]

I was trying to solve problem 11 but had issues on solving for the transient solution

The steady periodic (or particular) is correct according to the answer key, but the transient is wrong

Im sure the solution for y is correcg because i checked it online. So now when im solving for y_c my values for C1 and C2 are incorrect

Differential equations is such a pain dude i hate this class

[cedar kiln]

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[bitter nymph]

oh

[fallen heath]

$r^2 + 4r + 5 = 0 \implies r = -2 \pm i$

[wraith dagger]

Ansh

[bitter nymph]

oh you know what

i sent the wrong work

my bad

[fallen heath]

so uhh, $x(t) = e^{-2t}(c_1 \cos t + c_2 \sin t)$

[wraith dagger]

Ansh

[fallen heath]

and the particular soln. would be

[bitter nymph]

this is the correct problem but my work was on a diff sheet and yeah i also got that for the general solution. the particular would be -(1/4)cos3t + (3/4)sin3t

converting that to amplitude angle form and get y_p = 0.7905cos(3t - 1.8925)

which matches the solution in the book

[fallen heath]

lmao

then what is the issue?

[bitter nymph]

yeah the steady periodic solution is fine but i am having difficulty with transient

solving for C1 and C2

[bitter nymph]

in this equation

[fallen heath]

$$x(t) = e^{-2t}(c_1 \cos t + c_2 \sin t) + \frac{1}{4}(3\sin 3t - \cos 3t)$$

[wraith dagger]

Ansh

[fallen heath]

Now you can go ahead and solve

:|

[bitter nymph]

oh

[fallen heath]

bruh

did you just plug the values into the transient solutions?

lmfao

[bitter nymph]

lol yeah ill try maybe i can solve

[fallen heath]

the values are given for the complete solutions of x(t)

[bitter nymph]

oh

[fallen heath]

x(0) = 0 => c_1 - 1/4 = 0
x'(0) = 0 => c_2 -2(c_1) + 9/4 = 0

And you get your values for c_1 and c_2

enjoy

[bitter nymph]

yeah that gives the correct values for c1 and c2

and now the solution is the same as the one in the book

thanks ansh

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[unreal orchid]

can anyone help me with this please?

[crimson sedge]

Hmmm

If I tell you the angle $\theta$ can you find the coordinate?

[wraith dagger]

CatHashira

[crimson sedge]

If I just told u theta then do you know how to find the coordinate?

[misty spear]

((pi × r)/2)/7?

Then find the angle

[crimson sedge]

For radians yeah

Oh wait nezoku pfp

[misty spear]

Uwu yeah

[unreal orchid]

ill try 😭

[unreal orchid]

thank you!

thats the theta ?

[misty spear]

Wait that's just one of the step

Yeah

[crimson sedge]

Ya I don't wanna give the total answer

just help him/her figure it out

[crimson sedge]

You don't have worry about fishing theta now

[crimson sedge]

If I give u this angle how will u find the coordinate that's all

[unreal orchid]

mmmmm

[unreal orchid]

then is the coordinate for point e like (5, cos ((pi × r)/2)/7) , (5 sin ((pi × r)/2)/7) ?? :0

[crimson sedge]

Umm I said ignore theta

Just if I say angle how to find coordinate

Yea u right ig

But u made a typo

Or something u wrote three components

Also multiplied sin by 5

[unreal orchid]

I think i got it! Thank you so much :)

.close

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[carmine pilot]

hello good morning or good afternoon I have a question : Why is the terminal side in trigonometry inside the unit circle always positive

[fair plover]

$sine = \frac{y}{\sqrt{x^2+y^2}}$
$cosine = \frac{x}{\sqrt{x^2+y^2}}$
$tangent = \frac{y}{x}$
in quadrant I, x and y coordinates are both positive, therefore sine, cosine and tangent will always be positive in quadrant I

[wraith dagger]

exploseph

[carmine pilot]

since r^2=x^2+y^2 so r=±sqrt(x^2+y^2) why r need to be positive

let r be the terminal side

x be x corr y be ycoor

okay nevermind 😄

/close

.close

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[white schooner]

How will you solve word problem on composite figure?

