precalculus

[short sorrel]

thats still pretty vague

can you give a specific question you're struggling on?

[lament garden]

oh my bad lol

linear functions and slopes

[night spear]

Really? This is basic Alg then. Still more specific though

[urban vigil]

hullo can someone help me with a precalc question

it's a concept ig, i don't understand it

just need help with 9B

[viscid thistle]

@urban vigil what dont you get

[urban vigil]

how to do 9b @viscid thistle

[viscid thistle]

What dont you get from 9b

[urban vigil]

how to start and then do it

so basically: everything abt 9b

[viscid thistle]

Its telling you to write something for f(x) when its domain is unbroken

Hence its continuous right

So think about something

[urban vigil]

yea?

[pseudo sonnet]

can someone tell me

if f(x) = c, where c is a constant

is f(x) a polynomial function

e.g f(x) = 0, is f(x) a polynomial function

[acoustic geyser]

f(x) = Anx^n + An-1x^n-1 + . . . + A1^x + a0
Eqn of polynomial
An should be non zero

[viscid thistle]

A polynomial is of the form: $a_nx^n ... a$

[obsidian monolith]

rudy:

[acoustic geyser]

@rudy like this?

[viscid thistle]

Yes

But, an expression such as: 4x + 4m + 62xm + 5 is a polynomial

But it is linear as well

As this is all of degree one.

[pseudo sonnet]

my notes from class say it is lol

[acoustic geyser]

:(

[viscid thistle]

Did your teacher tell you that?

[pseudo sonnet]

yeah

[acoustic geyser]

Did you ask him/her how?

[pseudo sonnet]

my notes say

f(x) = 0 * x^0

[viscid thistle]

f(x) = c is a polynomial.

As long as c != 0

So the constant function such as this

is a polynomial of degree zero

[acoustic geyser]

Ohh so
F(x)=0×0x^2+0×0x+0 is a polynomial aswell???

[viscid thistle]

Yes.

[acoustic geyser]

Rudy can u answer my question in calculus chat?

[viscid thistle]

And a constant function is defined as a polynomial due to f(x) = k \implies f(x) = k(x^0).

[acoustic geyser]

Nice

[pseudo sonnet]

i got my first exam in precalc tomorrow

reviewing all my notes rn + hw

i just expect some random problem that requires PHD in mathematics of induction and differential calc etc 😂

always that 1 impossible question

[willow bear]

@viscid thistle that's a pretty horrible expression for the general form of a polynomial

[viscid thistle]

Youre supposed to fill in the blanks😳

[willow bear]

no like it doesn't make any sense

it doesn't parse as an algebraic expression

[viscid thistle]

Lets edit then

A polynomial is of the form: $a_nx^n + a_{n-1}x^{n-1} + \dots + a_0$

[obsidian monolith]

rudy:

[willow bear]

yes that's better

of course, that sum can contain one or zero terms

[viscid thistle]

Im on mobile so I write the tex that way as to give a faster summary

[willow bear]

no, what you wrote earlier is just incorrect

it's not abbreviated

it's just unreadable

[rigid sun]

@lament garden whats the problem

[solid reef]

Need some help

Pleasee

[zenith bough]

i am confused by his answer

why did he use the perpendicular equation for parallel

is this wrong?

[viscid thistle]

He did it wrong.

When he multiplied 3 on both sides, he didn't multiply the y.

[zenith bough]

i done this problem and want to check for correction

[viscid thistle]

Oh.

Parallel should be $y=3(x-2)+9$ I believe.

[obsidian monolith]

leviosa:

[viscid thistle]

Or y=3x+3.

[zenith bough]

i have the 2nd one

[viscid thistle]

K.

[zenith bough]

thank you for confirming that im not dumb

i thought my answer was wrong

[viscid thistle]

How about check your answers using SoupCalculator or something.

Yahoo answers don't cut it most of the times.

Or wherever you got that.

[zenith bough]

that's chegg

sadly

[viscid thistle]

Chegg?

[zenith bough]

yes

[viscid thistle]

Oh.

That's pretty stupid.

[zenith bough]

for quetion and answer in chegg, it's not realiable

[viscid thistle]

You can just check it yourself.

Make sure your slope is the same (or opposite reciprocal) then plug in x and y to see if they equal.

You don't have Chegg on a test after all :/

math is fun

[zenith bough]

right but i want to see if how i doiing it is correct

[viscid thistle]

There is no correct way in solving a math problem.

There is only an answer.

i just started learning reverse functions

[shrewd urchin]

Math is hard. Aswell

[viscid thistle]

#bots

[formal iris]

Can someone help me with a-d

[formal iris]

oh ok

😦

[reef rune]

$f(x) = \frac{1}{x + 1}$ \

$g(x) = \frac{1}{x - 1}$ \

$ f(g(x)) = \frac {x-1}{x} $ \

is the domain of $f(g(x))$ $x \in \R, x \neq 1, 0 $

[obsidian monolith]

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

[reef rune]

I'm confused because the answer key says x can't be 1 or 0, but when I graph in desmos there's a point at (1,0)

<@&286206848099549185>

[viscid thistle]

!15m

That's an F.

Ping Helpers 15 minutes after you post a question.

[reef rune]

oh oops

sorry bout that

[viscid thistle]

Also are you supposed to find the graph of that line or did they give you a line to graph.

[reef rune]

no graph is given

nor am I supposed to be graphing this

but I graphed it to check my answers

[viscid thistle]

Send the original question.

[reef rune]

but it doesn't check

the question is to find the domain of f(g(x))

that's it

[viscid thistle]

Okay do you really think I know what f(x) or g(x) is.

