precalculus

[jagged gazelle]

yeah i know

yeah

i mean for simple equations at least lol

[steady token]

well alright thx lol

[jagged gazelle]

obv youll still need to take the derivative for curves and stuff but

[steady token]

right

just curious

how long did you study each night on precal?

[jagged gazelle]

tbh...not that much. i studied more in trig/algebra 2

i studied the most in precal when it came to the stuff you're on and the stuff that dipped more into calculus. i was pretty much ok w everything else

[steady token]

oh

i cant seem to balance out my schedule i guess

[jagged gazelle]

we spent a good while on polar coordinates and like,, limits and trig identities...we spent more of the second half of the year on more calc based stuff

[steady token]

so calc stuff is the things ive never learned about in algebra right?

and the algebra side is things we've learned but harder?

[jagged gazelle]

yeah and the things that have more like...i dont know how to explain this esp bc im a noob but -- YEAH

and the algebra side throws in some other wack stuff like polar coordinates is literally that thing where u put paint on one side of a paper and fold it in half kdsfmdslfdmsl56tm4wl3mg

[steady token]

lol

[jagged gazelle]

uuhh you'll have to deal with a lot of degree/radian shit later

[steady token]

ah fuck

[jagged gazelle]

and you'll need the unit circle a decent amount

[steady token]

i hated that trig shit

[jagged gazelle]

yeah there's a lot of unit circle shit going on

[steady token]

i already forgot it

its over for me lmao

[jagged gazelle]

idk i find the hardest part of pre calc is just the complex math -- print it out

[steady token]

its really weird to me tho

[jagged gazelle]

ya i getcha

[steady token]

i barely studied in algebra 2 and i got an a

now i put in effort and recieve an bad grade

[jagged gazelle]

did you take algebra 2 and trig at the same time or?

[steady token]

yes

[jagged gazelle]

ah

[steady token]

i studied at most like 10 mins a day

1 hour for tests

[jagged gazelle]

uh matrices in precalc are also pretty fun

dude i honestly think your teacher's just dumping the hard shit on you all at once. like i really don't think later in the year youll have that much trouble w like...fucken matrices and the stuff that ties more to trig

[steady token]

weird lol i have the same teacher from last year

but yeah i feel like maybe i did rush through my last test a bit

[jagged gazelle]

you'll learn a lot of new ways to factor equations too

[steady token]

do you have any online resources for recommendation?

ive just been searching up random precal vids on youtube

[jagged gazelle]

tbh that's what i did

that and familiarizing myself w the calculator a lot

[steady token]

okay well thanks for the help

and sorry for wasting ur time with my questions

i hope ur doing well in ur currently class : )

[jagged gazelle]

ah no its fine it was a good refresher for me esp. since im just starting calc

[steady token]

nice nice

[jagged gazelle]

tysm!!!

=3

[steady token]

no no thank you

[jagged gazelle]

i really doubt you'll struggle later on though and im surprised she started off w something like that esp since it'd be more relevent the closer you get to calculus

[steady token]

i mean we are using larson precal books

lots of schools use it

shes just following the chapters by order

[jagged gazelle]

ah

[viscid thistle]

@viscid thistle So what differential calculus have you done?

ngl, it's pretty big.

Okay then

So you've seen derivatives of polynomials?

So, the biggest pointabout integration (and why it is sometimes called anti-derivation) is because of the fundamental theorem of calculus.

So.

Pretty much.

So yeah

We define the integral of f, F to be

[viscid thistle]

^

Dis.

FTC: fundamental theorem of calculus

The two things we really care about here though

Is that if g(x) is the integral of f(x)

Then d/dx (g(x)) = f(x)

And

This thing about bounds.

=tex \int_a^b f(x) dx = F(b) - F(a)

[low raft]

Yup.

[viscid thistle]

Moving the notation around.

but F is just g

Aaah, in derivation of FTC we use a parameter

But in this case we're using a,b to just mean constants.

Substituting in variables into our bounds will come up further down the line.

So

Let's do this for polynomials then.

By integrating a polynomial

I wish to find a function F such that F' = f

Try this for x^2 and see if you can arrive at something.

f(x) = x^2, F = ? such that F' = f

Whats' the derivative of x/2

Yep.

Yes

By integrate I mean, find a function F such that d/dx(F) = x^2

[viscid thistle]

Yees

Okay, partial solution

Since a constant always differentiates to 0 though

I add a constant term on the end of this to arrive at a general solution

[low raft]

$$\frac{x^3}{3} + C$$

[viscid thistle]

^

And that's the intergration of polynomials.

I'm sure you can find the general rule for integrating polynomials.

Actually this is a good exercise.

Find F such that F' = ax^n

[low raft]

@viscid thistle You will get into more methods of integration such as u-substitution and integration by parts.

[viscid thistle]

Eh, that's not so bad when you learn it.

[low raft]

Yep.

Try to grasp the jist of the method.

[viscid thistle]

Yup.

So what else is nice about integrals?

They inherit the linearity of derivatives.

This means that, scalar multiples and addition give us a way to seperate out integrals and prevent needless calculation

=tex \int af(x)+g(x) dx = a\int f(x) dx + \int g(x) dx

[granite stirrup]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[viscid thistle]

^ linearity

[low raft]

Good point.