Help

[white schooner]

.close

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[cedar shadow]

Do u guys do 4th grade math?

[cedar shadow]

Nvm

[broken mist]

What's your question?

Well I'm not in fourth grade so I'll let you guess

They haven't closed the channel

[cedar kiln]

@cedar shadow Has your question been resolved?

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[cedar shadow]

What is two and one fourth times eight fiths?

[cedar shadow]

Not eight fifths. I meant 5 eights

[trim sentinel]

What does and mean in math

And could you write it in numbers

[cedar shadow]

Sry

[rigid ridge]

I'm interpreting this as \[ 2 + \frac14\cdot\frac85 \]

[jaunty mural]

convert to improper fractions and proceed

[rigid ridge]

oops

[wraith dagger]

Lance

[cedar shadow]

2 1/4 X 5/8

[jaunty mural]

.

[cedar kiln]

@cedar shadow Has your question been resolved?

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[opaque gull]

<@&286206848099549185>

[solemn torrent]

1. Find the inverse of f
2. Find dom(f^(-1))

[opaque gull]

i'm not sure how to do that

[patent flame]

I mean this is probably not the fastest method nor would it be worth one mark, but you can find the domain of the inverse function, because the range of the original function is the domain of the inverse and vice versa

[opaque gull]

uhh sure lemme try

[solemn torrent]

In other words

[opaque gull]

i dont know the domain either

[solemn torrent]

Wdym

[opaque gull]

ok wait let me backtrack

[solemn torrent]

The domain of...?

[opaque gull]

right now the first step is to find the inverse of f

im stuck on that

[solemn torrent]

Ye

[solemn torrent]

Do these following images not make sense to you?

Okay well

In simple terms

[opaque gull]

they do but im struggling to apply them algebraically

[solemn torrent]

replace the variables x for y

Then solve for y

[solemn torrent]

ok now subtract -6x from both sides

Okay your objective is to get the variable you're solving for on the same side

either the RHS or LHS

then factor that variable

because you'll have the same variable in same expression on either side

[opaque gull]

alright give me a sec

[solemn torrent]

alright if i'm not responsive let the other dude take over cause atm i'm cooking

[opaque gull]

ok

[solemn torrent]

ye that's alright

but

why'd you subtract -6x?

all you had to do then was divide 3x+2

[opaque gull]

yea that was a mistake

sending another screenshot now

[solemn torrent]

So now you have y=(1-6x)/(3x+2)

ye

that looks good

wait

no

how'd you manage to simplify that

[opaque gull]

divided the 6x by 3x

?

wait that doesnt make sense

[solemn torrent]

I mean well certainly you could divide

but not like that

if that term is undergoing addition or subtraction you can't cancel them out or anything

[opaque gull]

alright yea so just leave it in that form of 6x and 3x

[solemn torrent]

Ye

[opaque gull]

now to find the domain of that function...

[solemn torrent]

Although if you applied polynomial long division

you could get

[solemn torrent]

But we'll leave it as a single fraction

[solemn torrent]

Ye

[opaque gull]

how do i go about doing that>

?*

[solemn torrent]

Well look at the form you've gotten

It's a fraction

just recall your indeterminate forms

[opaque gull]

ah so domain could be any real number as long as the denominator isn't 0

?

[solemn torrent]

ye

[solemn torrent]

And when it comes to radicands it can't be negative

[opaque gull]

yea

so the domain would be x E r, x=/ -2/3

[solemn torrent]

nice

how does this work

[solemn torrent]

there

[opaque gull]

thanks a ton for the help man

[solemn torrent]

what's xEr?

binary operation E?

[opaque gull]

x belongs to R

[solemn torrent]

oh right of course

R being the real field ye?

[opaque gull]

yes

[solemn torrent]

ok

Also x != -2/3

!= denoting not equal to over text

[opaque gull]

is that the official way of doing the does not equal to?

