precalculus

[serene heath]

I dont think they want you to solve it

[full garden]

this is the answer

doesn't this mean I have to solve for v(t)=65

and the second question is like super easy but I am getting it wrong

don't I just plug 67 in and solve for t?

[serene heath]

no they're not asking you to solve

just Express the question in function form

[full garden]

oh alright thank you man

but the can you clarify c for me

it's so embarrassing because the question looks tooo easy

i am getting this as the answer

[serene heath]

how r u getting that

[full garden]

idek man it's like embarrassing

[serene heath]

tell me what u did

[full garden]

i just plugged 67 in and tried to solve

[serene heath]

that's the position function tho

67 is a speed

[full garden]

the question says where is the car when it is going 67 mph that's why I thought that I should use the position one

because it says where

[serene heath]

yea but

you wanna find out the time when it's going 67mph

then use that time in the position function

and find out where it is

[full garden]

oh damn i got it correct bro

thank you bro

[neon garden]

Is the amount of turns in a graph always one less then the degree or is that just the maximum?

[rigid sun]

what?

what does that mean?

amount of turns in a graph?

tf

isn't a local max also a turn?

turn around

[fleet yew]

Just extrema ig

[rigid sun]

well

it is not always one less than the degree

x^3 has no local max

unless inflection points count as a turn?

tf

anyway

x^4 should break this rule

[fleet yew]

Root of the derivative

[rigid sun]

,w plot x^4

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[rigid sun]

this only has 1 turn

im not sure why that rule would even be useful

the amount of turns

its not like finding inflection points or roots was hard to begin with

[fleet yew]

Precalc

So yeah at most n-1 turns

At most

[rigid sun]

lol

actually no

,w plot y=5

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[rigid sun]

at most it would have -1 turns

that would mean at least it would have another -# of turns

and you can't have negative turns

so

n-1 when n>=1

[fleet yew]

What if n=1.5

Wheres my extrema

[gloomy vigil]

anyone know where i went wrong

nvm i forgot to divide by 0.001

[odd helm]

How would I approach this problem?

[vagrant stirrup]

optimization

[fleet yew]

x+2y=80, xy=A. y=1/2(80-x). xy=1/2(80-x)*x

then find the vertex of that parabola

[soft night]

Lol these are easy

Nice thing about them is that you already know what the right answer is

You just have to prove you got it

[full garden]

hey guys I have a very silly question it would be nice if someone can please clarify

for those which one would be shrinking the fastest

the ones that the b values is higher or lower

like does the 0.9 times mean that it's shirking faster than 0.87 times

[viscid thistle]

sinxcosx=-√3/4 How do I solve this?

[patent beacon]

@full garden
Yes. Note that if b > 1, then it grows instead. A higher value of b means it's growing faster.

@viscid thistle
Use 2sin(x)cos(x) = sin(2x)

[atomic talon]

does anyone know how to use pascals triangle to expand this binomial expression: (1 - i)^7

[patent beacon]

$(1)^7 + 7(1)^6 (-i)^1 + 21(1)^5 (-i)^2 + 35(1)^4 (-i)^3$

[obsidian monolith]

Kaynex:

[patent beacon]

Is the first half. See the pattern? Can you get the rest?

[vernal moon]

Can’t you just use de mourves tho?

[patent beacon]

Yeah that would be faster

[harsh cipher]

Hello all,

Question

In this question, why is tan theta relevant? Couldn't we find the answer without knowing tan theta is less than zero?

[viscid thistle]

How do I find a formula to expand (A^n)-1

[hard hornet]

@harsh cipher quadrants

that might help you

[gloomy vigil]

how do i find the asymptote for this

horizontal asymptotes are only when the leading coefficient on numerator is lower than denominator right

x^0/x^4 = horizontal = 0

so this problem would be x^3/2 over x^1/2 = over so no asymptote

but this one is x^3 and the other is x^4/2 so it's like x^2 so x^3 / x^2 = higher so no asymptote?

🤔

[harsh cipher]

okay...I get that tan theta is less than zero is quad Q2 and Q4.

and a is greater than 0

meaning our angle lies in quad 2

[lethal oracle]

Is anyone there to help me with something

[clever inlet]

🤔

[lethal oracle]

I have the ellipse

Would my vertices for the xy system be the ones below it

For when I graph it

[viscid thistle]

can someone explain this? i dont get it. the answer is c

[willow bear]

what don't you understand

why c is true, or why everything else isn't?

[viscid thistle]

yeah

[vernal moon]

(which one)

[viscid thistle]

why c is true

[willow bear]

how do you find it in you to reply with "yes" to a "is it this or that" question

ok well

suppose f isn't identically zero

[viscid thistle]

lol i misread sorry

[willow bear]

then it's positive in at least one point

i.e. there exists c in [a,b] such that f(c) > 0

since f is continuous, there exsits a positive number δ such that f(x) ≥ f(c)/2 for all x in (c-δ, c+δ)

so the integral of f over the entire [a,b] will be at least the integral of f over [c-δ, c+δ], which is at least f(c)/2 * 2δ = δ * f(c), which is positive, contradicting the fact that the integral of f over [a,b] must be 0

that probably wasn't the explanation you were looking for but ig i can't provide any other at the moment unless i am forced to be less rigorous than i'd like

[viscid thistle]

my 3 brain cells still dont understand lol

[willow bear]

suppose f is not identically zero

then it's positive somewhere

then by continuity it's positive in some interval around that somewhere

then the area under its graph will be at least the area under its graph over that interval

and that's positive

but the integral is supposed to be zero

so that's a contradiction

[viscid thistle]

ah ok thx

[harsh cipher]

can anyone help me

y= sin x/2 + cos x/2

what's the period of the above equation?

it says use a graphing calculator, but I'm completely lost with this question

[willow bear]

plot it in desmos lmao

desmos > graphing calcs

[harsh cipher]

omg Ann!

okay

i'll do that

hmmm I dont see what's going on even after plotting it on desmos

[willow bear]

go to settings and set the x-axis step to pi

you should see how long it'll take the thing to repeat

[harsh cipher]

on desmos?

ok done

okay.