You can factor out the coefficient in an integral expression.

[viscid thistle]

Yep yep

Integration inherits this by FTC from differentiation

So this was a brief thing on the indefinite integrals

Let's move on to definite integrals, a corrolary of the FTC.

For someone who doesn't know anything about integration, you sure know a bit about integration.

Fine - what would you like to do?

The zest of it

[low raft]

Triple integrals?

[viscid thistle]

Probably should cover u-sub and by parts method before delving into multivariable calc lmao

[novel monolith]

ez

you just learned how to integrate

[tulip quest]

Mathematica > Wolfram ALpha

[fading token]

Mathematica = WolframAlpha

[thick raptor]

@fading token Mathematica more like paid version of WA

also more code-y

[brave void]

tan(x)

over 5x

As x approaches 0

<@&286206848099549185>

Does it not exist

[viscid thistle]

$$tan(x) / 5x$$

[brave void]

?

[brave void]

yes

[viscid thistle]

that it?

[brave void]

Yes

[viscid thistle]

as x approaches 0 you get an asymptote

even if the value is miniscule

=pup tan(0.1)/0.5

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@plucky urchin

[viscid thistle]

hm

it exists

but you cannot compute for 0

=pup tan(0.01)/0.05

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@plucky urchin

[viscid thistle]

ah

for reasons I am too dumb to prove or realise

limit as x approaches 0 for tan(x) /5x is equal to 0.2

probably due to both approaching 0 and thus cancelling out leaving a 1/5 n_n

[brave void]

uh

It should be undefined?

No?

I put in 0

I got 0 in the denominator

Undefined

[wind igloo]

Not quite.

Because the tan(0) = 0 too

What you have is 0/0 which is called an indeterminate form.

Which means you have to do more work

[brave void]

gotcha

[rocky bison]

That lim ís 1

And it's a sandwich boi

[wind igloo]

It's not 1. It's 1/5

'cause /5x

[rocky bison]

Oh there's 5x

I didn't do that tho

Smh

[wind igloo]

You did (sin/cos)/x

cos -> 1, sin/x -> 1

[rocky bison]

Ye

I was proud of myself :(

[north bough]

Can someone help me with this

I basically have to get a sinusoidal function to model the behavior of the ferris wheel

Which would be cosine

I already know A = 19/2 and D=19/2 in the equation

y=Acos(B(x-C))+D

nvm I messed up the period

[civic plaza]

@north bough it would be easier if you use a sin function

since at t=0, y=0

[north bough]

oh

[civic plaza]

you wouldn't have to figure out a phaseshift or C in your case

[north bough]

true

[civic plaza]

Look at your A

what does that tell you?

[north bough]

I thought it was 3.75 but it was 7.5

[civic plaza]

No your A

[north bough]

o

[civic plaza]

that is your amplitude

[north bough]

A

yeah

A and D are 19/2

[civic plaza]

hmm

oh I see why you did that

ignore those for right now

if you have a sin function, then you wouldn't need a D either

since y starts at zero

[north bough]

oh wait

[civic plaza]

if you have a sin function without a phase shift then you need to find y=Asin(Bx)

[north bough]

I think I thought of it as a cosine function since the lower range of the wave intersects at 0

[civic plaza]

ehh

the lower range would be -19

[north bough]

why though?

[civic plaza]

since the ferris wheel cart starts at y-0

[north bough]

D = 9.5 and so does A

right?

[civic plaza]

if you do a sin function you don't need a D

it's not shifted up or down

[north bough]

A is (19-0)/2 and D is (19+0)/2

[civic plaza]

it starts at zero

[north bough]

o

[civic plaza]

also the bottom point is not zero

it goes down to -19

so the amplitude is (19-(-19))/2

or just 19

so we have y=19sin(Bx)

[north bough]

oh oof

wait

[civic plaza]

do you know the period of the ferris wheel?

[north bough]

how do u know the minimum is -19

since ur using sin?

15

[civic plaza]

from the picture

it starts at zero, goes up to 19, eventually it will go down to zero and past it

down to -19 and up to zero again

[north bough]

I thought you were supposed to use the two extrema and then subtract the minimum from the maximum value from the data given in the problem

which are 19 and 0

leading to (19-0)/2

[civic plaza]

the minimun is -19

look at your picture

[north bough]

oh

i see

[civic plaza]

it starts at y=0 which is the center of the ferris wheel

[north bough]

yeah

ok that makes more sense now

since that's only the radius

alright thx

[civic plaza]

what about the period

remember the period=2pi/B

so B=2pi/period

and your done

[stray obsidian]

2. ____ All repeating decimal numbers can be represented as ratios of integers.
3. ____ The set of real numbers is commutative for the operation of subtraction.

its either true or false

im pretty sure 2 is true but i dont really get 3

[wind igloo]

The commutative property for addition means that a + b = b + a

Similarly, for multiplication is means that a * b = b * a

What do you think the statement for subtraction would be?

[stray obsidian]

i know what it is

its a-b = b-a

but doesnt work for division

because order matters

so false?