[solemn torrent]

I suppose so

I assume that's what you meant with =/

[opaque gull]

$x /neq -2/3$

[solemn torrent]

!= is taken from computer science

meaning not equal to

\neq in latex

Use \neq

x\neq-\frac23

[wraith dagger]

taemed

[solemn torrent]

Go check #latex-help #latex-testing if you want to learn how to latex/tex

which would be beneficially when doing maths

[opaque gull]

sure

[solemn torrent]

or anything really, music, natural sciences, etc

[opaque gull]

idk what latex is

[patent flame]

$f(x) \neq -\frac {2}{3}$, $f(x) \in \mathbb{R}$ since it's the range you also have to go back to including f(x)

[solemn torrent]

I could explain the basics rn

[wraith dagger]

I can't believe you've done this

[solemn torrent]

But ye

ok i'm feeling nauseous af

brb

[opaque gull]

interesting

thanks a lot

[solemn torrent]

So you yeah may express your range using inequalities or interval notation

i prefer using interval notation

Thus

[solemn torrent]

For a function f:X->Y with domain X we denote the range of f using ran(f) or Range(f). This may also be used to denote the codomain (set of destination/target set) Y.

But of course in our context we're using it to denote the image set of our function

If you're done close the channel via .close

. is the cmd prefix for @cedar kiln

You there?

@opaque gull

[cedar kiln]

@opaque gull Has your question been resolved?

[cedar kiln]

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[dense wing]

Hint: Complete the Square

[opaque gull]

ah dammit

i forgot how to do that

what is the advantage of completing the square versus other methods of solving the equation

[dense wing]

Wdym

Nowhere does it say solve

[opaque gull]

ok yes that is true

[dense wing]

The advantage of my hint will be obvious when you apply it

[opaque gull]

i think i should brush up on how to complete a square then

brb

[opaque gull]

?

what now

[south dove]

you factored p(x)
f(x) g(x) h(x) are factors of p(x)

[opaque gull]

i dont understand what you're getting at

[south dove]

the question is asking you to find functions f g h

p(x) = fgh(x)

[opaque gull]

ok...

feel like there's an obvious connection im missingf

but i can't think of anything

[south dove]

fgh(x) = f(x)* g(x) *h(x) =p(x) = (x + 2)^2 -15

[opaque gull]

ok now how do i figure out what each function is

[south dove]

find 3 functions that multiply together to equal
p(x) = (x+2)^2 -15

[opaque gull]

but they dont multiply they get composited

i have to find functions that you can plug into each other to get that

i have the answers so don't worry about giving it away

i just want someone to explain what's the process behind it

[south dove]

if it were composite it would be f(g(h(x)))

[opaque gull]

ohhhhh

ok hold on

[jaunty mural]

wait f g h are multiplied together?

[opaque gull]

nope... still blank 😭

here are the answers

[jaunty mural]

I interpreted f(g(h(x)))
which one is it

I'm confused wut

So it is composing

[south dove]

must syntax error then in the original question, but makes sense

[jaunty mural]

So yes, Mosh' suggestion works

complte the square

Then can you see how the answer works?

[opaque gull]

i cannot

[jaunty mural]

f(g(h(x))) = f(g(x+2)) = ...?

Do you not know how to compose functions

[opaque gull]

i do

[south dove]

9 permutations, learn by trial and error

[jaunty mural]

So read this thing as a set of instructions

'Take x'

'Add 2'

'Square it'

[opaque gull]

but the logic of figuring out what functions compose into that is hard to comprehend

[jaunty mural]

'subtract 15'

And there you have your 3 functions

[south dove]

actually 6 permuations
given the answer given

[opaque gull]

oh my god

yes

that makes sense

thank you

[jaunty mural]

Well I have to say

I find this question a bit strange

[opaque gull]

this is some next level explaining

[jaunty mural]

f(x) = x+1
g(x) = x-1
h(x) = x^2 + 4x - 11

Surely this works

And yet the mark scheme is very specific

[opaque gull]

i think it's more geared towards testing the critical thinking that you described above

[jaunty mural]

Well I never would've thought to complete the square

without ^ hint lmao

[jaunty mural]

First thing that came to my mind

[south dove]

it could be that the class teaches a particular method to achieve a specific answer?

