precalculus

[viscid thistle]

:p

[willow bear]

thank you

[tawny stag]

https://gyazo.com/5f798ec5ad7a3af03b169bef410102c3

you turn tan

into sin/cos right

[willow bear]

that makes sense to do yes

[tawny stag]

then you make the demoniator the sane

[willow bear]

...you don't even need to do that

[tawny stag]

multiplying cos to 4sint

oh

[willow bear]

=tex 2\cos(t)\cdot 4\sin(t) - 2\cos(t) \cdot 3\frac{\sin(t)}{\cos(t)}

[willow bear]

distribute

and simplify, which i hope it's now obvious how

[tawny stag]

ohh okay i see it now

i have a question why is it -2cos*3 sin/cos

mostly on why -2cos

[willow bear]

uhh

because before the simplification, you had 2cos(t) multiplied by the parenthetical? i'm surprised you're asking that

[tawny stag]

i got it im sorry i confused myself sometimes

[viscid thistle]

^^ damn i am not that far in math yet

never seen that before, but i like it

https://gyazo.com/5f798ec5ad7a3af03b169bef410102c3

that what i meant

[tawny stag]

I don't

It's so confusing maybe I'm just slow

[still yew]

I have a random question, does anyone but me know the answer of $$i^i$$ and did the work?

[patent beacon]

Ya

[still yew]

K

[mental wedge]

hey

did i do this correctly ?

"Find all 8th roots of 1 and graph them on the complex plane. Be sure to show all work."

[red swan]

it seems good to me

[mental wedge]

ok

thank you

[willow bear]

yup :+1:

[mental wedge]

thanks

[timid jacinth]

https://gyazo.com/65adb10eb6fc07dd3a2a6e9f71ac71f0

not understanding the vertical asymptote rule

what does it mean the lines x = a

[willow bear]

vertical lines with x-coordinates equal to the zeroes of the denominator

[timid jacinth]

so if I factor the denominator and find the zeros, I've found the vertical asymptote?

[willow bear]

there need not be just one

also, that picture is missing an important detail

[timid jacinth]

what

[willow bear]

in order for x = a to be a vertical asymptote, a has to be a root of the denominator but not the numerator if the fraction is reduced to lowest terms

otherwise, you can make something like (x^2 + 3x)/(x+3), which has a hole at -3 rather than a VA

[timid jacinth]

why does it have a hole?

[willow bear]

because f(-3) itself is not defined, but the limit of f(x) as x approaches -3 exists (and is equal to -3)

(where f is my example function ofc)

and like, f(x) = x whenever x != -3

so it's like you've taken the graph of y = x and removed the point at x = -3, leaving a hole

[timid jacinth]

Ah I see

I kind of get it

I think I'm about to be introduced to what you're talking about

[timid jacinth]

This is a dumb question, and I can do it normally, but for some reason I'm having trouble factoring 2x^2+7x-4

how should I go about this

[red swan]

use the discriminant

[timid jacinth]

that seems a bit lengthy though

the discriminant is

-b sqrt(b^2-4ac) right?

I'd like to just look at it and be able to figure it out in like 10 seconds

[red swan]

that is one of the solutions, not the discriminant

[timid jacinth]

what's the discriminant

[red swan]

(and you forgot a sign)

b²-4ac

[timid jacinth]

so the discrim is just this b²-4ac

[red swan]

yes

[timid jacinth]

so with my problem, I'd do 49-4(2)(-4) = 49-32

=17

[red swan]

nop thats wrong

[timid jacinth]

where'd I go wrong

[red swan]

a sign

[willow bear]

49 - 4 * 2 * (-4) = 49 + 32

@timid jacinth

[timid jacinth]

oh yeah

so 81

so now what do I do to figure out the factors

[willow bear]

well i mean, the lazy way would be to just plug everything into the quadratic formula

get the roots

rewrite original expression as 2(x-a)(x-b), where a and b are your roots

[timid jacinth]

so what is the discriminant even showing me

what relevance does 81 have

[willow bear]

the 2, of course, is taken from the original leading coefficient

81 > 0 so you know your thing is factorable in the first place

[timid jacinth]

here's how I normally factor stuff, it's called the ac method

[willow bear]

it's also a perfect square, meaning the factors will be "nice"

[timid jacinth]

https://gyazo.com/3fe84a0fcfe975fb0fa7d5834625fa3c

so I'd normally try 2x-1 and x+8

[willow bear]

(2x-1)(x+4), surely?

[timid jacinth]

yeah

I'm trying to figure out

why my method isn't working

a*c = -8

then you figure out what multiplies to ac and gives you the b term when added together

-1+8=7

[willow bear]

i feel like that may not really work for a != 1 but i'm too asleep to try and come up with an Algorithm™

[timid jacinth]

ah

I think you're right actually

oh wait

[willow bear]

like, there's nothing wrong with simply finding the roots using the QF

[timid jacinth]

https://i.ytimg.com/vi/AYkaCZUT4O4/hqdefault.jpg

yeah but that always takes me forever

I'd like to be able to just look at it and figure it out in like 10 seconds like I normally do using the ac method

do you see why I'm confused now?

nvm

I'm really dumb

forgot a whole step

you have to group if a != 1

so 2x^2+7x-4 = 2x^2-1x+8x-4

(2x^2-1x) (+8x-4)

x(2x-1)4(2x-1)

(x+4)(2x-1)

[clever inlet]

looks right

but for the second last line

i would seperate with plus signs

[timid jacinth]

so let's say I get really stuck with factoring and I want to use the quadratic formula

to figure out the roots

https://gyazo.com/538c2c91a1f739924bee37c9bf05dc44

where did I go wrong?

oh wait

hmm

@clever inlet

by using the quadratic formula on the same polynomial, and I want to find out it's factored form, what do I do?

what is the 1/2 and -4 telling me?