[stone acorn]

@stray obsidian that doesn't work for subtraction either

It's a - b = - (b - a)

[stray obsidian]

oh yea my b

order matters

but what soes it mean by set of real numbers is commutative

having trouble with the vocab

[stone acorn]

Commutative means you can change order

That's it

[thick raptor]

baka

0 - 0 = 0 - 0

1 - 1 = 1 - 1

2 - 2 = 2 - 2

[stone acorn]

Hmm

[thick raptor]

so obviously subtraction is commutative

[stone acorn]

@stray obsidian

Did you arrive at the answer?

[stray obsidian]

ok

lol

thank you

[brave void]

How do I tell if a left or right side limit is going towards infinity or negative infinity?

[brave void]

<@&286206848099549185>

[stone acorn]

@brave void

If the number decreases as you move to one side, negative infinity

Otherwise infinity

[brave void]

wait woah

So lets say I have

1 over x+2

Limit from the left

How do I tell it's going to inifinity over negative infinity

I plug in -2.1?

As lim x approaches -2

From the left

What do I plug in?

-3?

-4?

@stone acorn

[stone acorn]

Pheonix

You mean this?

[brave void]

?

[stone acorn]

=tex \lim_{x \to -2} \frac{1}{x + 2}

[brave void]

Yeah

we know it goes undefined at -2

So I plug in like

-3

-4

And 1

-1*

[rocky bison]

I hate to say

That limit doesn't exist

[slender river]

lel pseudo

[brave void]

How to solve?

I know tan of pi over 2

It is undefined

but like

The book has a different answer

How would I know if it went to negative or positive infinity?

[rocky bison]

Ok

So one approach would be taking sin/cos

[fringe stream]

Or you could just see that it says $$\frac{\pi}{2}^+$$, so you're looking at values to the right of pi/2

[rocky bison]

That's also viable

[brave void]

to the right of pi/2

How would I do that without a graphing utensil

[rocky bison]

Plug values in

Closer to pi/2

[fringe stream]

The graph of tan is pretty standard, you should know what it looks like.

[rocky bison]

Like pi/2+0.0001

[fringe stream]

But you can always just plug in values.

[brave void]

aye

[fringe stream]

If you're just to the right of pi/2, you're in the second quadrant.

All tan values in the second quadrant are negative.

[brave void]

okay

[fringe stream]

Or, tan = sin/cos, sine is positive, cosine is negative, so tan is negative in II.

[brave void]

aye

Gotcha

tyvm

negative inf

[fringe stream]

Yup.

[brave void]

The thing on my tests

I rush

And sometimes I think that like

-4 is 4 from the left XD

I gotta slow down

[fringe stream]

Think of it like the number line.

• is the positive direction, which is right.

So when you say 4^+, you're looking towards the right side of 4

[rocky bison]

=tex \lim_{\delta x\to0}\left(\tan\left(\frac\pi2+\delta x\right)\right)

[rocky bison]

Very unnecessary but I remember when I was first doing I liked to think of it like this

[fringe stream]

I like this approach.

[icy fog]

The multiplication is whats messing me up, can anyone explain this better?

[zenith moat]

both sides of the equation are multiplied by sin(B) to isolate b

(any operation u do on one side of the equals sign has to be done on the other side as well)

[icy fog]

ya i got that part

but when i do it it doesnt look like that\

[rocky bison]

What does it look like when you do it

[icy fog]

b = a/SinA * SinB

[zenith moat]

thats the same thing

sin(B) is like sin(B)/1

they just multiplied a/sin(A) by sin(B)/1 which gives you a*sin(B)/sin(A)

and they pulled the "a" on the outside

[icy fog]

that i get its the pulling "a" to the outside that is twisting my brain

[zenith moat]

like the reasoning for it?

[icy fog]

this always was a weakness

[zenith moat]

there really is no reason. leaving it on the numerator is completely fine

[icy fog]

ok so help me understand this part

a*sinB/SinA

is the same as multiplying sinB/SinA by "a"

[zenith moat]

that is correct

sinB/sinA is just a fraction

[icy fog]

yes

[zenith moat]

think of it as anny other variable.

for instance say sinB/sinA is y. a*y is the same thing as ay

[icy fog]

but sinB is being multiplyed by "a"

the numerator

but by pulling it out you are now multiplying the whole fraction instead of just the top

omg this is painful

[zenith moat]

lol its hard to explain it

since "a" is not a fraction multiplying it by a fraction vs just the numerator is the same result

since a is technically (a/1)

1 * denominator = denominator

[icy fog]

right

ohhhhh

so a/1 * sinB/SinA

is the same

just dont need the 1

[zenith moat]

correct

does that make more sense?

[icy fog]

example

5*1/2

is the same as 5* 1/2

[zenith moat]

yep

[icy fog]

wow thanks for that , sometimes its the easy stuff

[zenith moat]

yw

[sweet viper]

Hello everyone I'm new to this server but I need Pre-Calc help asap!!

Can anyone help??

[leaden mulch]

Ok

I’ll see if I remember anything

[sweet viper]

Thank you!! Let me send a screenshot

I'm taking a quiz but Im lost

[leaden mulch]

Lol

[sweet viper]

how do i find the power function?