[opaque gull]

average IB question

anyway thanks guys

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[dire geode]

What have you tried

[opaque gull]

dont know where to start

[gray ice]

What formulas do you know that are related to quadratics?

Write them all down

See if any are helpful

[opaque gull]

completing the square
factorization
quadratic formula

[gray ice]

Any others?

[jaunty mural]

what is completing the square used for

[opaque gull]

creating a perfect square no matter the quadratic?

[jaunty mural]

What's the point of doing this though

$$f(x) = ax^2+bx+c$$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

$$f(x) = a(x-p)^2+q$$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

What's the point of this process?

[opaque gull]

i don't know

[jaunty mural]

$(x-p)^2$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

Can you tell me anything about this

x is a variable

p is a constant

[opaque gull]

yes...

it can square rooted

not sure what else there is to say

[jaunty mural]

uh no...

If I have a square number

the square of something

what must this always satisfy?

$k^2$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

No matter what k is?

[opaque gull]

im stumped

[jaunty mural]

hmmm

[gray ice]

What is a negative times a negative

[jaunty mural]

Let me plot a graph

[opaque gull]

a positive

[jaunty mural]

You can say something about all the y-values here

[dire geode]

just take a derivative

[opaque gull]

they go infinity but the min is 0

[jaunty mural]

The min is 0

$k^2 \geq 0$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

Does this make sense now?

[opaque gull]

haven't taken calc yet

[jaunty mural]

No matter what k is

this expression is always greater/equal to 0

[opaque gull]

ah ok

[dire geode]

Yea I figured. Shuri's giving great hints

[jaunty mural]

so lets backtrack

x is a variable

p is a constant

but its still something squared

It must be non-negative

[opaque gull]

yes

[jaunty mural]

So does this tell us something here 🤔

If you want, you can work with inequalities
$$(x-p)^2 \geq 0$$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[opaque gull]

so bracket must be positive

[jaunty mural]

right. (or 0)

[opaque gull]

yes

[jaunty mural]

what about a times it

a bit of a tricky question but maybe u can figure it out

A hint is there is no fixed answer --- it depends on something

[opaque gull]

well x is a variable

so obviously that'll affect the value of a?

[jaunty mural]

a p q

[opaque gull]

is that what you're getting out

[jaunty mural]

are all constants

[opaque gull]

ah this is in vertex form

[jaunty mural]

You told me (x-p)^2 >= 0

Now I want to know about a(x-p)^2

[opaque gull]

a could be a negative

[jaunty mural]

right, if a is negative

what happens

[opaque gull]

the signs inside the parentheses switch

[jaunty mural]

no...

[opaque gull]

it comes negative

[jaunty mural]

uh

$-2\times(-1)^2$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[opaque gull]

is =2

[jaunty mural]

The minus outside doesn't affect the sign inside the parenthesis

[opaque gull]

-2

ah..

[jaunty mural]

(the sign inside goes away because square)

[opaque gull]

yes

[jaunty mural]

So the squared term is always 0 or positive

If you multiply by something negative ... ?

[opaque gull]

it becomes negative

[jaunty mural]

(or 0)

what if you multiply by something positive?

[opaque gull]

it stays positive or 0

[jaunty mural]

exactly

so

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

And then we are one more step from saying something about f(x)

f(x) is the left side add q

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

Does this make sense?