[clever inlet]

your roots

[timid jacinth]

so how do I use these numbers to find the factors

[clever inlet]

well the second one is the easy one

x = -4

x + 4 = 0

wait

[timid jacinth]

ohhhhh

ok

[clever inlet]

and

2x - 1 = 0

that's just x = 1/2 right?

[timid jacinth]

yeah

but

where did you find the 2x - 1 = 0

[clever inlet]

well

x = 1/2

multiply through by 2

2x = 1

2x - 1 = 0

[timid jacinth]

ah

ok

so

when I got x = -4

from the quadratic formula

and I want to know the factor

what's my next step

x = -4

x + 4 = 0

this is the factor

but

I'm not sure how you're just coming up with it

once I find the -4, should I be telling myself, what will make this into a 0?

[clever inlet]

well you know intercepts form?

(x-a)(x-b) = 0

[timid jacinth]

sounds familiar

[clever inlet]

so

a and b are roots

[timid jacinth]

ok

so I'm right in saying once I find the -4, should I be telling myself, what will make this into a 0?

[clever inlet]

you can think of it like

(x-(-4))(x-b) = 0

(x+4)(x-b) = 0

[willow bear]

(x-(-4))(x-b) = 0

[clever inlet]

ah whoops

yeah

[timid jacinth]

and what is the b term

the other half of the quadratic formula?

so the 1/2?

[willow bear]

well, yes

[timid jacinth]

so I have (x+4)(x-1/2)=0

[willow bear]

and you'll also need to put your leading coefficient in front afterwards

[timid jacinth]

so my factors are (x+4)(x-1/2)

but they want (2x-1)(x+4)

[willow bear]

and you'll also need to put your leading coefficient in front afterwards

[timid jacinth]

hmm

how would I know that

why isn't x-1/2 acceptable

[clever inlet]

Expand it out

And while the roots are the same as the original quadratic

Its a different quadratic

[timid jacinth]

I'm a bit lost

https://gyazo.com/23e6f8dbff8b38f2a495c813be3e0189

I'm not understanding where that lead coefficient of 2 is coming from

[clever inlet]

What's a in your original quadratic?

[timid jacinth]

2

[clever inlet]

So if you expand it out with (x-1/2)

You'll only end up with x^2

Not 2x^2

[timid jacinth]

so I just have to go back to my original quadratic 2x^2+7x-4

and pull the 2 from the front?

that seems so arbitrary

[clever inlet]

You're basically taking the 2 in the denominator and putting it in front of x

[timid jacinth]

the denominator of the quadratic formula? The 2(a) or the 2(2) in this case?

[clever inlet]

Of

(x-1/2)

[timid jacinth]

Wow I'm really lost

again, that seems arbitrary

we're talking the 2 from (x-1/2) and just sticking it at the front?

what's the concept or math behind that?

[clever inlet]

Its the same as multiplying through by 2

[timid jacinth]

we're literally just multiplying the x-1/2 by 2

to get rid of the fraction?

Why is my question I guess?

because it's literally the exact same

2x-1 = x-1/2

right?

[clever inlet]

Yes

[timid jacinth]

so what's the point

[clever inlet]

well looks at the example of say

y = x^2 + 2x +1

and

y = 2x^2 + 4x + 2

same root right?

but different curves

actually that's a meh example

hmm

[timid jacinth]

so

the top one is (x+1)(x+1) right?

[clever inlet]

Yes

[timid jacinth]

how do I even factor the 2nd one lmao

the ac method doesn't work on it

I'd have to use the quadratic formula on it

[clever inlet]

2(x+1)(x+1)

[timid jacinth]

how did you work that out?

like what method did you use, out of curiousity

[clever inlet]

Factor out 2

[timid jacinth]

because for me, it's either the ac method of I'm doing the quadratic formula

[clever inlet]

2(x^2 + 2x + 1)

[timid jacinth]

that's factoring out a GCF though right?

[clever inlet]

Yeah

[timid jacinth]

but

there's a +1 at the end there

how are you pulling a 2 from it

[clever inlet]

Oh whoops

Did I write 1

It's supposed to be a 2

[timid jacinth]

ok

gotcha

if it was a 1 for example, it wouldn't be factorable right?

[clever inlet]

Yeah

[timid jacinth]

alright

so we have (x+1)(x+1)

so I see that if we don't stick that 2 from the original quadratic up front, then when we multiply the factors our we'd have a different quadratic than what we started with

[clever inlet]

Yeah

[timid jacinth]

https://gyazo.com/3c3e2404a5326e34a4c801870a8c106a

Where are they getting these additional values from

cause when I plug in -1 to the r(x) = 2x^2+7x-4 / x^2 + x - 2

I'm getting something completely different

ah nvm

[clever inlet]

are you doing your substitution correctly?