[reef rune]

yeah?

[pale kettle]

Think about what you can plug into g(x)

[viscid thistle]

Oh that was also you.

[reef rune]

so I could plug anything into g(x) except 1

[pale kettle]

That's why the domain can't include 1

[reef rune]

yeah but desmos shows point at (1,0)

is that a mistake

[pale kettle]

No

It's because of the way you simplified it

$f(g(x)) = \frac{1}{\frac{1}{x-1} + 1}$

[obsidian monolith]

Zopherus:

[pale kettle]

If you write it like this, without simplification, it's clear that x can't be 0 or 1

[reef rune]

so desmos shows undefined points

since it looks like it's defined at (1,0)

[pale kettle]

If you just straight graph $\frac{x-1}{x}$

[obsidian monolith]

Zopherus:

[pale kettle]

Then this will be defined when x = 1

[reef rune]

the thing is I didn't

[willow storm]

can someone tell me what to look for when doing this: (I just need a place to start, already tried setting equal and isolating k ) find all values of k such that the line y = 3x+k is tangent to parabola whose equation is y=2x^2-5x+3

[reef rune]

so that's where I got confused

I guess desmos literally likes to define undefined points lol

thats nice

[pale kettle]

In that case

No, Desmos doesn't define undefined points

It leaves that circle there

[viscid thistle]

Hole.

[reef rune]

the circle was filled in though

[viscid thistle]

I don't think desmos has open circles.

Or I haven't seen it.

[reef rune]

oh wait

nvm

I see the hole now

[viscid thistle]

[reef rune]

yeah it's because I labeled it

sorry guys

my bad

[willow storm]

nvm lmao i walked away from tha tproblem instantly but when i came back i realized how stupid it was; just had to use the discriminant properties, i was scared to calculate it wiht the k inside for some reason

[neon drum]

I can't figure out part iii for both methods

[toxic glen]

Guessing a bit I think a.iii is looking for b/a -1. Which is just another way of doing percentage change.

Errr maybe( b/a-1) *100 to keep it in percent

And I think b.iii is meant to be (b-a)/a * 100

[viscid thistle]

how can i solve this equation for x: 4-x=e^(1-x)+2

[pale kettle]

You can't

At least, not very nicely

[viscid thistle]

it's supposed to be 1 and it's for a high school class so it should be simple or i'm just doing it wrong

[pale kettle]

I guess one thing you can do is see that if x is an integer

then the left side is an integer

[toxic glen]

Answer is 1

[pale kettle]

so you want the right side to be an integer too

[toxic glen]

if you substitute it in.

err nvm I just realised you said the answer was 1, I just used trial and error <_<

[viscid thistle]

it's fine

um

[toxic glen]

this seems like one of those solve by inspection

[pale kettle]

Anyways, the only way to make the right side an integer

is to have e^0 which you can do by setting x =1

[viscid thistle]

oh ok then ty

[heady jewel]

e^t=t-1

[clear glade]

when it says from what the original parent graph is what do you they mean, should I use a.) as the original parent graph

[proud sparrow]

You can use any original graph

as long as you can transformit

[clear glade]

oh I see

[proud sparrow]

I guess

[clear glade]

so x^2 is good?

[proud sparrow]

Yeah

[clear glade]

ok ty

[proud sparrow]

So is y=254(x+2)^2 if you prefer that

[clear glade]

oh ok thanks

[scenic musk]

is this allowed

is this allowed

(adding 6 to both sides of the inequality)

[proud sparrow]

can you not abuse the equals sign?

and where are both sides?

@scenic musk

[scenic musk]

no the equal sign

is used to show

that i can get the right hand side

if i add 6 to both sides

like 1. then 2.

[proud sparrow]

Please don't abuse it

Use $\iff$

[obsidian monolith]

Element118:

[scenic musk]

i didnt know how to do that in word

but assuming i had done that

is my statement true

[proud sparrow]

no

[scenic musk]

how come

[proud sparrow]

because your inequality has 3 "sides"

[scenic musk]

how do i get rid of the -6 on both sides then

[proud sparrow]

you need to do the same thing to all 3 "sides"

which is adding 6

[scenic musk]

then my proof fails if i do that

to the middle

[proud sparrow]

then let's see your proof

@scenic musk

[scenic musk]

let me format it in word

one moment

wait am i allowed to take picture

and post picture from phone

im posting in calculus section

since not applicable here

[scenic musk]

@proud sparrow this is good tho right?

[proud sparrow]

yes

[scenic musk]

Yay ty

[proud sparrow]

you added 6 to all 3 sides

[scenic musk]

i figured out i was manipulating wrong inequality, tyvm

[clear glade]

Why does -(x-1)^3 (x+2) have an end behavior of falling on both left and right sides? I thought since it's negative and odd it would fall on the right and rise to the left?

[stuck lark]

$-(x-1)^3 (x+2)$ doesn't look odd to me

[obsidian monolith]

RokettoJanpu:

[clear glade]

since the leading coefficient is negative and the degree (3) is odd why wouldn't it be going down on the right and up on the left

[stuck lark]

because you're multiplying -(x-1)^3 by a linear term

[clear glade]

what do u mean

[proud sparrow]

check the degree of the entire polynomial

[stuck lark]

the degree is not 3

[clear glade]

4?

[stuck lark]

yes

[clear glade]

oh

why isn't it the same way when I do other equations sometimes?

[stuck lark]

show us what you're talking about

[clear glade]

well for example $f(x) = (2x+1)(x-3)^2$

[obsidian monolith]

Waterblade:

[clear glade]

hmm it might not make sense for this example but I'll say it anyways

[proud sparrow]

the degree is?