I see that its 2pi+ 2pi

[willow bear]

4π yes

[harsh cipher]

but could you show me how to do it algebraically?

[willow bear]

i mean you could write it as sqrt(2) sin(x/2 + π/4)

[harsh cipher]

these thing make more sense to me when written

[willow bear]

so it's just sin(x/2) but scaled and phase-shifted

[harsh cipher]

I think we can do P = Pi/1/2

1/2 is our "b" value

p= pi/b

[willow bear]

2pi not pi

[harsh cipher]

then b is 1

[willow bear]

no, b is still 1/2

[harsh cipher]

then we get 4pi?

[willow bear]

the period is 4pi yes

[harsh cipher]

we have cosx/2 sin x/2

that makes 8pi?

[willow bear]

no

the period of f+g is not the period of f plus the period of g

[harsh cipher]

why?

its two different graphs

[willow bear]

by your logic, sin(x)+sin(x), a 2pi-periodic function, would have a period of 4pi

[harsh cipher]

ok

[willow bear]

i mean

[harsh cipher]

then what did I get incorrectly

[willow bear]

do you really expect everything to be linear

[harsh cipher]

I actually think like that irl

lol!

[willow bear]

rip to you then

[harsh cipher]

where is my misunderstanding with the question

why is it 2pi..and not 4pi...

[willow bear]

what

cos(x/2), sin(x/2) and their sum are all 4π-periodic functions.

[harsh cipher]

@willow bear Thanks for the help.

[leaden stratus]

How do I find sin beta? This is a right triangle, and cos is adiacent side/hypotenuse

[willow bear]

consider which side is adjacent to alpha

and which side is opposite to beta

[leaden stratus]

This side, Ann?

[willow bear]

well which side is it

is it the side adjacent to alpha, or is it the side opposite to beta

[leaden stratus]

Both?

[willow bear]

that's exactly my point.

[leaden stratus]

Oh

So I assume that cos alpha = sin beta?

[willow bear]

you don't assume that because it's true regardless

[leaden stratus]

And when you've got a triangle like this?

[willow bear]

well consider now that alpha and gamma are the same angle

[leaden stratus]

Wait, why? How can I demonstrate that?

[willow bear]

they're both 90° - beta

[leaden stratus]

Oh

And then? @willow bear

[willow bear]

they're the same angle

did it not click yet

[leaden stratus]

So, cos gamma is the same

And if it was to find sin gamma?

[willow bear]

would probably have to use sin^2(γ) + cos^2(γ) = 1 to do that

sqrt(379)/20 is what you would get

[leaden stratus]

Oh thanks a lot

[full garden]

hey guys can someone please help me with finding the domain and range for this function that represents the area of a small circle that was taken from a big circle - small circle

I put the domain for r has to be bigger than 0 but I think that's wrong because if I put like 1, the equation is going to have negative radius times pie which is wrong

can someone please clarify the solution to this problem for me please

[full garden]

here is the question guys if it would make it more clear

[full garden]

can someone please help me with the domain and range

[serene heath]

you need to have r^2-16>=0

can you solve that inequality?

[full garden]

r.=+-4

[neon garden]

I’m supposed to use descartes rule of signs to find the number of possible solutions but I’m not sure what I’m doing wrong, any help would be appreciated

[full garden]

Panda bro please wait man

[neon garden]

oh sorry

[full garden]

@serene heath I see where this is going bro but like idk the actual r of the large one

like the small circle has the r of 4

[serene heath]

u dont need to

it says as a function of r

[full garden]

so is the domain -4<=x<=4

[serene heath]

how did u get that?

[full garden]

because you told me to solve the inequatlity and I got r>=+-4

[serene heath]

,w solve r^2-16>=0

[obsidian monolith]

Display results online and refine query

Requested by the one n only#3070

[full garden]

I thought like if you take the square root of 16 then +-4

ok i'm stupid mna

so is the domain r >=4

what about the range

[serene heath]

sketch the graph of r^2-16

and consider the domain

or just think logically

what values can an area take

[full garden]

0

[serene heath]

could be 0

[full garden]

=0

but let's take a second here...... HOW DO YOU KNOW HOW TO ANSWER EVERY SINGLE QUESTION LIKE YOU ARE THE TEACHER

[serene heath]

huh

[full garden]

you need to have r^2-16>=0
@serene heath how did you know this

[serene heath]

because otherwise youd get a negative area

[full garden]

like when I factored it i got (r+4)(r-4)

how did you get to the point that we have to solve r^2-16>=0

[rigid dune]

Just making sure is this right?

[dim jungle]

Yes

Idk what bottom thing is

But first 2 things yes

[rigid dune]

Tried distributing the +2 to x and h individualy

Didn't know if I should have just left it or not

Probably not actually

[fleet yew]

@rigid dune you dont distribute addition

Dont be dum

[rigid dune]

I was dum

now i am less

:3

[fleet yew]

(a+b)+c=a+b+c

Commutative property

[fluid shore]

[fleet yew]

^

[odd shuttle]

Is this true: 1-tany cotx = tany cotx-1

this is the problem I am trying to prove: sin⁡〖(x-y)〗/(cos⁡x sin⁡y ) = tan⁡〖x cot⁡〖y-1〗 〗

and I got down to 1-tany cotx

but don't know if I can just switch those around like that and say I can because of "Algebra"

I now think I did this problem completely wrong

I have no idea what to do

[fleet yew]

what's with those weird symbols

use latex

[rigid dune]

Okay now I feel like I over complicated this. It continues from my previous problem.