[leaden mulch]

A power function comes in the form of f(x)= kx^n

Do you have coordinates of two points?

[sweet viper]

I just have the x-intercepts and y-intercept

[leaden mulch]

What are they

x1,y1 and x2,y2

[sweet viper]

There's two x-intercepts: (-6,0) and (4,0)
And the y-intercept is (0,576)

[leaden mulch]

576 or 5.76

[sweet viper]

576

At least I hope that's right because I haven't understood this lesson very well

[leaden mulch]

I’m lost

Have you tried googling?

[sweet viper]

I've tried googling for like two hours and I'm just at a loss right now

And this quiz plus the homework are due by midnight

[leaden mulch]

Have you looking st class notes on how to find power sets?

Tried looking at*

[sweet viper]

I did but its an online class and the lessons are all videos (not even made by my teacher its from some random teacher from like the 80's not even kidding)

And it's just a horrible way for me to learn

They barely went into detail with power functions or most of the material actually

[leaden mulch]

Dang

It says to find power function from 2 given points

[sweet viper]

And the teacher in the videos uses a graphing calculator to find a lot of his answers but I'm not allowed to use that for this class

[leaden mulch]

I have no expertise, sorry

Try posting it in questions

[sweet viper]

Okay

Thank you for trying to help though

[olive meteor]

Hey guys I have a problem

so what if Im trying to calculate an angle but the problem throws minutes and seconds at me

example

[fickle moat]

can someone tell me if this is -6x^4+7x+7 is this even odd neither

[olive meteor]

@fickle moat Did you do the f(x) and f(-x) method?

[leaden mulch]

@olive meteor like a physics problem?

[wind igloo]

An even function is one where f(x) = f(-x)

And an odd function is one where f(x) = -f(-x)

[olive meteor]

@leaden mulch for example sec 48 degrees 12'

or 48 12'

Nevermind guys I got it

[fickle moat]

i did and i got neither

[viscid thistle]

can someone help me plss

For the question that asks about the olympic runner

can someone do B

[viral imp]

I'm trying tro graph y=x+|x^2-1|. I understand that the domain of the function is R, but after a quick check on wolfram I don't get how the range can be y>=-1.
As a first step I graphed both parabolas, i.e. y=x+x^2-1 and y=x-x^2+1, however I'm not sure how to obtain the actual graph of this function

[severe verge]

draw y=x too

and both parabolas

on the same graph

then erase everything below the y=x line

[viral imp]

why should I draw y=x? It's not clear to me

[iron rover]

Hello guys, so I was helping my little brother with his homework and we came across this, given S sub 5 = 45, find a3
Idk what to do lol

btw it's a question about arithmetic series

[limpid plover]

$$S_5=45$$ right?

[limpid plover]

Let the 5 terms be $$a-2d,a-d,a,a+d,a+2d$$

[limpid plover]

$$S_5=5a+2d-2d+d-d$$

[limpid plover]

$$45=5a$$

[limpid plover]

$$a=9$$

[limpid plover]

In this case $$a=a_3$$

[iron rover]

ahhh

let me read it again xd

[limpid plover]

Mention me if you don't understand something

[iron rover]

How did you know the 5 terms you used?

[limpid plover]

It's a general thing

[iron rover]

I mean where did youg get it form?

I mean where did you get it from?*

[limpid plover]

Intuition

[iron rover]

Ultra Instinct xD

but seriously, where?

and what general thing? I don't know any formulas regarding the above 5 terms

[limpid plover]

It would have been the same if there were 7 terms, a+3d,a+2d,a+d,a,a-d,a-2d,a-3d

[iron rover]

is there a formula about that? I want to know please

[limpid plover]

Nope

As you can see the common difference cancels out

Then you can use hit and trial to get the terms and make sure they follow an AP

[iron rover]

oh okay

I will just ignore how you got it, but I recognize the pattern you used so I think I'll just use it

[steady sleet]

@wind igloo do the 2's cancel out here?

[wind igloo]

Yes

[steady sleet]

ok

this is a bit confusing because of all the variables in play

@wind igloo i dont think this is correct

it should be 50/48 or 25/24 but i dont see where the arrithmetic is wrong

[wind igloo]

It's right at the end.

I think you got yourself confused where the division happens.

=tex \frac{1}{\frac{48}{50}}

[wind igloo]

That's 50/48

What I think you did is

=tex \frac{\frac{1}{48}}{50}

[wind igloo]

Do you see the difference? @steady sleet

[steady sleet]

oh

isee it

1/(48/50) and the inverse of that is 50/48... someone doesnt seem to get this through his head. 😃

ty

[slender river]

parentheses are fun

very important too

haha

[brave void]

is tanx over x 1

[slender river]

umm noo??

do you mean the classic limit as it approaches 0?

[brave void]

yeah

@slender river

[slender river]

well yes then

try to prove it with triangles and unit circle for better understanding

[brave void]

but not cos

cos x over x

As x approaches 0

Not 1

just sin and tan, right?