[clear glade]

1

[stuck lark]

no

[clear glade]

wot

3

?

[stuck lark]

yes

[clear glade]

so you have to multiply both of them?

if it's like that

if it's factored

?

[stuck lark]

imagine yourself expanding the whole thing and you get a degree 3 polynomial

expanding (x-3)^2 gets you a quadratic (degree 2), multiply that with a linear term (degree 1) and you get a cubic (degree 3)

[clear glade]

I see

Makes sense, thanks

[stuck lark]

no problem

[clear glade]

Roketto-senpai

need help with this first one

I got everything so far but not sure about the equation part now

$f(x) = -2(x+1)(x-3)/(x+2)(x-4)$

[obsidian monolith]

Waterblade:

[clear glade]

what do i do with the y-int now (-3/4)

[stuck lark]

how can you find the y int of f(x)?

[clear glade]

lol

$-2(x+1)(x-3)/(x+2)(x-4)$$

I got y int since it was 0, -3/4

[rose locust]

This is a dumb question , but in cosine law, does Labeling the triangle matter, as in does c always have to be the hypotenuse ?

[stuck lark]

why are you asking this?

what do i do with the y-int now (-3/4)

[clear glade]

?

Because I think I have to integrate into the equation now

[stuck lark]

@rose locust law of cosines works for all triangles

why don't you check to see if you even need to "integrate the y int"?

[clear glade]

oh

graph looks good on a website to check it

[stuck lark]

how can you check the y int of f(x) without looking at its graph?

[clear glade]

?

[stuck lark]

there's a way to find the y int of f(x) without graphing f(x)

[clear glade]

Sorry senpai, im so confused

well in that pic it says I have to give the formula for each graph and the y int in the first one is 0, -3/4

[stuck lark]

if i give you some random function f

$f(x) = 3x^3 - 2x + 2$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

what's the y int?

[clear glade]

2?

[stuck lark]

how did you find it?

[clear glade]

it's at the end...

[stuck lark]

fair

[clear glade]

but I can't just put -3/4 at the end of a rational function

[stuck lark]

$f(x) = (x+2)^3$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

what's the y int of this?

[clear glade]

0

[stuck lark]

no

and don't graph it

[clear glade]

ok

oh wait

i sub in 0 for x

so 8

[stuck lark]

good

[clear glade]

naisu

[stuck lark]

$f(x) = \frac{-2(x+1)(x-3)}{(x+2)(x-4)}$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

you came up with this function

[clear glade]

indeed

[stuck lark]

how can you check if you need to add a "vertical shift" or not?

[clear glade]

based on the y int?

[stuck lark]

the y int needs to be -3/4, so how do you check if f(x) already has that y int or not?

[clear glade]

0 for x and solve

[stuck lark]

plug in x=0, tell me what the y int is

[clear glade]

ok

sugoi

-3/4

[stuck lark]

any need to adjust the y int?

[clear glade]

I guess not

no

[stuck lark]

👍🏽

[odd helm]

Why is r(x)= 4 I thought the remainder was 4/x

[deft lodge]

Please use screencap. It's literally on your keyboard.

4/x isn't a polynomial

[uncut mulch]

it is asking for r(x) not r(x)/b(x)

a(x) = b(x)q(x) + r(x) where r(x) is the remainder

[odd helm]

Ok thank you

And sorry bad I got lazy

[viscid thistle]

but taking a picture is literally more effort

[clear glade]

I'm confused about a problem:

d.)

I got Domain: (5, infinity), Range (-infinity, infinity) and for Inverse: Domain(-infinity, infinity), Range:(5, infinity). When I checked my answer online though I got a different answer

Instead of negative infinity to infinity for range of function it says 0 to infinity but I'm confused why since domain of inverse is negative infinity to infinity

[flint vale]

the range for sqrt of above is not negative

how do you get negative as range if theres no form of negation outside the sqrt

so its 0 to infinity

you know this already im sure

[clear glade]

Oh I see

[flint vale]

@clear glade theres another way to look at this, the inverse is reflection over y=x

ex

so if you need an image that will help

[clear glade]

I see

but I thought range of the original function is the domain of the inverse

[flint vale]

well here it swaps

i dont know that as a hard rule but there could be a simple counterexample that flew over my head

btw, the domain of the inverse isnt even correct

[clear glade]

I see

?

[flint vale]

well you see the inverse function, if you got that then you can tell me that the domain doesnt bother with -infinity

yet you included it

i guess u just looked at the result and thought, i can plug in negatives here

[clear glade]

so for x^2+5 domain is not -infinity to infinity

[flint vale]

yeah i peeped that, i just thought about reflecting over y=x

as the inverse of the function, no

but alone it is defined with negatives

oh right

it needs to be one to one

so if you include the negatives its no longer one to one

@clear glade

bc lets say it isnt one to one and you go backwards in reflecting, its not a function anymore

bc you know vert line

[clear glade]

right

I understand now, thanks

[clear glade]

how do I this one? a.) for example

For Inverse

I changed x and y and got to x=-2^y + 3 but not sure what do now

$x=-2^y + 3$

[obsidian monolith]

Waterblade:

[willow bear]

well

you need to isolate y now

[clear glade]

oh ok

so x -3 = -2^y now

now I think I have to do something with log right?

but I can't do log base -2 so not sure how to do this part

[serene heath]

u dont have to

just move it to the other side

[clear glade]

move what?

[serene heath]

the -2^y

[willow bear]

it's not (-2)^y

it's -2^y

[clear glade]

so $x+2^y = 3$

[obsidian monolith]

Waterblade:

[clear glade]

im confused

[willow bear]

no!

multiplying both sides by -1 no bueno????