• = the same denominator, didn’t want to write it out

[odd shuttle]

sin(x-y)/cos⁡x sin⁡y = tanxcoty - 1

I have no idea why those symbols populated

[fluid shore]

$\sin(x-y) = \sin(x)\cos(y) - \sin(y)\cos(x)$

[obsidian monolith]

Abhijeet Vats:

[odd shuttle]

OOF Thank you @fluid shore !

[fluid shore]

You're welcome.

[sharp marsh]

How do I do this?

[blazing parrot]

do you understand the h(x) = {... thing

[smoky fractal]

Can someone tell me what i did wrong

Ik i cant have a log of a negative number

So i probably did this wrong

[serene heath]

,w solve (5^x-5^-x)/(5^x+5^-x)=8

[obsidian monolith]

Display results online and refine query

Requested by the one n only#3070

[serene heath]

@smoky fractal your working is fine

[willow bear]

@smoky fractal are you asked for real solutions only

[smoky fractal]

Oh, yeah i guess

What would i leave as a solution? Imaginary?

[willow bear]

i mean are you expected to look for non-real sols or not

if not then after 5^(2x) = -9/7 you could have just written "no real solutions"

[smoky fractal]

Alright thank you so much

[fading token]

[royal quest]

is this correct?

if the asymptote is a how can the last one be infinity?

wouldnt it not exist?

o wait nvm it approaches 0 but never is 0 oke

wait then how come

[willow bear]

the first limit is already enough to guarantee that there's an asymptote at -2

but also, 1/x^2 has a two sided limit at 0

[royal quest]

wait but how come it doesnt exist?

[stuck lark]

a two-sided limit exists only if the one-sided limits are equal. in cases where we're dealing with limits that don't evaluate to anything finite, ie things "blow up" to +infty or -infty, they must "blow up" in the same direction in order for us to say the limit is +infty or -infty

[royal quest]

wouldnt it be infinite cuz its not AT -2 but approaching -2

[sharp marsh]

@blazing parrot it = y right?

[blazing parrot]

yeah basically on the regions specified

then just use endpoints of the ranges and y values at those endpoints or shape of the lines to determine which graph fits the function definition best basically @sharp marsh

[sharp marsh]

got it

[proud sparrow]

@rigid dune looks about right, carry on

but yeah, perhaps you didn't need to expand that denominator

[rigid dune]

Was thinking, to subtract I would need to get the denominators to be the same

[proud sparrow]

yeah, just keep it as (x+2)(x+h+2)

[rigid dune]

So what order am I doing this in?

[rigid beacon]

this is why I hate that notation lmao

my guess is

g(f(h(x)))

[rigid dune]

That’s what I was gonna go with

[pale kettle]

That's correct

[rigid beacon]

I hate that notation tho

[pale kettle]

I mean, what would you rather have as notation

[rigid beacon]

I think g(f(h(x))) is better

[pale kettle]

I mean yeah sure, and that's what normally happens

but what if you want to talk about just the function g o f o h and not the function applied to any specific value

[rigid beacon]

call it g(f(h(x))) or what was common in problems I've seen in calc and stuff is just say "let j(x) = g(f(h(x)))"

[pale kettle]

The thing is that g(f(h(x))) specifically means this function applied to the value x

not really the function g o f o h itself

and sometimes you want to make these two concepts distinct

[rigid beacon]

like when?

I'm genuinely curious I can't think of a time when you would but I'm probably missing something

[pale kettle]

Like saying the function f is injective

You don't want to say that f(x) is injective because this is just a value of the codomain

[smoky delta]

@vital dome

[vital dome]

@smoky delta

[smoky delta]

Sniped

[rigid dune]

What would this be? 4?

Not a multiply symbol

[uncut mulch]

$(g \circ g)(3)$?

[obsidian monolith]

ramonov:

[rigid dune]

Wait would it be 3?

Cause I'd imagine we'd get 4 from (3,4) for the g(3) part

Oh wait scratch that

That would lead to g(4) which is the (4,0) which means its 0

[uncut mulch]

yes

[rigid dune]

Yes, i'm being the not dumb

[slow roost]

#1 is 3 right?

[neon garden]

@slow roost yes

[runic cradle]

anyone have any idea how to do part d? thank you in advance! (please ping me if you respond)

[vapid torrent]

what do i plug in here, a little confused

[vernal spindle]

I'm confused on this one

[willow bear]

sec(2 radians)?

is 2 radians the complementary angle of pi/6?

[vernal spindle]

I think I calculated the coterminal angle

I tried adding pi/2 to 1pi/6

[willow bear]

do you know what it means for two angles to be complementary

[vernal spindle]

ohh isnt it when they sum to 90*

[willow bear]

there's no such thing as degrees anymore.

what is 90°?

[vernal spindle]

pi/2

[willow bear]

okay

so pi/6 + what = pi/2

[vernal spindle]

pi/3 ??

[willow bear]

well there you have it don't you

[vernal spindle]

This is my last problem

I thought cos would give the adjacent / hypotenuse

[willow bear]

b/c = 3/5 does not mean b = 3 and c = 5

[vernal spindle]

would side a and side b be the same since they are both oppposite

[willow bear]

"opposite" by itself means nothing

it's always "opposite to <angle>"

[vernal spindle]

I can't figure this one out I'm looking through my notes and tried googling for help but im stuck

[willow bear]

maybe actually draw the triangle

[vernal spindle]

like this?