[slender river]

yeah i think just sin and tan

[hexed ermine]

Yeah, cosine creates an asymptote because as x approaches 0, for the numerator (cos(x)), the numerator converges to 1

while the denominator approaches 0

[limber heath]

How do I approach / solve this problem?

[hexed ermine]

write it in sigma notation

[elfin night]

=tex 4+4\cdot\frac{2}{5}+4\cdot\frac{2}{5}\cdot\frac{2}{5}+4\cdot\frac{2}{5}\cdot\frac{2}{5}\cdot\frac{2}{5}

[limber heath]

oh that involves sigma? okay nvm I'm not suppose to know that yet

[elfin night]

Its doable

If you know geometric progressions

@limber heath

=tex S_{\infty}=\frac{a}{1-r}

[limber heath]

that bot is cool

does it use php web image generation or something like that?

[elfin night]

Idk sends it off to some server to interpret it as LaTeX (I'm not a CS person ;p)

[limber heath]

I only ever did cs in unity

every language is very similar (for the way i see it)

[wind igloo]

I have a similar view, but I also have a suspicion that my position is naive.

[pearl peak]

is anybody good with physics here?

[wind igloo]

[prisma scroll]

that's what he did though

he asked if anybody is good with physics xD

[fickle moat]

inverse function of x^5-2 is x^5+2?

[bold hatch]

the inverse of a function should get you back to x, when you apply it to f(x)

[fickle moat]

so you dont switch out - to +?

[bold hatch]

a useful way of thinking about it is, if you imagine f(x) as performing a series of "steps" on x, the inverse function is reversing those steps

[fickle moat]

i am bit lost..

[bold hatch]

so let's call the function f

f(x) = x^5 - 2

what does f "do" to x?

[fickle moat]

you divide?

i mean i get the basic (a,b) inverse is b.a

[bold hatch]

right

so currently the function is (x, x^5-2)

the inverse would be (x^5 -2, x), but we want some way of describing it in the form (y, f^-1(y))

if i told you to calculate f(2), how would you do it?

[fickle moat]

x^5-2 solve 2?

plug x into 2?

[bold hatch]

yeah

[fickle moat]

2^5-2?

[bold hatch]

ok, what's the first step?

if you want to calculate the number

[fickle moat]

the thing that i dont get how back of the book gives the answer as f-1(x) = 5root x+2

i m just lost...

[bold hatch]

we're getting to that, don't worry

[fickle moat]

ok s 2^5-2

30?

[bold hatch]

right, so when you're calculating it, what's the first thing you'd do?

yep how did you get that?

[fickle moat]

2^5 = 30

32""

32-2 = 30

[bold hatch]

right

so we raised 2 (the input) to the fifth power, then subtracted 2

if we generalise, how would we calculate f(x) for any given x?

[fickle moat]

the way you explain makes sense but i am still lost

[bold hatch]

ok well so to calculate f(x) you have to raise x to the fifth power, then subtract 2

does that make sense so far?

[flat blade]

is this the place to ask about matrices?

[fickle moat]

can i ask you something actually can i take picture of note and follow this step ?

is this samething as x^5-2?

the concept of transition i am confused i mean when you solve f o g (x)

f-1(x)?

[bold hatch]

you can do the same thing as that, yeah

in both cases, what you're doing is "going backwards"

[fickle moat]

so for x^5-2 i can not use backwards

[bold hatch]

you can

=tex x = y^5 - 2

[bold hatch]

then solve for y

[fickle moat]

how did you make mathbot do that

ok

um

[bold hatch]

"=tex" lets you type in latex

[fickle moat]

interesting;; ok back to questions thank you

wait for f(x) you changed it to y?

[bold hatch]

yeah so for the inverse we need:

=tex x = f^{-1}(y)

[bold hatch]

erm, i guess that's a bit confusing

when we swap x and y, you solve for y, but the y is really the x because we swapped them around

[fickle moat]

bout to slam my head into wall

i am sorry i am really lost

i get x^2

but x^5..

even you root it u have to y^2 = 5rootx-2

[bold hatch]

hmm ok let's walk through it

we've got

=tex x = y^5 - 2

[bold hatch]

we're solving for y

let's add 2 to both sides

=tex x+2 = y^5

[bold hatch]

can you see how to solve for y from here?

[fickle moat]

x+2 = y^5

idk how to read or write perfectly but the equation should be both side 5root and cancels the ^5

[bold hatch]

right, yeah

[fickle moat]

leaves y = 5root x-2?

[bold hatch]

=tex y = \sqrt[5]{x+2}

[fickle moat]

where did that +2 came from

we started as -2

so inverse is 2?

[bold hatch]

we did, but we added 2 to both sides when we were solving for y

[fickle moat]

so not just only we effected the x and y switch the real number change too?

OH

OH

OKOKOK

[bold hatch]

hahaha 👌

[broken mirage]

If f(x)= (x+6) (x−3)x+k and f(1)=13then find k.

what is k?

[hoary yoke]

Fucking plug it in

[gritty blaze]

k = 27.

[hoary yoke]

@gritty blaze do you know what spoils a kid?

[gritty blaze]

No.

[hoary yoke]

Yep

[fickle moat]

Y= |x-5| inverse function ? How do you cancel out abs?