[clear glade]

oh yea

does log base 2 (-x+ 3) = y look good

[stuck lark]

@clear glade 👍

[clear glade]

Arigatou senpai

Roketto-senpai

why is this true?

I thought range would be positive 1 to infinity

since inverse has domain of 1 to infinity

[willow bear]

no, the domain of the inverse is not (1, +∞)

[viscid thistle]

Look at the transformation of the graph.

[clear glade]

Oh I see

[graceful wing]

Suppose that cosx=−3/10 and that m is in the third quadrant. What is the value of tanx?

[proud sparrow]

what did you try?

[graceful wing]

I found sinx by 1-cosx

then did sinx/cosx

[proud sparrow]

Why is sin x found by 1-cos x?

[graceful wing]

idk if the answer should be positive or not since the answer is tan in the 3rd quadrant

my bad I meant squared

[proud sparrow]

okay is sin positive?

[graceful wing]

yes

[proud sparrow]

in the 3rd quadrant?

[graceful wing]

Yeah

oh I forgot that the sqrt is plus or minus

that would explain why my final answer is wrong

well ty for pointing out where I went wrong

[upbeat prairie]

Could someone help me with this, how would I convert this to a standard form.

[proud sparrow]

type an equation?

oh, the standard form is given above

a(x-h)^2+k

@upbeat prairie

[upbeat prairie]

@proud sparrow I see

Also, isn't the answer supposed to be y= -(x-2)^2-14

[proud sparrow]

looks right

wait

[upbeat prairie]

Since 10-4(-1)

would equal 14

I'm going off of this

[fresh schooner]

hey can somebody help me parse what this statement means in plain english?

[pale kettle]

It's basically written in English, what are you confused about?

[normal sundial]

lmao

for all elements v and w in V' v+w is also in V'

basically that V' is closed under addition

[fresh schooner]

cool, I just wanted to be certain

I wasnt sure if the comma meant "and"

[noble isle]

im struggling with finding the area under a curve

[short sorrel]

could you be more specific?

what sorts of integrals are you struggling with?

[noble isle]

like in general

integrals?

[short sorrel]

(also, probably better suited for #calculus )

oh wait

you're not working with integrals?

[patent beacon]

You gotta put a lot of rectangles in there

Lots

[short sorrel]

in that case, yeah, you're just dealing with tons of rectangles/triangles

[noble isle]

yeah were doing these draw rectangles under curve

[short sorrel]

view it as a geometry problem

[noble isle]

its acc precalc so idk calc

[short sorrel]

yeah, sorry, i thought you meant integration (the more sophisticated calculus version of what you're doing)

[noble isle]

if yall know the CPM book i can guide you to the area im at

[patent beacon]

Oh. You likely just need the area of triangles, rectangles and circles

[noble isle]

mk is it ok if i post some problems here later?

[patent beacon]

Ya

[uncut bronze]

why cant we graph sin x sin y = a

"a" being a number

[hardy abyss]

,w plot sin(x)*sin(y) = 1/2

[obsidian monolith]

Display results online and refine query

Requested by Seoin#0692

[heady jewel]

damn

[hardy abyss]

@uncut bronze looks like we can to me?
though if |a| > 1 and x, y are real there will be nothing to plot

[heady jewel]

,w plot sin(x)sin(y)=e

[obsidian monolith]

Display results online and refine query

Requested by Lionel#5497

[heady jewel]

lol

,w plot sin(x)sin(y)=e^x

[obsidian monolith]

Display results online and refine query

Requested by Lionel#5497

[sonic rivet]

only works for constants <=1, no?

since you can't really multiply two sine outputs and get a number larger than 1

[uncut bronze]

i tried putting it in desmos and didnt work

[sonic rivet]

works in desmos for me

[uncut bronze]

huh

[uncut bronze]

nvm i forgot brackets

[upbeat prairie]

Could someone explain to me how Xv=1 and not -1

It is supposed to be -1

[willow bear]

no it's not

[uncut mulch]

It is supposed to be -1
wdym?

[viscid thistle]

What is the problem

[uncut mulch]

@upbeat prairie

[upbeat prairie]

How though

Oh wait...

well that was embarrassing lol, apologies folks

[rigid sun]

[viscid thistle]

@stiff basalt this channel

[stiff basalt]

@viscid thistle ye tyty

[rigid sun]

[charred stratus]

test for convergence; i used the integral, but its getting stuck at gaussian integrals

[willow bear]

$\sqrt{\log(n)} \leq n^{1/2}$ \ so your series can be compared termwise with $\sum_{n=2}^{\infty} \frac{1}{n^{3/4}}$

[obsidian monolith]

Ann:

[charred stratus]

o i c 😄

but will that fetch me marks :/

[willow bear]

why would it not

it's just an instance of the comparison test

[charred stratus]

the professor insists on using integrals as much as possible

[willow bear]

it's not worth it here

[charred stratus]

agreed, it gets too shabby afterwards

[willow bear]

all you want is to test a series for convergence, diving balls deep into a complicated integral isn't worth your time for that

[charred stratus]

😄 i wish my professor understood that

anyways thanks man!

[willow bear]

does the professor actually give you 0 marks for a solution that doesn't use the integral test or what

[charred stratus]

for instance this question was for 4 marks, if i didnt use the integral workings, he'll only give me 2 marks for the answer, and cut the rest for 'insufficient' explanation

[rigid sun]

well

what was your explanation?

did you use words to explain?