[timid jacinth]

I missed a sign

or a few

why did they pick those x values though

they seem a little obscure

[timid jacinth]

https://gyazo.com/d4c5ab25cb84a7f92ab2efc924dc4d34 why do they pick these numbers for additional vlaues?

@willow bear Are you available?

[willow bear]

these look arbitrary

[timid jacinth]

is it because they're inbetween the two verical asymptotes?

because if they didn't pick -0.5, I never would have known how far the graph of inbetween the two asymptotes goes down

[willow bear]

seems so

[timid jacinth]

or how deep the U is so to speak

[willow bear]

if you wanted to figure out exactly how deep it is you'd need calculus

[timid jacinth]

ah

so it's more of an estimate

[willow bear]

p much

[timid jacinth]

Ok great, and for this one could you clear up the domain for me

they've got the domain as x | x is < 1/2 or x > 7.5

first, shouldn't it be y, not x?

sorry

I'm talking about range!

[timid jacinth]

Anyone around?

[willow bear]

first, shouldn't it be y, not x?
formally, it doesn't matter

{ x | x < 1/2 or x > 7.5 } and { y | y < 1/2 or y > 7.5 } describe the exact same set

[timid jacinth]

So I did my exam, I got a 97/100 but the questions I struggled with was finding x-intercepts

E.g finding the x intercepts of a polynomial likef(x)= 6x^2-3x+7

[willow bear]

"x-intercept" here is a synonym for "zero"

[timid jacinth]

So I should have just factored?

[burnt vigil]

When it crosses the x axis the function equals 0

So you want to find the values of x that make the function equal 0

[willow bear]

@timid jacinth you should've used any method you wanted for finding the roots of that polynomial

be it factoring or using the QF

[timid jacinth]

Ah, so simple. So if a function is presented in standard form, should I rewrite it in ax^2+bx+c form before factoring?

[willow bear]

define standard form?

[timid jacinth]

@willow bear for example, if I'm given a polynomial on ax^2+bx+c form and told to put it in a(x-h)^2+k form

And later, I'm told to find the intercepts of a polynomial in the form a(x-h)^2+k, would I want to convert it back to the original ax^2+bx+c form?

[willow bear]

not necessarily?

i mean there's no Formal Requirement™ on which form you Have™ to use

it's possible in both

i'd wager it's even easier in vertex form since that's basically a completed square

[timid jacinth]

How do I do it in vertex form?

[willow bear]

a(x-h)^2 + k = 0
a(x-h)^2 = -k
(x-h)^2 = -k/a
x-h = ± sqrt(-k/a)
x = h ± sqrt(-k/a)

nothing complicated honestly

[unique scarab]

Why is the square ±

[willow bear]

because (-t)^2 = t^2

so an equation like t^2 = s necessarily has two solutions if s is positive

[hallow ember]

Does anyone know how to find the range of this

[clever inlet]

Do you know what the graph looks like?

@hallow ember

[hallow ember]

Yes

[willow bear]

well, what does the graph of this function look like?

[hallow ember]

it's a small u shape with x intercepts at 2 and -2 and y intercept at -2

[willow bear]

it's not just any ol' U shape

it's a semicircle

[hallow ember]

o yeah

[willow bear]

so it should now be obvious what y values it spans

[hallow ember]

yes

but

my math book says that y is less than or equal to 0

which confused me

[willow bear]

=tex -2 \leq y \leq 0

[hallow ember]

yes that's the one I came up with at first

[willow bear]

and that's correct

[hallow ember]

the book was wrong then

thanks for the help

[viscid thistle]

I forgot something i think

Is this correct ? $$ (a -bi)^2 = (a^2 - 2abi -b^2) $$ or this $$ (a -bi)^2 = (a^2 - 2abi +b^2) $$

[willow bear]

the first one

[viscid thistle]

Thanks Anto

== -196-96

[granite stirrup]

-292

[willow bear]

fyi, the bot can do things with complex numbers

== i^2

[granite stirrup]

(-1)+0i

[willow bear]

== (2+3i)^2

[granite stirrup]

(-5)+12i

[viscid thistle]

Okay thanks

😄

[viscid thistle]

is this correct ?

=tex z^2 - (5 -14i)z - 2(12 + 5i) = z^2 + (-5 +14i)z + (-24 - 10i) = 0

[granite stirrup]

Failed to parse equation: Invalid token at position 4

tex \begin{equation}     z^2 - (5 ...
    ^

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

[signal fable]

The equation logbx = logbx has two solutions in x whose sum is 10. What is the value of b?

HELP PLEASE @everyone

[viscid thistle]

log like natural log right?

[signal fable]

I sent it wrong

wait please

https://gyazo.com/a110dd8951baa30cb68abc2d94401c3d

Yes

btw

[viscid thistle]

but it's log base b

it's only natural if it's

log base e

[signal fable]

yeah

I mixed up

You think you can solve it?