My question is how do you get x=y-5?

I got y= x+5

[zealous shell]

When you have y = |x - 5| , that is essentially,

y = x - 5 for x > 5
y = 5 - x for x < 5

[fickle moat]

X= y-5?????? How

....

[zealous shell]

Wait, sorry, I was somewhere else

[fickle moat]

I don't ger it. Inverse function says x=y-5

[zealous shell]

So you are trying to invert y = | x - 5 | ?

[fickle moat]

Yes

Inverse function

[zealous shell]

y = x - 5 for x > 5

Think about this function

How would you go about inverting that

[fickle moat]

U solve fir xn

[harsh pollen]

hi one question where can ask for help in inverse power law calculation 😅

[zealous shell]

Yeah, so you basically switch x and y

So you get an inverse for a particular range of y

[fickle moat]

Then it wouls be 5+x=y its not x=y+5

[zealous shell]

So you have y = x + 5 as an inverse in the range y > 5

I'll get a graph, that'll help

See, essentially, in the end you'll have the inverse turn out to be x = | y - 5 |

[fickle moat]

Someth8ng like this correct?

[zealous shell]

Yeah, that's basically it, to invert a function in two variables, you can just switch the variables with each other

[fickle moat]

What do you mean by variables

[zealous shell]

x, and y are variables

Something which is not constant, it's what you use to define functions so that they are easy to understand

[fickle moat]

Still lost...

So do you cancel out abs -5 +5?

[zealous shell]

Let's start from the basics

what is the inverse of y = x * x

[fickle moat]

X=y*x

[zealous shell]

See, here, if you wanted to take the inverse of a function which squares what you put into it

You would want to take the square root

[fickle moat]

Root of x = y

[zealous shell]

Exactly

You can arrive at that in the same way, by changing x with y and vice verca in y = x * x

[fickle moat]

So when you inverse abs it cancels out?

[zealous shell]

You get x = y * y

Which gets you to y = sqrt(x)

Which gives you the inverse

In your case you are starting with y = | x - 5 |

So the inverse is x = | y - 5 |

I'm not really good at explaining things, if that still doesn't make sense, I suggest asking in the questions channels, and someone else can probably help out better

[teal fiber]

Alright I’m having trouble with this
I’ve already completed the square on this equation so now I have
x^2 - 2x + 2 + y^2 = 5
But I need to put it in radius form to find the center and radius of the circle it makes
I eventually gave up and went to desmos.com and put it in and that told me the center of the circle is (1,0), so presumably that would mean the (x - h)^2 part would be (x + 1)^2, but that does not yield x^2 - 2x + 2

[brave void]

x cot 2x

As x approaches 0

[viscid thistle]

@teal fiber (x-1)^2 + y^2 = 5

[brave void]

I know cot is 1 over tan

or cos over sin

<@&286206848099549185>

[teal fiber]

But (x - 1)^2 gives x^2 - 2x + 1

[viscid thistle]

= 4 sorry

[brave void]

Rip

[teal fiber]

How’d you get it equal to 4? Completing the square on the original = 3 makes you add 2 to 3 giving 5??

[viscid thistle]

@teal fiber just subtracting 1 from both sides

[teal fiber]

Ohh, the 1 you get after doing (x - 1)^2?

[viscid thistle]

yes

[teal fiber]

Ok, thank you tons

[teal fiber]

Just realized I’m an idiot and made that problem way harder, 1^2 isn’t 2 🙄

[steady token]

How do you find relative maximum and minimum on an ti84?

I'm so confused

<@&286206848099549185>

[slender river]

second calc

then there should be a min max function in the menu

then itll ask for left and right bounds

and then itll brute force compute and give you the min/max you need

[steady token]

when i tried to find the min for y=(x^2-4)^2 it gave me ( -2 , 2.722E^-11 )

[rocky bison]

What if he needs it in exact?

Brute force won't give that

And that is actually just wrong

Wait no it's not

It's giving you one of them @steady token

But y is just 0

[steady token]

i guess that number is very close to zero

[rocky bison]

Yea

That's the problem with brute force

Like if it was something less obvious

Like root3

You won't notice it

[steady token]

yeah, i guess thanks for helping

[rocky bison]

Find it the good old fashioned way with differential

[rare zephyr]

Uhh I don't know if I'm in the right chat but

Anyone can help me with 10 and 11?

<@&286206848099549185>

[fringe stream]

You're in an appropriate channel.

What have you tried?

[rare zephyr]

For no. 11 I tried SOR and POR. Stuck after c/a

I didn't answer the second question in 11 yet

In no.10 the answer wants no coeffiecient x in any of the terms

I tried using the substitude method. Ended up wring multiple times

Also tried to make sense with it using SOR and POR. Tangled lines is the result

[fringe stream]

So for question 10.

If one of the roots is a, the other is 2a.

Sum of roots = a + 2a = 3a = -q/p

Product of roots = a*2a = 2a^2 = r/p

$$3a = -\frac{q}{p}$$\$$2a^2 = \frac{r}{p}$$

[fringe stream]

You can rewrite the second equation.

$$r = 2a^2p$$

[fringe stream]

And the first one too.

$$a = -\frac{q}{3p}$$

[fringe stream]

Then sub that into the first one.