[pseudo sonnet]

so u guys know the zero product formula

A * B = 0

i had a true/false question on my test

something about A * B = 1

can the zero product formula be applied if you set it equal to 1

[rigid sun]

uh

what

the first requirement for a zero product formula

if im understanding this correctly

is that it has to equal 0

[pseudo sonnet]

good

thats what i was thinking too

and literally in the name zero product lol

[short sorrel]

the zero product property is that $ab = 0$ implies $a = 0$ or $b= 0$ (or both)

[obsidian monolith]

Namington:

[short sorrel]

this obviously doesnt apply to anything other than 0

for example, $3 \cdot \frac{1}{3} = 1$

[obsidian monolith]

Namington:

[short sorrel]

but neither 3 nor 1/3 are 1

[fresh marsh]

For the quadratic ax^2 + bx + c

What do people call the motion which occurs to the graph when you manipulate b?

[proud sparrow]

It's definitely a translation

@fresh marsh

[fresh marsh]

I see

thanks, I got a hit on wikipedia

[tepid basin]

f(x) = {x^(2)-a x<-1
{sin^(-1) x -1<=<x<=1
{-bx^(2)+2x 1<x

what would be the domain of f(x)?

[viscid thistle]

All reals.

Also never notate it like 1<x.

[tepid basin]

sorry

[odd helm]

Guys I’ve been struggling with this for so long and I just can’t figure it out

And not just this question specifically it’s just the whole thing

Because khan gives only 1 video on this that doesn’t go over any problems really it just goes over what y=asin(bx+c) +d means and the written explanations are hard for me to understand

So if someone could link me to something that can help me learn this or explain it to me I’d appreciate it thank you

Bro this is why I hate math like this stuff is so confusing makes me want to just punch a wall sometimes

[hazy patrol]

I never remember going over Sinusodial functions let me see if I can look something up

[odd helm]

Ok thank you

[hazy patrol]

Oh, I just wasn't the familiar with the term. From my understanding of this it appears that Sinusoidal functions simply refers to transformations of the sine function

[odd helm]

Yeah and cosine

[steady plaza]

is reflection over x axsis anodd or an even function

[sonic rivet]

neither

[odd helm]

I have no clue

[sonic rivet]

reflection across x axis wouldn't be a function (of x)

[steady plaza]

yeah right

[odd helm]

So how do I do this

[steady plaza]

cus thats 2 y values for an x value

[sonic rivet]

yes

it could be even as a function of y

@odd helm do you know what a cosine looks like

[odd helm]

Yes it doesn’t touch origin and goes down

After hits y right

[sonic rivet]

what is the y-intercept

and what are some x-intercepts

[odd helm]

It’s 1 isn’t it

Uhh

[sonic rivet]

yes

y-intercept is 1

[odd helm]

Idk the x intercepts

[sonic rivet]

where does cos(x) = 0

[odd helm]

Maybe pi/2?

[sonic rivet]

yes

[odd helm]

Ok

[sonic rivet]

and -pi/2 as well

you have x/2 instead of x

so these intercepts become pi and -pi

then you multiply everything by 2

so the y-intercept is 2 now

the x-intercepts (also midpoint values of the cosine) are still at pi, -pi

then take what you have

and shift it one unit down

and you have it

[hexed bolt]

Hey guys, how could I do this?

[sonic rivet]

if you have two intercepts a and b

then the quadratic is of the form k(x-a)(x-b)

[hexed bolt]

Mhm

[sonic rivet]

for whatever constant k you want

k>0 upwards parabola

k<0 downwards parabola

[hexed bolt]

Yep

[sonic rivet]

ya, so just do that

pick whatever positive and negative k you want

[hexed bolt]

Oh ok

What about this?

[sonic rivet]

you have two numbers x and y

[hexed bolt]

Yes

[sonic rivet]

x+3y = 54

x = 54-3y

you are maximizing the product

so maximize xy, or (54-3y)*y

it's a downwards facing quadratic

so find the maximum

then find the value of y which gives you that maximum

then use x = 54-3y to find the value of x that gives you that maximum

[hexed bolt]

Alright

[vestal lake]

can y’all help me with #29 please and thanks my brain hurts

[obsidian monolith]

Gryff:

Gryff:

[viscid thistle]

If niether then it's not odd nor niether.

@vestal lake

[vestal lake]

thanks @viscid thistle

[viscid thistle]

👍

[willow bear]

no

NO!!!!!!!!!!!!!!!!!!!!!!

@viscid thistle you do realize that the only odd function according to what you wrote is the ZERO FUNCTION, right???

[viscid thistle]

What.

[willow bear]

-f(x) = f(x)

[viscid thistle]

Oh.

Shit.

Fuck.

f(-x)=-f(x).

I'll DM him.

[worn leaf]

can anyone give me a hand with a limit question?

[willow bear]

no because you haven't posted it

[worn leaf]

lim x-->pi/2= (x-pi/2)sec(x)

[willow bear]

what's that = doing there

[obsidian monolith]

RokettoJanpu:

$\lim_{x \to \frac{\pi}{2}} \left(x - \frac{\pi}{2} \right) \sec(x)$

[willow bear]

thank

[stiff basalt]

Anyone know the answer to this?

[stuck lark]

what have you tried so far?

[stiff basalt]

make it equal to like 0 but that just finds roots does it not

[willow bear]

yes

and that isn't what you want

[stiff basalt]

im not sure how exactly sure how to start with this

[willow bear]

have you ever done any questions of the form "here's a function defined by a formula, find its domain"?

[stiff basalt]

only one with fractions

[viscid thistle]

Shouldn’t you graph that function to determine the domain

[stuck lark]

graphing is one way; i think the question wants you to do it algebraically

[stiff basalt]

yes

[stuck lark]

notice the function has ln

[stiff basalt]

but putting it in desmos, shows it as less than 3

[stuck lark]

start by asking yourself "what numbers are ok to put inside ln?"