[opaque gull]

somewhat yes

[jaunty mural]

So the point of vertex form is to find the maximum and minimum of a quadratic

(also known as the extremum)

The last thing to check is when the maximum/minimum happens

If you reverse the steps

You should see f(x) = q (the minimum or maximum)

When this original squared term is 0

[opaque gull]

yea im lost

[jaunty mural]

👌 which bit

[opaque gull]

here

[jaunty mural]

check this inequality

I claim q is the minimum if a > 0

the maximum if a < 0

[dire geode]

Plug in some numbers for ^ if it doesn't make sense @opaque gull

[jaunty mural]

$$g(x) = (x-1)^2+2$$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[dire geode]

It's the equivalent of parabolas opening up or down

[jaunty mural]

We have 'something positive or 0' plus 2

[opaque gull]

yes

[jaunty mural]

The smallest g(x) can be

[opaque gull]

2

[jaunty mural]

is when the 'something positive or 0' equals 0

If the square equals 0

we have x-1 = 0

So I can immediately tell from vertex form

(1, 2) is the minimum in this case (since the coefficient of x^2, a = 1, the quadratic is a u-shape)

[opaque gull]

yep

[opaque gull]

so that's the point of the vertex

[jaunty mural]

👌

[opaque gull]

why not -1

[jaunty mural]

for x?

x = -1

gets me
(-1 -1)^2 + 2 = 4 + 2 = 6

I want the bracket to be 0. So I solve x - 1 = 0

[opaque gull]

ah so it must be positive one

okay

[jaunty mural]

$$h(x) = -2(x+4)^2+5$$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

So quickly try this to make sure you got it

Find the coordinates of the extremum of h(x)

and determine if it is a minimum or maximum

[opaque gull]

minimum is -4, 5?

[jaunty mural]

yes

no

(-4, 5)

not a minimum

(x+4)^2 is always non-negative
So -2(x+4)^2 is always non-positive

[opaque gull]

maximum since its opening downwards?

[jaunty mural]

yes

If (x+4)^2 isn't 0

you are adding something negative to 5

[opaque gull]

yes

[jaunty mural]

https://www.desmos.com/calculator/nthpa4lrsc

You can play around with the slider and check examples with something like this

[opaque gull]

oh ok thanks

yea it helps seeing it visually

[jaunty mural]

So to begin the question, you should be able to deduce something from the 2nd sentence using vertex form

[opaque gull]

so the graph is upwards opening and the minimum point is at (-1.5, ?)

[jaunty mural]

yep

you already knew its facing upwards

[opaque gull]

yea

[jaunty mural]

Standard form
$$f(x) = ax^2+bx+c$$
Vertex form
$$f(x) = a(x-p)^2+q$$
Factored form
$$f(x) = a(x-r)(x-s)$$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

^ these are the forms you need to know in general (the letters other than x are constants)

[opaque gull]

alright

i thought the question was referring to plugging -1.5 into the x of the equation

so that threw me off

but it seems like i was supposed to convert it to vertex form for it to make sense

[jaunty mural]

Also remember the x-coord of the minimum/maximum is always exactly halfway between the x-coord of the roots (if they exist).

[opaque gull]

yea

AH

GOT IT

YES SO HALF THE 9

IS 4.5

and when added to x demonstrates that roots are at 3 and -6

[jaunty mural]

Oh I just noticed you don't need vertex form then lel

But it definitely is helpful for a good number of Q's

[opaque gull]

[opaque gull]

yes i have encountered it alot

ok so now question b

should be intuitive

since we know all the values

q is -1.5

but idkt he r

[jaunty mural]

Notice all these 3 forms must be equal

You can convert either of the bottom ones to standard form and equate coefficients

[opaque gull]

equate coefficients?

[jaunty mural]

$$f(x) = 2x^2+(k-3)x-44$$
$$f(x) = (k+2)^2x^2 + jx - s$$

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

If I tell you these 2 are the same quadratic

then

[opaque gull]

wait

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[opaque gull]

i don't think i've taken this yet

nevetheless

[jaunty mural]

I chose a bit of an extreme example

[wraith dagger]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[jaunty mural]

If these two are the same quadratic

we need a = 3, b = -5, c = 2

Otherwise they will be different

[opaque gull]

yes

[jaunty mural]

The same applies here

x is a variable, the rest are constants

And finally the same applies here

If you multiply out the vertex or factored form

[opaque gull]

yes each one has three constants that correspond to a b and c

[cedar kiln]

@opaque gull Has your question been resolved?