[willow bear]

...

no. angle B is not three fifths of a degree. also, C is a right angle, so this diagram is all kinds of misleading.

[vernal spindle]

Is the cosB=3/5 side b?

[willow bear]

those sure are some words but they don't mean anything

[vernal spindle]

I'm lost then

are 5 and 3 the lengths of the other two sides?

[vapid torrent]

can someone help me solve this

[lusty gorge]

Can anyone help me with 1?

[vapid torrent]

<@&286206848099549185>

[valid violet]

It looks like it has 3 roots and 2 distinct roots

Anyway gn @lusty gorge that's my hint but im off to bed

[lusty gorge]

[vapid torrent]

how can i find U from these two equations: Ω _T*U = 0.0754992/2 = 0.0377496 and 0.993 = U/(Ω_T30)

[valid violet]

@lusty gorge im awake,did you figure it out?

[lusty gorge]

Not yet @valid violet

[sterile kernel]

Hi, i have a question about basic derivatives. The given problem is to be simplified by using derivatives. I have heard about the method listed below in the image to be shorter and easier. I asked my math teacher twice to help me with this but he didn't seem interested. He just looked at my method and said "he took the long route". Now I'm not good at maths so i need some help with how to work with the method below. Just applying the quotient rule and simplyfing will solve the problem or something else needs to be done? Thanks in advance!

[valid violet]

@sterile kernel the equation after "- the other method -$ should have du/dx not d(u)/du

It's the chain rule d/dx[f(u(x))]=df/du du/dx

Which would be -1/sqrt(u) du/dx here with u=(a-x)/(a+x). This second method doesn't look any easier

@lusty gorge when a polynomial touches the x axis from above (or below) and immediately moves away like that

It's an indication that a root has multiplicity >1

If i draw the x axis at height 10 units

Then i want a root at x=0 and a multiple root at x=60

So that suggests x and (x-60)² are both factors

Then you need to multiply the polynomial by a constant to make the maximum height correct

@sterile kernel you have a sign error too, whats d/dx sqrt(x)?

Tagged you in #calculus bc this isn't precalc, continue there

[lusty gorge]

What is a constant

[valid violet]

A number

Not a variable

Like it might be a variable but it doesn't change

You might write f(x)=ax³+2x

and say a is constant, it's just a number

[atomic talon]

I need help with this question. "Consider the expansion of (5x + 2y)^6. Determine the coefficient for the term that contains the expression x^5y. Enter your response as an integer."

[uncut mulch]

what have you tried?

[fleet yew]

pascal triangle

[remote veldt]

@vernal spindle the trig ratio rules (sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent) only work in a right triangle.

as for cos(theta) = 3/5, that tells you the ratio of the sides is 3 and 5. But, the sides could be 30000000 and 50000000, or 6 and 10, or 3/1234567 and 5/1234567

One of the great things about cos (well, all the trig functions) is that cos(theta) is the same no matter how big the triangle is. The only thing that matters is the angle.

Since this is a right triangle, you can use the pythagorean theorem. a^2 + b^2 = c^2. You have that a = 18, and that b/c = 3/5, or b = (3/5)c.

18^2 + (3/5 * c)^2 = c^2

can you solve from there?

[neon garden]

Can anyone help with 33?

[patent beacon]

@neon garden
Name two of the sides. Call the bottom x, call the side y. Can you express the area? The hypotenuse?

[neon garden]

Ok well area is (xy)/ 2 and the hypotenuse is x^2 + y ^ 2 = (x+1)^2

So y equals square root of 2x + 1

[patent beacon]

That equation results in y = √[2x + 1], that's right. You've got the area equation as well, xy/2 = 30

[neon garden]

Yeah working on the area rn

Wait

The area equation simplifies to part a

[patent beacon]

Sick, looks like you got it

[neon garden]

alright cool thanks

@patent beacon I’m sure there’s a better than to solve part b then guessing and checking using synthetic division. What would that be?

[patent beacon]

Note that part b isn't asking for the solution, but just that there is a solution less than x = 13

Let f(x) = 2x³ + x² - 3600
What's f(0)? What's f(13)?

[neon garden]

f(0) is -3600 and f(13) is 963

So using the intermediate value theorem

there must be a solution between there

correct?

[patent beacon]

Yus. Mention that f is continuous, and you're done.

[neon garden]

thanks

[quick rune]

question
im trying to simplify an expression using trig identities
its: cost(1+tan^2t)

any help?

[tardy crescent]

Hey @quick rune, look at what's in the parentheses

[quick rune]

yeah i know

i am dumb

Pythagorean identities...

[tardy crescent]

1 + tan^2 = sec^2

[quick rune]

i got it thanks anyways

[tardy crescent]

ah cool np

[full garden]

hey guys can someone help me find the equation to this problem

I don't think I Can do the y2-y1/x2-x1 becuse it's like not linear

[clever inlet]

You're finding the slope of the line segment

Which is linear

The function itself is not linear

That is true

[full garden]

oh

[clever inlet]

But not exactly important

[full garden]

but how do I find like the y values

[clever inlet]

What's f(5)?

[full garden]

so the delta x will be 4

[clever inlet]

Sure

That's your denominator

[full garden]

yeah bro

what about the top one

[clever inlet]

What's f(5)?

[full garden]

idk bro I am too stupid 4 * b^5?

[clever inlet]

Yes

That's it

[full garden]

oh

[clever inlet]

And obviously f(3) is that but with 3

[full garden]

yeah bro

[clever inlet]

Those are your two y values

[full garden]

b^5-b^3

[fleet yew]

nice

[full garden]

/4

nooo sorry

just b^5-b^3

[fleet yew]

nice

[full garden]

wait is that it?