[stiff basalt]

ahhhhh i get it

i just didn't know about ln too much, so ln(0) is not possible

therefore x cannot be 3

[stuck lark]

algebraically how would you find it though?

so yeah, ln(0) is undefined

so can you write some kind of inequality that lets you solve for all values of x where ln(64-4^x) is defined?

[viscid thistle]

Bruh

$ ln(x)$ s.t $ x > 0 \implies x\geq3 $ is undefined

[obsidian monolith]

rudy:

[viscid thistle]

$ x = {x | x < 3}$ or $(-\infty, 3)$

[obsidian monolith]

rudy:

[neon drum]

How do I do this? I don't know what to plug in

[stuck lark]

do you know what $\Delta f(x)$ means?

[viscid thistle]

You got them all wrong?

[obsidian monolith]

RokettoJanpu:

[neon drum]

No I checked an answer to see if I was doing it right

I wasn't

[viscid thistle]

Watch the KhanAcademy playlist for limits..

[neon drum]

delta f(x) is change in x right?

[stuck lark]

nah, it means a certain change in the value of f(x)

[stiff basalt]

u would make the equation = 0?

[stuck lark]

@stiff basalt make exactly what equal 0?

[stiff basalt]

the ln(64 - 4^x) = 0

is there such thing as inverse ln?

[stuck lark]

yes

[stiff basalt]

it should still be 0 on the right side

[stuck lark]

but that equation is not helping here

[neon drum]

okay so would i do f(0)/0 and f(1.5)/1.5

and put in the difference?

[stuck lark]

@neon drum that's not how to do it

@stiff basalt do you know the domain of f(x) = ln(x)?

@neon drum delta f(x) means a change in f(x) not f(x) itself

[viscid thistle]

I literally sent step by step

Solution

[stuck lark]

look at part A... they're asking for the ratio [delta f(x)]/[delta x] for the interval x=[0, 1.5]

[viscid thistle]

With interval notation and set builder notation

@neon drum

🤔

[neon drum]

oh @stuck lark got it, thanks

i didnt see @viscid thistle

i still dont see it

[viscid thistle]

Look at this picture

Do you see it now

[stuck lark]

rudy, did you mean to ping red glass cups instead of trinomial?

[viscid thistle]

Oh fuck

😳😳😳

Big whoops

Yeah Trinomial you just have to check KhanAcademy as I suggested

First 3 vidyas

[stuck lark]

lmao, btw be easy on the brainlets, talk em through solving it without posting the solution in one go

[stiff basalt]

yea im not sure

[neon drum]

how can I find the ratio? @stuck lark

[stuck lark]

first, what's delta x for part A?

[stiff basalt]

i know if u excuse the ln part

u can make it equal 0 and solve for x

but other than that im not too sure

it will end up being x = 3, but the actual domain is x < 3

[neon drum]

8.48972

[stuck lark]

nah, do you know what delta x means?

[neon drum]

change in f(x)

[stuck lark]

nope, delta f(x) means change in f(x)

[viscid thistle]

Hello

[neon drum]

okay so then a change in x

[viscid thistle]

Having issue with this equation

[stuck lark]

yes indeed

[rigid sun]

you're having...

[stuck lark]

@viscid thistle sorry, this channel is a bit crowded atm

[rigid sun]

problems

[viscid thistle]

|1-2x| <= -1

[rigid sun]

what is that symbol?

<=

[stuck lark]

@neon drum yes, delta x means change in x

[viscid thistle]

less than or equal to

[stuck lark]

and what is delta x for part A?

[rigid sun]

ok

what's the problem?

o

[viscid thistle]

there shouldn't be a set of numbers...

mainly because of it being <= -1

[rigid sun]

wdym there shouldn't be a set of numbers

oh

nvm

[uncut mulch]

was the problem written correctly because yes, what you posted has no real solutions

[rigid sun]

im sped

[neon drum]

1.5? 😭

[rigid sun]

uhhh

[viscid thistle]

yea

[rigid sun]

there are no solutions to your problem

[viscid thistle]

@uncut mulch it was written correctly

[rigid sun]

there are no solutions to that

[stuck lark]

@neon drum yes

[uncut mulch]

were you looking for confirmation or was there an answer key saying otherwise?

[stuck lark]

@neon drum now we have to find the delta f(x) that corresponds to the delta x

[neon drum]

so from the x = 3 , x = 5 delta x would be 2?

[stuck lark]

uh no

we're still on part A, yes?

[viscid thistle]

looking for confirmation

[neon drum]

yes, I was just trying to figure it out

[stuck lark]

what are the endpoints of the interval in part A?

[neon drum]

like (x,y)?

[stuck lark]

the interval you were given is x = [0, 1.5]

what are the endpoints of that interval?

[neon drum]

[0, 1.5] right?

[stuck lark]

@neon drum that's the interval, but i want you to tell me the endpoints of the interval

[neon drum]

i dont know what is meant by endpoints

[stuck lark]

@neon drum x = [0, 1.5] means the set of x values between x=0 and x=1.5

[neon drum]

right so wouldnt the endpoints be x = 0, x = 1.5 since we cant go any higher

[stuck lark]

the endpoints are indeed x = 0, x = 1.5, it's just because we were GIVEN the interval x=[0, 1.5]

the endpoints are the leftmost and rightmost x values of the interval

[neon drum]

So after finding the endpoints, how would we go about finding the ratio?

[stuck lark]

first repeat to me what delta x is

[neon drum]

change in x

[stuck lark]

and what exactly is its value for part A? @neon drum

[neon drum]

1.5

[stuck lark]

ok, now how do you find the delta f(x) that corresponds to the delta x?