[rare zephyr]

Anyone mind helping me with question 11.? <@&286206848099549185>

[fringe stream]

It's the same logic.

Use the given equation to find equations in alpha, beta, m and n.

Then solve those.

[north bough]

When graphing a tangent function, how do you show the vertical dilation on it? For instance, how do I show the difference between tan(x) and 2tan(x)?

[hexed ermine]

uh just use some points

notice that with tan(x) 2tan(x), they are both 0 at x=pi, so just use some example x values to show the comparison before and after the dilation

notice that at x=pi/4, tan(x)=1 and 2tan(x)=2

[sacred falcon]

Aaah doesnt looking at words that you cant even begin to understand feel so refreshing?

[sullen shoal]

Don't

[brave void]

If a limit goes to positive or negative infinity
Like I got
5x -3 + (10/x^2)
How would I know if it went pos or neg inf

<@&286206848099549185>

[fading token]

try values that get closer and closer to the vertical asymptote

[brave void]

well like

What about this one

5x^2

Over x+3

[fading token]

Alright

[brave void]

x goes to -infinity

[fading token]

oh

I thought you were trying to move closer to the asymptote

But instead you want the variable to go to -infinity?

is that correct?

[brave void]

Thats the problem

[fading token]

oh ok

That's easier actually

you just put in a negative number that has ridiculously large magnitude

like -100000000000

so then you get 5*x^2 which is very large and positive

divided by x + 3 which is still a ridiculously large negative number

positive/negative = negative

[brave void]

oksu

cool

thx m8

[fading token]

🍮

[brave void]

i love me some custard

Last thing @fading token

So I got this 2x + 1

Over root (x^2 - x)

x approaches - infinity

how would I solve

[fading token]

I would first divide top and bottom by x

[brave void]

x^2?

[fading token]

No, just x

[brave void]

okay

so 2 + (1/x)

But what about denominator

[fading token]

sqrt(x^2 - x)/x

But we can simplify this

x = -sqrt(x^2) if x is negative, which is important because we're going to -infinity

[brave void]

ok

[fading token]

so sqrt(x^2 - x)/(-sqrt(x^2)) = -sqrt((x^2 - x)/x^2) = -sqrt(1 - 1/x)

[brave void]

so let x = - sqrt(x^2)

[fading token]

yeah we rewrite it like that

[brave void]

okay

so we got this fraction

Now we can take x from the numerator of that

[fading token]

What do you mean?

[brave void]

That fraction simplifies to -((x-1)/(x))

wait that doesn't work

I don't fcking know

please help xD

[fading token]

-(2 + 1/x)/sqrt(1 - 1/x)

yes?

[brave void]

yes

[fading token]

you don't have to simplify

you can now take the limit x goes to -infinity

1/x terms will be very very small

so in the limit you get -2/sqrt(1) = -2

[brave void]

so 1

[fading token]

Hmm?

[brave void]

yeah

-2 over -2

[fading token]

No

-2 over 1

[brave void]

-2

[fading token]

yes that's the value of the limit

[brave void]

okay

so then if we just had x in the numerator

For another problem

and we had the same denominator

I do the square root thing and simplify?

[fading token]

yeah

[brave void]

gotcha ty

[fading token]

sqrt(🍮^2)

[obtuse fulcrum]

ehhh not sure how to go about this using this v / s concept... can anyone help?

[rocky bison]

It's velocity displacement

not velocity time

[viscid thistle]

Oops

[rocky bison]

@obtuse fulcrum

Where do you think the point of highest acceleration is

oh it says minimum

oops

but where do you think the minimum acceleration is is

[main shore]

Hellllllllllp

Plzzzzzzzz

[smoky delta]

Lol I think the person who said he'd help you is coming

[faint acorn]

Alright

[main shore]

Hopefully

[faint acorn]

So it has zeros at -2, 0, 1 and 3

Which means when you substitue any of those number into the polynomial you'd get zero

which means (x - 3) for example is a factor since if you substitute in 3 no matter what the other factors are you'd get 0

[main shore]

Tru

Yeh correct

[faint acorn]

So it would have the facotrs (x+2), x, (x-3), etc

So try writing out those 4 factors and multiplying them together

[main shore]

Ok

Hey

[faint acorn]

I'm gonna guess the leading term isn't 2x^4

[main shore]

I finished but I can't get the 2 infront of the leading term

Yeh

Lol

[faint acorn]

Ok so that means one of the xs must be a 2x

Which means you'd have to change it to (2x - a) but one of your factors is just x so if you change that to a 2x it'll still equal 0 at X=0

[main shore]

Yeh

[faint acorn]

And will multiply out to a 2x^4

[main shore]

Do you know how I can go about doing this?

[faint acorn]

(2x)(x+2)(x-1)(x-3)

That satisfies the 4 zeros given and the 2x will become a 2x^4 term once expanded

[main shore]

Omg you're right

[faint acorn]

Since if you multiply 2x by each of the first terms of the factors you'd get 2x^4, and then any other terms will have a lower degree

[main shore]

Yeh

Thanks

[faint acorn]

So that's generally how you solve that type of question. List out the factors and then modify it to make the term they give you a part of it

[main shore]

Yeh I think I understand now

Legend

[fading token]

👏

[viscid thistle]

Cool

[valid vector]

Is this correct?