[clever inlet]

Yes

[full garden]

love you bro thank you so much

💙

[clever inlet]

There's no more /4 once you simplify

[full garden]

I was so stupid I thought I was supposed to get numbers

[clever inlet]

But yeah, that's it

[full garden]

thank you brooo

[vagrant stirrup]

hey can anyone help me with the 2nd deriviatve test on how to go from your first deriritive to your second when you derive it again? pls dm me

i will love u forever and ever

its a quick question i promise 🙂

[glass ledge]

can someone help me with interval notation

im buggin rn

if the function is just a straight line and the range is pi/2 is it just [pi/2 , pi/2]

idk how mathxl wants me to input

[stuck lark]

no need to write intervals if the function takes on a single value over its domain. keep it simple. just $\brc{\frac\pi2}$ is enough

[obsidian monolith]

RokettoJanpu:

[glass ledge]

ty

[stuck lark]

np

[harsh cipher]

where would it be appropriate to ask questions about discrete math?

i mean

ap statistics 12

I don't know why I said discrete math lol

[patent beacon]

It pretty much is discrete math. Go ahead and ask here I'll bite

[harsh cipher]

I haven't started yet will register tomm

haha

[patent beacon]

Oic, well we do have the #discrete-math channel and #probability-statistics channel

When those do come up

[harsh cipher]

thank you!

[full garden]

guys can someone please tell me like the first major step into like isolating for those variables to the form of f(t)=ab^t because I tried like 3 times but then I keep on getting back to this form

[serene heath]

can you get rid of that 5 first?

[full garden]

like the 5.2^t/11.2 or just the five

[serene heath]

the 5 at the bottom

[full garden]

would it be correct to say 10^t/11.2

[serene heath]

no?

how did you get that

[full garden]

i'm stupid I just 5 * 2.. that bottm

[serene heath]

$\frac{60}{5} \cdot \frac{1}{2^{t/11.2}}$

[full garden]

do you want me to get ride of five

[obsidian monolith]

the one n only:

[serene heath]

what about now

[full garden]

12/2^t/11.2

like 12 times that fraction

does that work bro

[serene heath]

yes

$\frac{1}{2^{t/11.2}}=2^{-t/11.2}$

[obsidian monolith]

the one n only:

[full garden]

I see bro yess

I think that is the answer bro

thank you so much

[full garden]

can someone please just give me a tip for getting these right

I can't use like a calculator so I am assuming that I have to answer the question from just looking at those equations

[serene heath]

solve them and see if the x value u get is positive or not

[full garden]

like by plugging x, y values or by solving for x, because I tried solving for x in the first equation and there's no solution for f(x)=0

[serene heath]

no, solve for their intersection

[full garden]

man I know that I always cringe you with my stupid retarded questions but will I need to use log to solve for x

[serene heath]

yes

[full garden]

piss off man this book be always giving me questions that need log before it even introduces it

is using log that hard

[viscid thistle]

I guess so

[deep umbra]

Hum noh basicaly ln(x) is defined for only positive x values without 0

ln(a^b) = bln(a)
ln(ab) = ln(a) + ln(b)
ln(a/b) = ln(a) - ln(b)

ln(e) = 1
ln(1) = 0

lim( x -> 0) ln(x) = -inf
lim( x -> +inf) ln(x) = +inf

Thats pretty much everything u need to know

To use it

And log is just ln(x)/ln(10)

Actualy when i think about it, i realise its not that simple to assimilate in 1 take

Hum

[sturdy haven]

If you put sin or cosine of something into a calculator what does that tell you

In terms of triangles

[raw night]

If you put cosine it tells you the length of the contiguous side of the triangle of that angle and hypotenuse 1 (when he angle is positive and less than π. And if you put sine it does the same but with the opposed one

It is, it sets hypotenuse=1

The best way to understand this is understanding and having in mind the goniometric circumference

[uneven dawn]

highlighted the local extrema but I don't know what im missing as far as absolute?

I thought it was f and d but I guess not?

[fleet yew]

@uneven dawn that last one is not an extrema

It is a zero

[uneven dawn]

idk why I circles that... I put b, d, and f at local

I was hoping for some info about absolute extrema? I thought it was just basically the highest and lowest points but I'm wrong.

[fleet yew]

what degree function is this

odd degree functions never have global extrema

because of the way they look on the graph just tending towards infinity at either direction

only even degree functions have global extrema (and only 1 global extrema) which is either a min or a max

[uneven dawn]

all im told is it is a continuous function

how do you know if the global extrema is supposed to be the max or the min?

[fleet yew]

it is either the lowest or the highest point on the entire graph

think about parabolas for a sec

the vertex of the parabola is always its extrema

because there is no point either higher or lower

[uneven dawn]

so its f because there is no higher point, but not d because there is a lower point

[fleet yew]

from the information in this graph yes

also it is a quartic

a negative quartic

[uneven dawn]

i dont think we've covered that yet

[fleet yew]

dude there's nothing you shouldn't know about determining the degree of a polynomial from its graph

you can see that there are 4 zeros

and that it goes towards negative infinity at both directions

[uneven dawn]

i just didnt know the word quartic tbh

i know what you mean now that i looked it up lol

Im going to get back on the grind, thanks for the help 🙂 this discord the best

[deft briar]

how would you graph (x^2 - 5x)^2 - 2(x^2 - 5x) - 24 without using calculus or rational root theorem? (i dont need to find the y-int or vertex btw)

[willow bear]

the vertex? this is a quartic, it doesn't have a vertex.

surely, though, you would want to find its x intercepts?

[deft briar]

yeah i do

[willow bear]

so what exactly is tripping you up on that front?