I want extra credit so bad

This is a extra credit question

[viscid thistle]

i will try

[signal fable]

thanks habibi

[viscid thistle]

and also besides ln there is log that is base 10

called log

on calculator

[signal fable]

Idk bro

its an extra 10 points on the test

homework*

would it be that simple

[viscid thistle]

Ask someone else I have no clue how to solve

soz

[quartz garnet]

uh

=tex \log_b \sqrt{x}=\sqrt{\log_b x}. \
\Leftrightarrow \frac{1}{2}\log_b x=\sqrt{\log_b x}~\rightarrow~\sqrt{\log_b x}=t.\
\Leftrightarrow \frac{1}{2}t^2=t. \
\Leftrightarrow t=0,~2. \
\Leftrightarrow \sqrt{\log_b x}=0,~2. \
\Leftrightarrow \log_b x = 0,~4. \
\Leftrightarrow x=b^0,~b^4. \
b^0+b^4=10~\Leftrightarrow~b^4=9~\Leftrightarrow~b=\sqrt{3}.~(\because b>0)

salutes

[viscid thistle]

Why did you draw a conclusion in the end, im talking about the therefore

[quartz garnet]

what

[viscid thistle]

You wrote therefore b>0

[quartz garnet]

no, because b>0

[viscid thistle]

Oh did I mix up

[quartz garnet]

$$\therefore$$ = therefore, $$\because$$ = because

[viscid thistle]

soz

[quartz garnet]

@signal fable

[viscid thistle]

Physics is lucky tonight

@quartz garnet what is your main source for learning math?

[calm whale]

will he finally reveal his secret :?

[quartz garnet]

uh
by myself

[calm whale]

wherefrom

like you can't magically conjure the knowledge out of thin air

[viscid thistle]

Tell that to ramanujan

[quartz garnet]

My dad and my math academy, mainly

[viscid thistle]

says myself then says my dad

[calm whale]

lol

the legend of ramanujan

[viscid thistle]

Math academy?

What is that

school?

[calm whale]

something like khan academy

I presume

[viscid thistle]

oh

[calm whale]

a mooc (massive open online course)

[viscid thistle]

Do u use it

[calm whale]

I used khan academy

for all my math needs :^)

and physics

[viscid thistle]

Khanacademy is simple

Let's be honest

[calm whale]

khan academy is great

because its simple

[viscid thistle]

It does give knowledge

but it doesn't go complex

You can use khanacademy to learn the basics because you will need to learn the basics anyway khan just explains good

[quartz garnet]

uhm
I meant cram school lol

[calm whale]

was Ramanujan the one that tried and almost solved squaring the circle problem

🤔

[viscid thistle]

Is cram school like irl school

Where you go and they teach u

[calm whale]

cram school afaik are intense courses on a certain subject

or aimed at a certain test

SAT/ AP etc

[quartz garnet]

"Although the South Korean educational system has been criticized internationally for its rigorousness, it remains common for South Korean schoolchildren to attend one or more cram schools after their elementary school-day is finished. Some types of institutes include math, science, art, and English. English-language institutes are particularly popular."
🤔

[calm whale]

I'm not sure if he's us based or not

hm are you korean

[quartz garnet]

yeap

[calm whale]

interesting

inserts cringy K-pop/drama lol

[quartz garnet]

ehw

[viscid thistle]

Says myself then lists 2 sources in which he interacts with other people

[quartz garnet]

get that away

[calm whale]

eww > ehw

[viscid thistle]

Is number theory hard?

[quartz garnet]

sure

[viscid thistle]

did u do it

[quartz garnet]

a little bit maybe

[calm whale]

why are you asking 🤔 ?

[viscid thistle]

because I am going to start learning it soon

[calm whale]

what are you doing?

[viscid thistle]

?

[calm whale]

high-school ?

[viscid thistle]

middle

8th grade

[calm whale]

what grade is that

oh

hmmm 🤔

[viscid thistle]

basically HS

[calm whale]

khan academy really is great

[signal fable]

thank you habibi @quartz garnet

[calm whale]

especially for middle/highschool

what do you mean by number theory?

[viscid thistle]

not sure

All I learned so far to get prepared for it was logic and set theory

[signal fable]

thank you too

[quartz garnet]

no problem physicsc

[viscid thistle]

Khanacademy is made for people to think they are good at math while it is just teaching basic stuff

[ocean cove]

could you justify your first claim

[calm whale]

I must disagree D

[ocean cove]

i dont think anyone's going on khanacademy to boost their ego. rather, they're willing to learn new things

[calm whale]

exactly

[viscid thistle]

i agree

[calm whale]

also some of the things on khanacademy are far from basic

[viscid thistle]

If you learn it step by step it is not complicated really

[calm whale]

if you learn anything step by step

it doesn't seem complicated

[ocean cove]

does the difficulty of the lessons matter? if you're learning more advanced stuff, it's not a bad thing to have it explained simply

[viscid thistle]

Not saying khanacademy is bad

It's great

I do most of my math there

[calm whale]

number theory: a branch of pure mathematics devoted to the study of integers

I'm confused cuz im studying divisibility theorem under discrete mathematics

🤔

along with induction and some other methods of math proof

soo number theory I guess would be

pythagorean triplets

[quartz garnet]

Cus that helps solving some problems of discrete mathematics

[calm whale]

along with fermat's last theorem

[quartz garnet]

Not only that lol

[calm whale]

all the sequences

[quartz garnet]

Things like calculating the remainder when \
$$\Large 2004^{2003^{2002^{2001^{2000^{1999^{\cdots}}}}}}$$ \
is divided by 1000 and stuff

[calm whale]

might run into diophantine equations

[quartz garnet]

that's also a huge part of number theory

[calm whale]

how old are you?