[neon drum]

plug in 1.5 to f(x)?

[stuck lark]

that's part of it

do you remember finding the slope of a function from an earlier algebra class?

[neon drum]

yes

point slope formula right>

[stuck lark]

no, just slope

remember calculating delta y, aka y2-y1?

[neon drum]

yes

[stuck lark]

same thing here

x1 is 0 and x2 is 1.5, so what is y1 and y2?

[neon drum]

y1 is 6 y2 is 14.48972

[stuck lark]

@neon drum last step... [delta f(x)]/[delta x]

[neon drum]

got it, thank you

[stuck lark]

you're welcome

[fresh marsh]

How does one describe the geometric consequence to the graph of a quadratic like ax^2 + bx + c when you manipulate the a coefficient?

Is it a combination of "vertical" and "horizontal" scaling?

[short sorrel]

it's fixing the value of the y-intercept, while changing its distance from the vertex.

this isnt exactly accurate, but its the rough intuition

[fresh marsh]

ahh thanks

[short sorrel]

sorry, b value is the derivative

a value is distance from the vertex

(and again, this explanation isnt perfect, but its the intuition behind it)

[rigid sun]

wdym

just do it

on a graph

use geogebra

[fresh marsh]

I have

I used desmos

[rigid sun]

then there's no point in asking

lol wtf

[fresh marsh]

there is a point

I don't know how to describe it

or am I supposed to go, "oh, it kinda goes like this?"

In fact, isn't that when you'er supposed to ask a question?

Surely one doesn't ask a question just to receive a response of, "Why don't you just graph it..."

[rigid sun]

i mean

we can't help you put it into words

you saw what it does like you said

lol one sec

[fresh marsh]

I don't know if it's in fact a combination of vertical and horizontal scaling

so I wanted a confirmation

[rigid sun]

well

[fresh marsh]

I mean, isn't that called putting it into words?

I'm sure someone can do better

[rigid sun]

lol

it technically

is

just say vertical scaling

one sec

[cyan marsh]

I'm not sure if this is the right channel, but how do I simplify this?

[willow bear]

consider writing -4x as (-3x) + (-x)

[cyan marsh]

hmmm alright

so would the x² -3x on the top and bottom cancel out?

[willow bear]

no

they would not

$\frac{a+b}{a} \neq b$

[obsidian monolith]

Ann:

[cyan marsh]

oops

there goes a good chunk of my hw progress

[rigid beacon]

Well now you know not to make that mistake again, because you'll have the memory of having to redo all those problems

[scenic musk]

if S'(t) is speed as a function of temperature, then the units for S'(t) would be speed per temperature. Correct?

[shrewd urchin]

What?

[violet nebula]

Can someone explain how i would solve for a function in an exponent?

[willow bear]

consider rewriting 1/49 as a power of 1/7

[violet nebula]

oh, i get it, thanks

[shrewd urchin]

So

f(x)=6x+13

[hybrid night]

does anyone know how to factor x^2+25?

ik its (x+5i)(x-5i) but idk how

like why isnt it (x+5i)(x+5i)

[willow bear]

(x+5i)(x+5i) = x^2 + 10ix - 25

tfw 10i != 0 and -25 != 25

[hybrid night]

wouldn't (x+5i)(x-5i) = x^2-25?

[stuck lark]

foil it out yourself and see

[quick dagger]

when its not a unit circle, why is sin = y/r rather than just y?

[stuck lark]

use the circle to draw a right triangle and you'll find out soon enough

[random tulip]

when defining range or domain in a function in interval notation do you round down? a problem i have has a maxima of -0.333 on one of the lines and but the answer is (-inf,0) instead of (-inf, -0.333) is this just how its done or am i missing something

i have the domain correct but i cant understand what is going on with answer to the range

[vague zephyr]

the homework site says at least one is wrong but im not sure which one

[limber bone]

sqrt(4x^2+9) is even

cuz sqrt(4(-x^2)+9) = sqrt(4x^2+9)

x^2+6x+10 is not even cuz x^2-6x+10 != x^2+6x+10

x^2+6x+10 is neither

[vague zephyr]

oh crap, thanks for point that out

i havent been taught how to do this yet, not in the book im using or in the video my teacher sent me

nvm saw a video

[noble isle]

can someone help me with math

im v confused

I'm having trouble with 2-108 and 2-110 and 2-111

[heady jewel]

#❓how-to-get-help holy

[solid reef]

Can I get some help w simplifying that?

[viscid thistle]

Factor out common terms in numerator.

[solid reef]

i tried that but i only got lost

so i tried multiplying top and bottom by ((x^2)-5)^2/3

[viscid thistle]

Nononono.

[solid reef]

but i got lost again

[viscid thistle]

Factor first before cancelling.

Simplify the numerator first.

[solid reef]

honestly dont know how

[viscid thistle]

Bruh.

Can you multiply (2x)(4)(x^3) together?

[solid reef]

yea

24x^4

[viscid thistle]

;-;

[solid reef]

wait

no

sorry

8x^4

[viscid thistle]

$3x^2(x^2-5)^{1/3}-8x^4(x^2-5)^{-2/3}$

So the numerator looks like this.

[obsidian monolith]

Gryff:

[solid reef]

ok

[viscid thistle]

So what can you factor out?

[solid reef]

x^2

and

the stuff inside the brackets

[viscid thistle]

$x^2[3(x^2-5)^{1/3}-8x^2(x^2-5)^{-2/3}]$

[obsidian monolith]

Gryff:

[viscid thistle]

How to factor the stuff inside the parantheses?