[deft briar]

when i factor it it becomes ((x^2 - 5x) - 6) ((x^2 - 5x) + 4), so i guessed the x-intercept would be -4 and 6 but desmos shows that the x-intercepts are 4 and 6 not -4

[willow bear]

you guessed the x intercepts would be -4 and 6??

why not actually be honest to yourself and sit down and solve the equation (x^2 - 5x - 6)(x^2 - 5x + 4) = 0 properly?

[deft briar]

okay

i tried to solve (x^2 - 5x - 6) = 0 and (x^2 - 5x + 4) = 0 separately oops

[willow bear]

and what did you get

you should get a total of four roots from these two equations.

[harsh cipher]

Question

[willow bear]

answer

[harsh cipher]

Hi Ann!

Why add 2pi to pi/3?

Period is pi/3 for tangent

[willow bear]

what the hell is that mess of colors

[harsh cipher]

lol

ratio is the same for tan pi/3 and tan 3x

so what I did was solve for x...

x = pi/3 and 4pi/3

then all of the sudden 7 pi/9 and 16pi/9 came up...I see that that's 2 pi from the reference angle..but why?

start of pi/3 and 4pi/3 i mean

[willow bear]

...

tan(3x) = sqrt(3) gives you 3x = π/3 + kπ

for k in Z

[harsh cipher]

oh no I was wrong

yes..then we get 7pi/9 , 16pi/9...

I looked the last section of the lesson.

No need multiply by 1/3 if we do it your way Ann

[willow bear]

actually that's what i was gonna do anyway to get x = π/9 + k π/3

$0 \leq \frac\pi9 + \frac{k\pi} {3} < 2\pi \
0 \leq 1 + 3k < 18 \ -1/3 \leq k < 17/3 \ 0 \leq k \leq 5$

[obsidian monolith]

Ann:

[harsh cipher]

didn't understand the last part, but thank you as always.

[willow bear]

k is an integer

it cannot be strictly between -1/3 and 0, or strictly between 5 and 17/3.

[harsh cipher]

okay, thanks! I need to sleep for now

[plush violet]

Idk if this is the correct place to ask this but I've got a circle question where I have to find the relation ship between their radi but i have gotten their radi to be root(80) and root(68)

Am I missing somthing or have i made a mistake somewhere

[serene heath]

huh

can you post q?

[rapid lance]

@plush violet what's the question

[worldly crag]

can anyone on here help me out with a quadratic equation question?

i am struggling SO bad

[soft night]

Sure

[rapid lance]

@worldly crag sure

[worldly crag]

it should read y=ax^2+bx+c, but c equals 0

[soft night]

Just do what it says

what does a quadratic look like?

[rapid lance]

@worldly crag where does it intersect with the x axis?

[worldly crag]

thats my problem...there are no a values or b values, just that they cant = 0

so im not sure considering there are multiple possibilities

what i have done, is set a and b to 1

[rapid lance]

Do you know what a quadratic looks like?

[worldly crag]

where i found the vertex to be (-1/2, -1/4)

yes i do, it should be a parabola

[soft night]

ok sure

Then draw it

[worldly crag]

i know, but i need to know the vertex, axis of symetry, and y intercept to graph it

[rapid lance]

How can we use c=0 to our advantage @worldly crag

[worldly crag]

im just stating that the equation for a qaudratic would normally read y= ax^2 +bx + c

but since there is no c, this means c=0

[rapid lance]

Yes so that means that there is no shift

[worldly crag]

correct, the y intercept equals (0,0)

but how to i find the necessary vertex, axis of symetry to graph this equation

how do*

[soft night]

Just do -b/2a

lol

[worldly crag]

ok, but what are the values of a and b lol

[rapid lance]

-b/2a looks familiar from deriving the quadratic formula so just use that, I don't get what your missing

[worldly crag]

its says in the problem a and b can not equal zero....thats all numbers other than 0

[soft night]

Dude

Did you not pick values for a and b?

[rapid lance]

It's a sketch

[soft night]

ik that

But he said he picked 1

bruh moment

[rapid lance]

You need to prove that for any a,b it works @worldly crag

[worldly crag]

ok not sure why you guys are being assholes, thought this was a channel where people help other people

im struggling and need help, thats all im asking for, not asking for people to tell me "just do it"

[serene heath]

[rapid lance]

@worldly crag the a and b are irrelevant in the problem you need to work with the variables

[worldly crag]

you did read the problem right? there are no variables

[soft night]

But this is a literally just do it problem

Like

There is no thinking involved here

literally take your hand and draw a parabola

[worldly crag]

i would understand how to graph y=2x^2+4x

oh yeah, let me just draw a random porabola on the page lol

[soft night]

That’s literally what they are asking

[worldly crag]

what is the point of this discord if noones going to actually want to help

yes sketch the graph

its going to be graded upon being CORRECT

[soft night]

moments of bruh

[serene heath]

what graph are u trying to sketch

[worldly crag]

i already asked my professor if i can pick any numbers for a and b, he said it has a specific answer

hope you feel dumb for being an asshole

[serene heath]

a?