[viscid thistle]

Turned 14 today

you?

[calm whale]

18

[viscid thistle]

my bd today

[quartz garnet]

heh

happy birthday x'D

[viscid thistle]

ty

[calm whale]

h birthday

for middleschool you would look at divisibility

dunno

@quartz garnet what's calc I?

is it just normal calc?

[quartz garnet]

Calc I is uhm
learning how to take derivatives, identifying continuity and differentiability, learning how to take the definite/indefinite integral, and applying it to various situations
in korean standard, calc I only deals with polynomials

[calm whale]

you don't do parametric equations

or polar cordinates?

[quartz garnet]

Parametric equations are also dealt with
However polar coordinates is not in the high school curriculum over here

[signal fable]

Boi are you in highschool?

[quartz garnet]

Hehe take a guess

[signal fable]

No

Im guessing you are not

But you may be my friend Jacob

either or

[quartz garnet]

Okey then I'll stay with that =w=

[signal fable]

are you jacob?

shit dude I didnt know you were teaching kids here your genius

[calm whale]

jacob? :
https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

[signal fable]

No

jacob my freind

friend

Idk anymore

[quartz garnet]

Wait what

Who is this jacob x'D

[signal fable]

IS YOUR NAME JACOB?

xD

[quartz garnet]

n o

[signal fable]

oh okay then xD

[calm whale]

gnight

[ocean cove]

o/

[quartz garnet]

G'night!

[viscid thistle]

Bye bye

[willow bear]

i wouldn't call that precalculus

not in the American sense anyway definitely

[quartz garnet]

🤔 where are those from

[main flume]

0⁰ ≠ 0

👀

[willow bear]

damn you and your fancy Unicode symbols

[quartz garnet]

0⁰ = undefined

👀

[viscid thistle]

@quartz garnet

Desmos is w👁ke

lol

[willow bear]

👀

[quartz garnet]

Lol

[viscid thistle]

now, they do have a point however, its undefined but its a good idea to think about, its like the same idea as 0/0 equals 1

[quartz garnet]

0^0 = 1 conveniences many things, but that doesn't mean it must be defined

To be f o r m a l to the nth degree

For $$f(0)=g(0)=0,$$ the limit $$\lim_{x\to0}f(x)^{g(x)}$$ doesn't tend to any specific value

[quartz garnet]

Even though $$\lim_{h\to0}(1+h)^{\frac{1}{h}}=e,$$ we don't say $$1^{\infty}=e,$$ right?

[quartz garnet]

The same thing

[viscid thistle]

You are correct, I would believe so

however, I do find it funny how desmos does try to "define" it as one

not saying its correct but its funny

[frigid hearth]

I need help on a question

" Write the equation of the tangent function with a period of 3pi, phase shift of -pi/4, and a vertical shift of 2 "

I don't think it's possible because the " b " part for phase shift and the period is different :\

but I'm not sure

[floral rose]

so
f(x) = tan3π(x-(π/4)) + 2
?

if i recall correctly

[frigid hearth]

is 3pi the amplitude?

[floral rose]

err no

f(x) = tan [ 3π(x-(π/4))] + 2

can't really recall for sure

the amplitude would be multiplied to the tangent function, so it would have to be on the outside

[frigid hearth]

what would the stuff be in terms of y=d+atan(bx-c)?

because phase shift is c/b

[floral rose]

well i know for sure that d is vertical shift, a is amplitude

i thought c was simply phase shift, and i always think of b as "how curvy the function is"

[frigid hearth]

phase shift is c/b

so it confused me lol since the b found was different

[floral rose]

ah ok

b is factored out of the function

atan(bx - c) + d

see that b? it's factored to the outside to get
ab tan(x - c/b) + d

the resulting c/b that you have after you factor out is your phase shift

the period in this case, since its tangent, is pi/b

@frigid hearth

[frigid hearth]

wait factored to the outside??

[floral rose]

yes

what's troubling you

[frigid hearth]

wait how do u factor to the outside .-.

[willow bear]

errrr yeah

=tex a\tan(bx - c) \neq ab \tan\left(x - \frac{c}{b}\right)

[frigid hearth]

yeah im confused lmao thats what i though

thought*

[leaden karma]

Does anyone know much about professor leonard's courses

should I take his calc courses or TTP videos

if I want to get better at prcalc

[floral rose]

or rather

uh

atan [b (x-c/b) ]

that should be right this time, mb

[frigid hearth]

wait

but then that means

ohhh

okay i see what u mean

[floral rose]

yeah the "original equation" would be in the form a * tan(bx-c)

and if you just factor out that b (but still within the function itself) it's easy to see what the phase shift is

and then the period of your function divided by that b is the new period

" Write the equation of the tangent function with a period of 3pi, phase shift of -pi/4, and a vertical shift of 2 "

since the new period is pi/b, im assuming you're supposed to find b by equating it to 3pi

[willow bear]

y = 2 + tan([x-pi/4]/3)

[floral rose]

if thats not how, im lost 😛

[frigid hearth]

yeah so b should be 1/3 right?