[solid reef]

you can right?

[viscid thistle]

Yes.

[solid reef]

umm

[viscid thistle]

Hint: When you factor, you factor out the lowest factor.

So out of the two in the brackets what would be the lowest factor?

[solid reef]

so -2/3?

[viscid thistle]

Yes.

Do you know how to factor it

Or would you like me to give an example.

[solid reef]

i have no clue

[viscid thistle]

Aight.

So if we have $x^{1/2}+x^{-1/2}$

[obsidian monolith]

Gryff:

[viscid thistle]

We can factor it like: $x^{-1/2}(x+1) \ = x^{1-1/2}+1(x^{-1/2})$

[obsidian monolith]

Gryff:

[viscid thistle]

Do you understand what I did?

[solid reef]

yea you subtract the exponent value

right

[viscid thistle]

No.

[solid reef]

umm

[viscid thistle]

Here.

I think this'll help.

$2+4=2(\frac{2}{2}+\frac{4}{2}) \ x^{1/2}+x^{-1/2}=x^{-1/2}(\frac{x^{1/2}}{x^{-1/2}}+\frac{x^{-1/2}}{x^{-1/2}}) \ = x^{-1/2}(x^{1/2-(-1/2)}+1) \ = x^{-1/2}(x+1)$

[solid reef]

ooohh

ok

[viscid thistle]

Yep.

[solid reef]

so the exponent value in the first parentheses is just 1?

[viscid thistle]

Yes.

[solid reef]

oooohh

[obsidian monolith]

Gryff:

[viscid thistle]

Wait.

[solid reef]

im talking about the original question

[viscid thistle]

Yeah I ain't answering your original question.

You do that.

[solid reef]

yea ik

[viscid thistle]

I mean you can use what I just did for your case.

Gl.

[solid reef]

aight thanks

[viscid thistle]

👍

[scenic musk]

i have a function: 6.2 = x^3 - 3x + 4

when i solve for x in quadratic formula i get way different number than the graph online shows me

could anybody help me please

[stuck lark]

6.2 = x^3 - 3x + 4
is not a function of x

[scenic musk]

oh, well im trying to find X when Y = 6.2 for that function

how would i assemble the equation?

[stuck lark]

er... also it's not a quadratic function

[scenic musk]

oh

sorry for the mistake

how would i solve for x then, would u happen to know

[stuck lark]

to solve this?

6.2 = x^3 - 3x + 4
looks a bit ugly, solve graphically

[scenic musk]

i did,

so i guess i can asssume prof wont ask me to solve for exam

without graph

[stuck lark]

you should expect him to give you nicer numbers so there's some hope of solving by your usual methods

[scenic musk]

oh ok, shouldnt be an issue with nicer numbers

ty for ur time

[stuck lark]

np

@scenic musk are you sure you have to find x that satisfies f(x) = 6.2?

[scenic musk]

yes because im need to find X when y = 6.2 and x when y = 5.8

so i can find the delta for the limit of the function

[stuck lark]

right, yeah, do it graphically then

[viscid thistle]

Does business count as maths

[proud sparrow]

@viscid thistle Depends really. Mostly accounting and stuff rely on putting numbers into the right buckets.

[viscid thistle]

Hello! I'm really confused here

I checked the direct rule, everything, and it seems to be correct, but when I do it manually I hurt s whole different result and I don't know why

<@&286206848099549185>

[patent beacon]

The question was fine lol

Just, no helper ping

[cold jackal]

sorryyy I saw the person above do it

bad choice

[patent beacon]

After 15 mins is the rule

That's all. What's the q?

[cold jackal]

was unaware

Hiii, does anyone know the CORRECT order of transformations on a graph? They are:
a) horizontal shift
b) vertical shift
c) vertical stretch/shrink
d) horizontal stretch/shrink
e) reflections

I’ve been told first you do vertical shifts, after that I don’t know but I know it matters

[patent beacon]

Nope, shifts are last

[cold jackal]

my teacher does this thing where she gives us the chapter and tells us to teach ourselves and we did one practice problem and that’s what I got from it

[lament garden]

I need help with slope

[cold jackal]

Do you know the rules about how slope changes when it is perpendicular vs parallel?

i know I was asking him to make sure he knows

[brave obsidian]

oh

[cold jackal]

I don’t wanna give him the answer

...

[lament garden]

No I don’t

[brave obsidian]

my bad dude

[cold jackal]

AHHAHA it’s fine

[lament garden]

My teacher dosen’t really teach

[cold jackal]

what does perpendicular mean?

[lament garden]

Isnt it 90 degrees to a given line

[cold jackal]

Yes

It intersects the line at a 90 degree angle

So imagine shifting the line 90 degrees

What would the slope be?

Use y=2x for this example

it’s easier visually but try to think of it in your head

[lament garden]

Isn’t the slope -4?

[patent beacon]

You'll never guess if you don't know it

[cold jackal]

No, so if the slope is positive at first, and you turn it 90 degrees, the slope will become negative

[lament garden]

Oh

[cold jackal]

and it will also become the reciprocal which means the opposite

so the new slope will be - 1/2x

[lament garden]

Wow I really feel stupid 😂

[cold jackal]

In short the slope of a line perpendicular to the original is the negative reciprocal

I didn’t mean to make u feel like that lolll

It’s okay

You said your teacher doesn’t teach and you didn’t even learn it so it’s fine

Was just trying to work through it

[lament garden]

Ah alright

[cold jackal]

now do you know how you would find the line that goes through those points?

[lament garden]

Yeah

[cold jackal]

You need to plug the given values into x and y

so far we have x, y, and m