[cedar kiln]

@opaque gull Has your question been resolved?

[cedar kiln]

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[gusty lotus]

Hi

[gusty lotus]

im only doing the odd ones but im not sure about 1 and 3

how would I find when the tangent line is horizontal

[hollow osprey]

well what does it mean for a tangent to be horizontal

what slope does a horizontal line have

[gusty lotus]

0

[hollow osprey]

correct

so set dy/dx = 0

[gusty lotus]

what's dy/dx

[jaunty mural]

f'(x)

[hollow osprey]

are you in calculus

[gusty lotus]

yes

[hollow osprey]

or precalculus?

[gusty lotus]

calc

[hollow osprey]

what

[gusty lotus]

so I find the derivative of f(x)

[hollow osprey]

yes

and set it equal to 0

[gusty lotus]

and set it = 0

[hollow osprey]

cuz we wanted the slope of the tangent line to be horizontal (0)

[gusty lotus]

then solve ?

[hollow osprey]

yes

[gusty lotus]

oh and then those should be the answers?

[jaunty mural]

Do you understand what f'(x) represents

[gusty lotus]

yes

[jaunty mural]

if you do, then you should be certain that these are the answers.

[gusty lotus]

derivative

of f(x)

[jaunty mural]

no not just that

[gusty lotus]

oh

[jaunty mural]

f'(a) represents the derivative of f(x) at x = a

[gusty lotus]

oh yea

[jaunty mural]

Geometrically, it also represents the slope of the tangent at x = a (or you can say at the point (a, f(a)) )

[gusty lotus]

at a certain point (a)

yea

oh shoot sorry I gotta go. Thank you for the help guys

.close

[cedar kiln]

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[mint coyote]

Hello, I have no idea how they went from the first one to second one, could anyone help explain to me what did they do here? Thanks in advance!

[safe abyss]

Are you familiar with
root(x) = x^(1/2)
3rd root(x) = x^(1/3) ?

[mint coyote]

yeah!

but im having trouble with

x^1/2 - 1/3

why did they do that

[safe abyss]

you can break out 1/x^(1/3)

and that is equal to x^(-1/3)

(it's generally true that a^(-1) = 1/a)

does that make sense? i'll explain the rest, just making sure you're following so far

[mint coyote]

yeah i understand so far

like 16/x^2 = 16x^-2 right?

[safe abyss]

yep yep

so then we get x^(-1/3)*(x^(1/2)-2)

let me format it better so you can see it easier 1 sec

[mint coyote]

okay!

[safe abyss]

,, x^{-1/3}(x^{1/2}-2)

[wraith dagger]

PixzleOne

[safe abyss]

does that make sense too?

[mint coyote]

yeah

you multiply to get rid of the denominator right?

[safe abyss]

,, \frac{x^{1/2}-2}{x^{1/3}} = x^{-1/3}(x^{1/2}-2)

[wraith dagger]

PixzleOne

[mint coyote]

👍

[safe abyss]

then just multiply and get

,, x^{1/2}\cdot x^{-1/3} - 2x^{-1/3}

[wraith dagger]

PixzleOne

[safe abyss]

which is equal to

,, x^{1/2-1/3} - 2x^{-1/3}

[wraith dagger]

PixzleOne

[mint coyote]

thanks for breaking it down for me!

I appreciated it

[safe abyss]

was that last step confusing? or was it just the whole thing :)

no problem though

[mint coyote]

Nope!

Tysm!

[safe abyss]

no worries! good luck have fun :)

[mint coyote]

.close

[cedar kiln]

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[azure magnet]

So the objective was to find p and q

Constants

Did I do it correctly?

<@&286206848099549185>

could Q posibly be 9 actully if I messed up my -/+ signs?

[cedar kiln]

@azure magnet Has your question been resolved?

[modern compass]

your signs are definitely messed up

$10\cosh(x)-8\sinh(x) = 10[\frac12(e^x+e^{-x})] - 8[\frac12(e^x-e^{-x})]$