[floral rose]

yes

anto i think that's a +2 there

[frigid hearth]

yeah its +2, not 2tan

[willow bear]

good point

i was intending to type that plus sign lol

[frigid hearth]

yeah so b is 1/3 and phase shift is c/b

lmaoo XD okay

that means phase shift is like c/4 .-.

which doesn't make any sense because the b for period is 1/3 and the b for phase shift is c/4

c would be positive pi

[floral rose]

not quite

it says phase shift is -pi/4

so c/b = -pi/4

you have b, solve for c

[frigid hearth]

ohh okayy

ty!!!

Lol i originally thought that -pi/4 is c/b XDDD

ahhhhhhhhhhhhhhhhhhhhhhhh

i realized my mistake now ty XD

[floral rose]

you're welcome!

though i'm not sure what you're on about, -pi/4 is c/b 🤔

[frigid hearth]

i mean like

i thought -pi was c

and 4 was b

XD

[floral rose]

oh, haha 😛

[obtuse salmon]

wait whats the difference between precalculus and calculus

[mental maple]

Generally the transition from precalc to calc is when limits are introduced

But it differs per country

[calm whale]

https://www.khanacademy.org/math/precalculus
check out calc I, II, III for the differences

[gleaming schooner]

Hi guys

I have a problem

=tex x^2+y^2+2x-6y = -7

[gleaming schooner]

I have to find the center and radius for the circle represented by that equation

So, I complete the square and get:

=tex (x^2 + 2x + 1) + (y^2 - 6y + 9) = 3

[gleaming schooner]

Is this correct?

[calm whale]

seems correct? -7+10 = 3 :^)

=tex (x+1)^2 + (y-3)^2 = 3

[gleaming schooner]

How did you get to that?

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

@calm whale

[clever inlet]

(x^2 + 2x + 1) factors to (x+1)^2

@gleaming schooner

Basically it's a perfect square since you made it to be

Then the 1 comes from half of b

[gleaming schooner]

So what would the radius of that circle be?

Would it be sqrt(3)?

@clever inlet

[clever inlet]

Yep

[languid wind]

i need to prove that cos(x + y)cos(x - y) = cos^2(x) + cos^2(y) -1

[calm whale]

=tex cos(x + y)cos(x - y) = cos^2(x) + cos^2(y) -1

[languid wind]

yes that

[calm whale]

which trig identities do you know?

https://gyazo.com/a7c950d5630354249f3b54c1c74df18e?token=8262f4a0a976002399fb1d5bd693b2fa

[languid wind]

i know the pythagorean identities, quotient, reciprocal identities

[calm whale]

are you aware of this one?

[languid wind]

cost(a+b) = cosabosb -sinasinb?

*cosacosb

*cos(a+b)

[calm whale]

=tex cos (a+b) = cos(a) cos(b) - sin(a) sin(b)

[languid wind]

yes

[calm whale]

yeah just do that

[languid wind]

alright

[calm whale]

and I'd assume you'd get it

[languid wind]

let me see

i got it

thanks

[viscid thistle]

Can anyone explain to me how log5 = 1 - log2?

[sour hemlock]

log2 = log(10/5)

log(10/5) = log10 - log5

log10 = 1; log10 - log5 = 1 - log5

so log 1og5 = 1 - log2

[viscid thistle]

thank youu

[languid wind]

A circle with centre at (0,2), radius 2, rolls along the positive x-axis. The centre
moves to the point (5,2). Point P, initially at the origin will roll along the
circumference, as the circle turns, to the point Q(x,y). What are the co-ordinates
of Q, accurate to two decimal places?

[calm whale]

draw it

plot it and ull see

[sour hemlock]

wait

almost got it

[calm whale]

how the fuck does a point at the origin roll along the circumference what

[sour hemlock]

is the answer (5.19875, 9.9 x 10^-3)

?

[calm whale]

to two dec places

[sour hemlock]

nvm I can tell you my method so you can try this too

so

our wheel makes 5 / (2xpix2) revolutions

[calm whale]

I'm just curious how does the point at the center roll along the circumference

😄

[sour hemlock]

which is equal to 0.398 in terms of distance traveled by point P on circumference

now you need to find the angle it forms to cover this much distance

So, (angle/ 360) x (2 x pi x radius) = 0.398

this will give angle = 11.408

Now you need to use geometric manipulation

You will have a sector with 11.408 angle

Find the length of the straight line from (5, 0) to (x, y)

aaaarrrrghhh it's so hard to explain here and my writing and calculation is messed up

You basically have to use simultaneous equations

Like the length from (5, 0) to (x, y) is one equation and the distance from (5, 2) to (x, y) will form another equation (This is where I made a mistake because I took 0,0 instead, dumb me). Solve these two equations to get x, y

[languid wind]

whatever

how about sec^4(A) - tan^4(A) = 1 + 2tan^2(A)

[calm whale]

what of it

tehres once again an identity

that will allow you to simplify that

[gleaming schooner]

@willow bear

[willow bear]

mhm

okay so

there's a template that all exponential graphs' equations follow

y = a * b^x

and in your graph, increasing x by 1 made y increase by a factor of 3/2

so b = 3/2

and from there it's kinda trivial that a = 1

[gleaming schooner]

By looking at it, how do I quickly identify that y increased by a factor of 3/2

[willow bear]

@gleaming schooner (27/8)/(9/4)

[gleaming schooner]

Isn't that 11/6?