[worldly crag]

y=ax^2 + bx

where a and b can not equal 0

[serene heath]

ok that's gonna have 2 different shapes depending on the sign of a

but since it's a parabola u only really need 3 things

[soft night]

inf many different shapes

but ok

[serene heath]

the x and y intercepts

and the vertex

[worldly crag]

right, i get that, (and thank you for helping!! )

[serene heath]

can u find the x intercepts

should be str8 forward

[flint star]

since c is zero the y intercept would also be 0

[worldly crag]

yes thats what i said above that y intercept = (0,0)

[serene heath]

shh

[flint star]

oops sorry didn't see that

[serene heath]

@soft night yes but the question seems to be just asking for a general idea of the shape I guess

it's only a sketch

[worldly crag]

there are no a values or b values, just that they cant = 0
so im not sure considering there are multiple possibilities
what i have done, is set a and b to 1
where i found the vertex to be (-1/2, -1/4)

[soft night]

That’s literally what I just said

[serene heath]

ok what about the x intercepts

[worldly crag]

i dont know

i dont know if i should be setting a and b to 1

or 2

of 5

or -6, whatever

[serene heath]

you shouldnt be giving them values here

just work in terms of a and b

[flint star]

try with a and b first and see what you get

[worldly crag]

but if there are no values to be plugged in for a and b...and none for x...then i dont know what to do lol

im just looking for an explination

[serene heath]

$x(ax+b)=0$

[rapid lance]

@worldly crag the question is asking for a general shape is it a parabola or something else

[obsidian monolith]

the one n only:

[flint star]

sketch the graph but put the values of the intercepts in terms of a and b

[serene heath]

what about now

[worldly crag]

im sorry i guess im the stupid one, or you guys are confused, there are NO GIVEN VALUES

[serene heath]

sure there are

[worldly crag]

this is due tomorrow, ive worked on it for over a week

[serene heath]

the values here are a and b

dont be put off cuz they're letters and not numbers

just pretend they're numbers

[worldly crag]

but if they are not numbers...i simply cant understand how i can use those letters to find an equation to graph

[rapid lance]

It's algebra!

[serene heath]

(internet is dying on me so delays here)

[worldly crag]

its ok, appreciate your help and patience

[serene heath]

ok from the equation above we have x=0 or ax+b=0

giving us x=0 or x=-b/a

that's it

those are your x intercepts

[worldly crag]

okay, thank you, how would i graph that?

[flint star]

just graph a normal parabola

imagine the values are 1 and 2

but label them as a and b

[serene heath]

ok it's a bit weird cuz u dont know where -b/a is gonna lie

since u dont know their signs

[worldly crag]

exactly lol

[serene heath]

so like I said q is weird

[worldly crag]

such a strange problem i feel...

cant even google ANYTHING relatable to this problem

[serene heath]

it says equations so maybe split it into different cases I guess?

idk I would get clarification

[flint star]

that would yield 4 graphs right?

[rapid lance]

Since a and b aren't zero you know that the signs are either positive and negative maybe use that with -b/2a. But honestly I think this problem has something to do with the proof for the quadratic formula

[worldly crag]

Thanks for all the help

Im still lost as hell, but its ok lol

[rapid lance]

Since with a negative a and b the parabola is upsidedown it doesn't matter at all

You can then assume they are possitive

[flint star]

I'm pretty sure the question is meant to show that the x coordinate of the vertex would be at -b/2a

[rapid lance]

@flint star I do agree but the only other place with a quadratic and -b/2a was the proof for the formula, so maybe they should try to isolate for x

[flint star]

yeah but I think the question is looking for a graphical proof so I don't see why going into the quadratic formula is required

you can solve for x without using it because there is no c

[rapid lance]

I don't think so it's just an isolation for x

[flint star]

hmm ig

[full garden]

hey guys can someone please just clarify why P = 2Po like why two

[uncut mulch]

population doubles every 20 years

[full garden]

like why is there 2Po one on each side

ohhh I see I am so stupid

thank you guys a lot

I am so stupid I can't believe I asked that question

thank you guys

[harsh cipher]

Hi

question

[clever inlet]

🤔

[harsh cipher]

Hello

[clever inlet]

🤔

[harsh cipher]

Left section in red.

Pi/4+ npi where n is an integer

why divide npi by 2?

It's such a simple concept I don't get it....

[clever inlet]

i mean

it just follows from the previous line

[harsh cipher]

I seriously need to start thinking logically

but I can't

lol

I can see that...

because we multiplied by 1/2 because of 2 infront of theta

[clever inlet]

yes

[harsh cipher]

but I can't wrap my head around how it moves on the unit circle

which makes it MORE confusing

[clever inlet]

the multiplier on theta makes it go around the unit circle twice as fast

[harsh cipher]

exactly that's the part I don't get

[clever inlet]

what about it?

[harsh cipher]

ok so by that red equation

we will end up at 3pi/4?

period is pi!!

we need to end up at 5pi/4

its really confusing me because the way they taught me in the lesson video is....

cos theta is 0 at pi/2 and 3pi/2

so what you do is 2x = pi/2 , solve for x and x= pi/4

and

${frac3pi/2}/2}$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

oh no

2x = 3pi/2....I get 3pi/4

but when writing general solution.....

is the part where I seem not not understand

to*

[remote veldt]

@harsh cipher for fractions it's $\frac{a}{b}$. To get a pi, put a backslash before it so $\frac{3\pi}{2}$

[obsidian monolith]

Nicholas:

[remote veldt]

note: I'm not checking your work just your latex

[harsh cipher]

okay thanks

[valid violet]

\frac a b yeet

[sharp marsh]

how do I

Foil (X^2+3x + 9/4)

[fleet yew]

@sharp marsh nothing to foil

Unless you mean you want to factor it

[sharp marsh]

oh

er so the problem is x^2+3x+1/4

And I'm suppose to make it into standard form?

@fleet yew

I'm kinda confused

[fleet yew]

Wait 1/4 or 9/4?

You're giving me two different numbers here

@sharp marsh

[sharp marsh]

yg

Ok so the

Problem itself is

@fleet yew

[fleet yew]

Lmao

Do you know what standard form is

[sharp marsh]

WrONG EMOTE

uh

https://cdn.discordapp.com/attachments/375398467930947585/678465038570225674/fc754164b6015099bfb49dbe2aca5f18.png

This?