[gritty blaze]

No.

[limpid plover]

$$2y^2-y-10=0$$

[limpid plover]

$$2y^2-5y+4y-10=0$$

[limpid plover]

You individually take common 2y and 5 to get the factors

[valid vector]

Oh right, fuck I forgot!

Thanks!

[charred rune]

@limpid plover im not realling understanding the concept of it

[limpid plover]

You need to expand it first

Then do exactly what you did in the first one

[charred rune]

ok what do u mean by expand?

the fraction their is whats confusing me msotly

mostly*

[limpid plover]

=pup expand (-1/2)(x-4)^2+3

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@limpid plover

[limpid plover]

Basically, multiply everything, and add/subtract the stuff

[charred rune]

oh ok

[opal belfry]

so for this quad equation you get x=4 and x=-3/2

[slender river]

yes.. and?

[opal belfry]

but why do they switch the negative symbol and positive for the two when factoring in the form??

[slender river]

oh

wym where

[opal belfry]

the original prob was 2x^2-5x-12 < or equal to 0

at the bottom

x=4 turns into x-4

and -3/2 turns into x+3/2

[slender river]

yeah

they flip the signs because those are just zeroes

put back into factored forms

i dont understand the actual location of the confusion tho

perhaps im missing something

[opal belfry]

i guess ill just remember the process

ty tho

[slender river]

ok

[opal belfry]

another thing if you don't mind

do you know the table for inequalities pretty well?

[slender river]

t a b l e?

sorry for delay

was travelign

[opal belfry]

No rush

This is the table in question though. Just having a hard time understanding the process of constructing it

[slender river]

woah

i have never seen anything like that in my life

[opal belfry]

And I’ll probably never ever see it again hopefully

[slender river]

hopefully

erm

[opal belfry]

No prob man

[slender river]

i dont really know what to do with it

sorry

[opal belfry]

I’ll figure it out eventually

[slender river]

ok

good luck :)

[spring thunder]

@opal belfry got it?

[night plover]

Can someone explain question D to me

Even with answers i dont get it lol

[gentle hedge]

what do you not understand>?

[night plover]

So they add everything together to make 3pq

And the step after that is taking P away and multiplying by P

And then they just add

idk it just seemed rly difficult to me lol

[gentle hedge]

what?

[hearty plinth]

Topic :hyperbola
3x^2-2y^2-42x-16y=-67

What dis

[fading token]

@night plover Nederlands?

[night plover]

@fading token yep 😃

[fading token]

Nog hulp nodig?

[opal belfry]

@spring thunder sort of. instead of using the box I'm just going to check them w/ the equation by plugging in

[valid vector]

In this problem: (x - 2)^2 + (y + 1)^2 = 4

If I wanna use the table of values to draw the graph of this equation, do I have to solve the equation before substituting the values for x so that y ends up alone on one side of the equation?

[limpid plover]

Nope. But sometimes, simplifying can make it easier.

It may not here though

[valid vector]

OK

[viscid thistle]

Someone please provide me some valuable erudition on this fine contest question here

Pardon

Isn’t NSFW content prohibited here?

Unfortunately, I’m not quite satisfied with a final answer

[limpid plover]

Boi.

@viscid thistle Ask questions in the question channels below

[viscid thistle]

Hey, I was just extrapolating from previous messages and assumed I could ask here

Oh well

[limpid plover]

This is more for discussing stuff and not solving homework questions

[severe swift]

What is that called? Ignore the limit.

I can't find anything on Khan Academy to help me factor something like this

There isn't any examplea under quadratics or factoring polynomials

Sorry realised I need to post in questions my bad

[gentle hedge]

@severe swift

u dont even need to divide tbh

manipulate it so you can see where each term converges to

[viscid thistle]

Limits at infinity you can take the highest powered term in the top, and divide it by the lowest term in the denominator

so -5t^4 / 3t^4

@severe swift

This can be simplified

-5/3

that is the answer

The theory is that, whenever you get to infinity, -5(infinity^4) will grow MASSIVELY more faster than infinity^2 in the next term

Thus; the t^4 dominates

in bot top and bottom above all other terms

[spring thunder]

@viscid thistle first try to express the volume and surface area of the container (in function of height and radius\diameter)

$$A=\pi r^2 +2\pi r h \newline V=\pi r^2 h$$

[spring thunder]

ie $$\pi r^2 h = 25$$

[spring thunder]

And you want the volume of the container to be 25 ft^3

ie $$\pi r^2 h = 25$$

[spring thunder]

Sorry my connection is shit

[dusky vale]

rip

[spring thunder]

The idea is to get the height in function of the radius, so that we can get an expression for A which depends only on the radius, in order to differentiate it

Here, $$h = \frac{25}{\pi r^2}$$

[spring thunder]

So let's inject this expression into the first eq

$$A=\pi r^2 + 2 \pi r \left(\frac{25}{\pi r^2}\right)$$