@willow bear

[willow bear]

ahem?

how in the world are you getting 11/6 out of that?

[gleaming schooner]

Maybe I did it wrong

27/8 * 4/9

[willow bear]

yes

=tex \frac{27}{8} \cdot \frac{4}{9} = \frac{3 \cdot 9 \cdot 4}{2 \cdot 4 \cdot 9}

[gleaming schooner]

108/72

Right?

=wolf 108/72

[granite stirrup]

Query made by @gleaming schooner
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=108%2F72
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[willow bear]

i mean if you insist

[gleaming schooner]

Ah

[willow bear]

but ya know, starting from what i had you could've canceled out the 4 and 9 multipliers

[gleaming schooner]

Ohhhh

Wow

@willow bear That makes sense

Thank you

Wish I were as good at math as you are 😛

[willow bear]

i mean, in my head i did it more like

=tex \frac{27}{8} \Bigg/ \frac{9}{4} = \frac{27/9}{8/4}

[gleaming schooner]

Ooh

So with that now

[willow bear]

so

you know y = (3/2)^x is your graph's equation

[gleaming schooner]

(3/2)^-3

[willow bear]

yes

[gleaming schooner]

Wouldn't that be (2/3^3)?

[willow bear]

(2/3)^3

[gleaming schooner]

Interesting, alright

So it's B

8/27

I'm actually preparing for a precalc test, but I'm taking practice exams on the individual topics online.

My professor refuses to give review sheets.

[willow bear]

yikes

do they at least give you a list of topics to study?

[gleaming schooner]

Kind of, but she does so in a convoluted way

She gives us a chapter range in the text book

But it's such a large range

So I'm just taking a practice exam on each chapter topic

Exponential functions, logarithmic functions, trig functions, inverse trig functions

Solving inequalities

I have today and tomorrow to study :<

[willow bear]

"solving ineqs" is a pretty broad name

any particular type?

[lost pawn]

You guys learn inequalities at the same time as exponential functions?

Or what Anto said

Is it proving inequalities?

[gleaming schooner]

I think so

Yea

Proving inequalities

[willow bear]

so things like prove that x^2 + y^2 ≥ 2xy for all real x and y?

[gleaming schooner]

Solving polynomial and rational inequalities

[willow bear]

right

that's what i thought

honestly, it's a bit algorithmic, if anything

the hardest part is factorization

i don't think it follows any country's curriculum

it's supposed to be international

and comprehensive

[deft kettle]

So

[viscid thistle]

the way id use sth like that is just look up topics themselves

ie if i am shaky on polynomial long division, i look up a vid on that

curriculums may vary depending on your institution i guess

[willow bear]

grade numbers are also anything but constant

:p

and countries have different amounts of grades in schools

US has 12, Russia has 11

[deft kettle]

Oh gee

[viscid thistle]

us > russia proved

😅

[deft kettle]

😬

But you guys

Is it worth learning there?

[willow bear]

definitely

[deft kettle]

Ok, thank you

I will start 😃

[gleaming schooner]

How do I determine an exponential function's asymptotes

[willow bear]

1. an exponential function never has any vertical asymptotes

2. it always has a horizontal asymptote, which is the value that y approaches as the actual exponential term goes to zero

[gleaming schooner]

Well this function has a vertical shift 2 units down

So there would be a horizontal asymptote at y = -2

[willow bear]

yes

[gleaming schooner]

I'm lost on this

[willow bear]

that's

not

how you do that

at all

okay so it looks like k = -3 @gleaming schooner

if the drawing is to be trusted, that is

argh!

[viscid thistle]

Drawing in american?

[calm whale]

laughs in british

[viscid thistle]

Yes, I forgot, the rotation of our flat earth angles the Western sector of the flat earth so that tons of radiation causes American drawings to not be trustable at all

Einstein was right

[blazing raven]

About?

Just wondering. Ellen Degeneres says he was wrong about black holes.

[gleaming schooner]

How do I do (b)

[willow bear]

get everything to one side

=tex \frac{1}{x-2} - \frac{1}{x+3} \geq 0 \ \frac{x+3-(x-2)}{(x-2)(x+3)} \geq 0

[willow bear]

=tex \frac{5}{(x-2)(x+3)} \geq 0

[calm whale]

this can't be = 0 🤔

[willow bear]

sure

i'm just retaining the non-strict inequality sign lest confusion happen

[gleaming schooner]

How'd you get to the second one

(x+3)-(x-2)/(x-2)(x+3)

[willow bear]

surely you know how to add fractions

yknow

common denominator and all that jazz

[gleaming schooner]

What would be the common denominator between 1/(x-2) - 1/(x+3)?

[calm whale]

(x-2)*(x+3)

[willow bear]

that

the simplest option

[gleaming schooner]

I don't get how 😮

[calm whale]

whats the common denominator between 5 and 7

5*7

🤔

[gleaming schooner]

1

[willow bear]

seriously?! this is algebra II stuff at best?!