[fleet yew]

Yes

[sharp marsh]

I'm wondering why this worked then

[clever inlet]

you lost your x for one

[patent beacon]

Ugh when schools use a dumb "standard form"

[fleet yew]

Dude that original question literally gives you it in standard form

It's literally just asking do you know what standard form is

[sharp marsh]

o

Wait why did the other one worked then?

[fleet yew]

Khan academy?

[patent beacon]

Some sources take standard form to actually be vertex form

Make sure you know what your teacher wants

[sharp marsh]

Web assign

[fleet yew]

It probably accepted it because it is algebraically equivalent and it can't actually tell the difference

Or maybe what kaynex said

Know your teacher's definition

[sharp marsh]

aaaa

[fleet yew]

Computers are really smart

Except when they're dumb

[sharp marsh]

oh

[sharp marsh]

So the one that worked

[fleet yew]

Youre missing an x here

[sharp marsh]

Is a quadratic function or no?

[fleet yew]

It is a quadratic

Lol how could it not be

[sharp marsh]

ck

Idk

So the answers (3/2 x)^2-2?

[fleet yew]

No

Check your math

Just start the problem over and you'll see your mistake

[sharp marsh]

er

[fleet yew]

compare this to what you got

[sharp marsh]

Am I turning it to vertex form then or?

[fleet yew]

Yeah ig that's what it wants you to do

[sharp marsh]

oh

So I make it (x^2+3x+9/4)+(1/4)-(9/4) right?

[fleet yew]

Yeah just turn it into vertex form

[sharp marsh]

(x^2+3x+9/4)-2?

[fleet yew]

You do know how to convert to vertex form right?

[sharp marsh]

um

not really

Don't I just factor the x^2+3x+9/4

and then include the -2?

[fleet yew]

Nope

Let me just tell you how it works

h=b/2

k=c-(b/2)^2

[sharp marsh]

o h

[fleet yew]

Assuming that there is no leading coefficient

[sharp marsh]

So was there any point in doing the (x^2 +3x + (3/2)^2)) +(1/4) -(3/2)^2

?

[fleet yew]

There's different ways of doing it

I've never really seen that method

[sharp marsh]

o h

Yep I'm back

[fleet yew]

Yo whats up

[sharp marsh]

So what was the graph and stuff

About the forms and how it worked or something?

[fleet yew]

What graph

[sharp marsh]

oh nvm

You just posted the triangle thing and I got confused

[fleet yew]

Lol that was a meme

[sharp marsh]

lmao

[fleet yew]

So you need help with smth or what

[sharp marsh]

uh

Vertex (2,-1) and point (0,7) I guess

Write the standard form of the equation of the parabola that has the indicated vertex and passes through the given point. (Let x be the independent variable and y be the dependent variable.)

[gilded mirage]

whats the command for texit?

[fleet yew]

Go away

Channel is busy

[gilded mirage]

oh

[fleet yew]

Ask ann

She loves latex questions

[sharp marsh]

lol

Does a = 7/5?

7 = a(0+2)^2+1?

wait no

[fleet yew]

Youve actually almost got it

Watch your signs

[sharp marsh]

3/2

cause order of operations

[fleet yew]

It is not +2

For h

[sharp marsh]

oh

OHH

[fleet yew]

Also check k

[sharp marsh]

Isn't k = -1

but the form has - k in it?

nvm

it doesn't

[fleet yew]

Lmao

You flip the sign for h but not for k

[sharp marsh]

a = 2

why am I like this

ok so

IF a = 2

7 = 2(0-2)^2-1

but now what .-.

Do I keep the x and y instead?

[fleet yew]

7 = a(0-2)^2-1

7=a(-2)^2-1

Yeah a is 2 just checking your math

[sharp marsh]

alright cool

Now what do I do?

[fleet yew]

Just put x and y back into the equation

Where they belong

[sharp marsh]

oh?

erm

y = 2(x-2)^2-1

[fleet yew]

Yes

[sharp marsh]

y = 2(x^2+4)-1

y= (2x^2 + 8)-1

er

Do I need to distribute or no?

[fleet yew]

Wtf are you doing man

[sharp marsh]

I DONT KNOW

[fleet yew]

I expected better from you

[sharp marsh]

[fleet yew]

Didnt you learn PEMDAS in the 7th grade

[willow bear]

what is happening in here

[sharp marsh]

Isn't it exponents first

[fleet yew]

Yes

y = 2(x-2)^2-1

Do the exponent first

[sharp marsh]

2(x^2+4)

no?

[fleet yew]

y = 2(x-2)^2-1

y = 2(x-2)*(x-2)-1

[sharp marsh]

wait

I forgot I can't just

square everything

oops

I've always wondered though

why doesn't it work

like I just know it doesn't work

[fleet yew]

Ooh i have a really good image for this

Hold on

Think of squaring as literally SQUARING

Imagine (x+a) as the side length of a square

And you must find the area

[sharp marsh]

so it wouldn't make sense cause uh

u doing wonky stuff

ya

Wait my brain just expanded

Is this brain expansion jutsu

[fleet yew]

,rotate

[sharp marsh]

OHHH

oh?

Oic

It's like the punnet square thingy

[fleet yew]

Uh yeah kinda

Thats an interesting connection

[sharp marsh]

So it's actually 2(x-2)(x-2)-1

and then 2(x^2-4x+4)-1

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

[fleet yew]

Yes

Just one more step

[sharp marsh]

do I

Subtract the 1

now

[fleet yew]

Yes

[sharp marsh]

2x^2-8x+7

Do I then factor it

foil?

factor

[fleet yew]

No you got it

Question asked for standard form