[gleaming schooner]

My algebra isn't very good =[

[calm whale]

🍭

so how do you add up 1/5 + 1/7

:)))

5+7/5*7

say 1 and I'll be done

[gleaming schooner]

Don't you find the lowest common multiple

[calm whale]

dude how do you add up 1/3 + 1/2

I couldn't care less about the terminology

[gleaming schooner]

Let me write it out hold on

[calm whale]

🤔 🤔 🤔

[willow bear]

@gleaming schooner yes, and the lowest common multiple of (x+3) and (x-2) is precisely (x+3)(x-2)

it's often a lot easier to find it with polynomials since there's a lot of coprime elements in that ring

[calm whale]

I think he's still adding 1/3 + 1/2

[gleaming schooner]

Back, sorry about that lol

But yea, you would just find the lcm which is 6, divide the denominator into 6 and then multiply the numerator by the quotient

So 2/6 + 3/6

5/6

Hmm okay

I understand

Thanks @willow bear @calm whale

So once I have (x+3)-(x-2)/(x-2)(x+3)

What do I do

[calm whale]

well

(x+3)-(x-2)= x+3-x+2 = 5

[viscid thistle]

nvm im retarded i read it wrong kek

[willow bear]

@gleaming schooner i showed

you

[gleaming schooner]

@willow bear How'd you get 5

At the top

[willow bear]

.........

x + 3 - (x - 2) = x + 3 - x + 2

[gleaming schooner]

Ahh alright

[willow bear]

i'm

[gleaming schooner]

Apologies

[willow bear]

to say i'm surprised you're asking this question would be an understatement

[gleaming schooner]

I'm a bit out of it

Cramming for a test and a bit overwhelmed

[willow bear]

take a break

[gleaming schooner]

Test in a few hours e.e

[willow bear]

seriously take a damn break

oh crap

well in that case

you should honestly still stop cramming

it's gonna do more harm than good

[gleaming schooner]

Yea

Very true

So for that proble, it would just be x >= 2, x >= -3

RIght?

problem*

[willow bear]

uh

no?

[calm whale]

😄

[willow bear]

x > 2, x < -3, surely??

[calm whale]

how do you get the fat < ?

[gleaming schooner]

5/(x-2)(x+3)

[floral rose]

** like this **

oh

[willow bear]

**bold**

[floral rose]

put two asterisks

[calm whale]

o thats bolded lo

[willow bear]

anyway i'm going to go to sleep since i'm in a horrendous mental state right now

so

[calm whale]

how come

you alright :?

[gleaming schooner]

Alright, feel better

[willow bear]

studying for a cs theory test is driving me crazy for no apparent reason

[calm whale]

🤹

gl

[gleaming schooner]

:3

@calm whale Would you be able to help me understand this inequality problem?

I'm up to:

[calm whale]

ye the way I do them

is I draw a line

then I draw all the roots

[gleaming schooner]

=tex \frac{5}{(x-2)(x+3)} >= 0

[calm whale]

and then it goes like - + - + -+ etc

and you just look where your ineq satisfies the equations

[viscid thistle]

Oh 🅱oi!

Partial fractions!

fun stuff

makes me giddy inside

[calm whale]

ugh what is a partial fraction 🤔

is that a partial fraction 😄

cause it is just a normal fraction

putting a fancy name infront of it does not make it complicated

[late haven]

=tex \frac{5}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3}

[late haven]

partial fractions

[calm whale]

very nice

[late haven]

intimidating stuff

[calm whale]

it does look pretty insane

[late haven]

It's just occured to me that A and B probably have units associated with them

[calm whale]

they do?

[late haven]

er, dimensions

[calm whale]

o

[late haven]

If x were, say, meters

then A and B would have to be per-meters

[calm whale]

🤔

[late haven]

Assuming that 5 was dimensionless

[calm whale]

but A and B themselves

are x +- smth

[gleaming schooner]

@calm whale What do you mean you draw a line and draw all roots?

Is there a video I can watch that explains this

[calm whale]

that's how I do it

I getmyself an x axis

and I plot all the roots

[gleaming schooner]

So in this case, there are two roots, right?

-3 and 2

[calm whale]

and then decide where my inequality is true

yes

[gleaming schooner]

So I drew a number line with those points.

Now what do I do

[calm whale]

now you go from right to left

and well fuck 😄 not sure how to explain it

you write - + - + , change signs when you encounter a root

[gleaming schooner]

Is there a name for this

I'll just look it up

[calm whale]

nah fuck that

thats how I did them

anyway the mathsy way

is to do 2 simultaneous equations

[gleaming schooner]

Cause that's similar to what my professor showed us

I just didn't understand it fully

The - + - +

And test points

[calm whale]

is that uni maths

when you say professor 🤔

[gleaming schooner]

Yea

[calm whale]

what

uuh anyway

|x-2 > 0
|x+3 > 0

thats one

|x-2 < 0
|x+3 < 0

thats two

=tex \frac{5}{(x-2) \times (x+3)} >= 0

[calm whale]

it can't be equal to 0

[gleaming schooner]

So how do I figure out the domain

Or solve it

[calm whale]

uh it has to be positive

cause I mean that's what it says

and the only way for it to be positive is to have

=tex - * - = +

or

=tex + * + = +

[gleaming schooner]

Isn't it > or = to 0 though?

So x can be 0, no?

Nvm, I have an example here that I'm gonna follow