precalculus

[opaque olive]

so we evaluate leading degrees of x if we r dealing with int/-inf?

[viscid thistle]

With limits*

Yes

[opaque olive]

yes limits

nice

didnt know that

and thats 1?

[viscid thistle]

As long as they are fractions tho

and thats 1?
@opaque olive yes

So the slope is 1

[opaque olive]

okay

[viscid thistle]

y=1x+b

Now the b

Evaluate $$\lim_{x\to \infty}{\frac{x²}{x-1}-x}$$

[opaque olive]

im unsure why you initially divided by x though

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Its how you find the slope...

I said it

Thats why i ask you to read everything...

[opaque olive]

sorry i suck at english

lol

carry on please :)

$\frac{x}{x}$

[obsidian monolith]

Yes:

[opaque olive]

so 1

and thats y = x + 1

and thats the oblique asymtote

[viscid thistle]

Yep

You are done now

You have everything to sketch it

[opaque olive]

did we even need the oblique asymptote

[viscid thistle]

Yeah

[opaque olive]

is it just a line of symmetry ?

[viscid thistle]

But it was mentioned before so i assumed you already knew it

Not symmetry at all

Don't call asymptotes lines of symmetries bc they don't have to be symmetric

[opaque olive]

ok

[viscid thistle]

An asymptote is when the function gets closer and closer to it without touching it

[opaque olive]

what means an asymptote oblique ?

[viscid thistle]

Its like diagonally-positioned

Let me take a pic

[opaque olive]

so a asymptote that isnt vertical or horizontal

?

[viscid thistle]

Look how the oblique asymptote is

Its not dotted sadly

[opaque olive]

yeah i see it at x + 1

[viscid thistle]

@opaque olive also look how you said it touches 0,0 and it had a vertical asymptote at 1

Look how i said it increases fastly

[opaque olive]

yeah..

[viscid thistle]

All good now

?

[opaque olive]

yep

should be

thanks

sorry for taking alot of your time haha

[viscid thistle]

Hehe np

[opaque olive]

https://gyazo.com/37c9315c75aeafd5d0420812666cc107

@viscid thistle

in case you wanted to see it with the dotted asymptotes

[viscid thistle]

Oblique asymptote?

Beautiful

[tulip jay]

what is the relationship between the equation and the asymptote?

basically what is the slope of the asymptote

[tulip jay]

for x^2/(x+b) it is simply x-b

but whenever the x coefficient is not 1, it doesn't work

x^2/(ax+b)

[wide sequoia]

How do this??

[kindred whale]

Having trouble piecing this together anyone help?

Number 11

[uncut mulch]

getting crowded here

[tulip jay]

um does anyone look before they post?

[uncut mulch]

@wide sequoia @kindred whale please move

[past meadow]

@wide sequoia try taking a log on both sides but both you and crow need to move lol

[uncut mulch]

for x^2/(x+b) it is simply x-b
do you know how that works?

[wide sequoia]

What do you mean move?

[past meadow]

to another channel. this one is occupied. one of the channels in #❓how-to-get-help

[uncut mulch]

@tulip jay

[tulip jay]

um no

I'm thinking about how that would work because I don't want to be dumb

lol

[uncut mulch]

$\frac{x^2}{x+b} = \frac{x^2 - b^2}{x+b} + \frac{b^2}{x+b}$

[obsidian monolith]

ramonov:

[uncut mulch]

consider the above

[tulip jay]

I see how they are equal

I don't see how you simplify it tho

[uncut mulch]

x^2 - b^2 is a difference of 2 squares

[tulip jay]

oh

so x-b + b^2/(x+b)

[uncut mulch]

yes

[tulip jay]

how do you just delete the second part

[uncut mulch]

as x→ inf or -inf, b^2/(x+b) → 0 , (is negligible) and x^2/(x+b) will behave like x-b

[tulip jay]

doesn't that make x-b = x by logic

inf-b=inf and -inf-b=-inf

[uncut mulch]

b isn't negligible compared to b^2/(x+b) at extreme values of x

[tulip jay]

???

[graceful ledge]

???

[tulip jay]

b seems pretty negligible when you subtract it from inf

[uncut mulch]

i mean as x→inf, x-b does approach infinity

but that isn't what that the asymptote is about

[tulip jay]

oh ig that makes sense

[uncut mulch]

similarly: $\frac{x^2}{ax+b} = \frac{a^2x^2}{a^2(ax+b)} = \frac{a^2x^2-b^2}{a^2(ax+b)} + \frac{b^2}{a^2(ax+b)} \ \
=\frac{(ax + b)(ax-b)}{a^2(ax+b)} + \frac{b^2}{a^2(ax+b)} \ \
=\frac{(ax-b)}{a^2} + \frac{b^2}{a^2(ax+b)}$

[obsidian monolith]

ramonov:

[tulip jay]

my brain hurts lol

but that makes sense

so the 2nd part goes to 0 leaving (ax-b)/a^2?

omg this is beautiful. I love math

I love this discord too

that totally makes sense how we can ignore the second part now that I think about it because we are thinking about what happens as x approaches inf. That is what an asymptote is

thanks for the explanation

[viscid thistle]

why do i keep getting a math error whenever i type this into my calculator

[uncut mulch]

limitations with its output

|x| <= 9.999999999E99
for most hand-held scientific calculators

[viscid thistle]

is there any way to remove those lmitations

[uncut mulch]

wdym by remove?

the calculator is unable to display a value that big
but you can get it to do smaller calculations and do parts of it by hand

[harsh smelt]

or write program

haha

[uncut mulch]

$=-2 \cdot (2^{180})^2 \
\approx -2 \times (1.3532495541 \times 10^{54})^2 \
\approx -2 \times 2.348542583 \times 10^{108} \
$

[obsidian monolith]

ramonov:

[viscid thistle]

oh

i've seen my calculator output answer in that format but idk which mode it's called

[uncut mulch]

scientific notation

alternatively standard form (in context)

[viscid thistle]

i chose mode Sci but its making choose from 0-9

nvm i got it

[uncut mulch]

i believe that mode lets you specify number of significant figures

[viscid thistle]

@wide sequoia still need help?

[past meadow]

is it about the question with the 5^whatever? I answered that one

[viscid thistle]

Oh okay

Free channel now

[wide ocean]

the horizontal translation is -4(pi/6) correct?

= 2pi/3 units left?

[small creek]

you can check with desmos

[past meadow]

no it is not 2pi/3 units left

[wide ocean]

it looks like it

but not sure

@past meadow what is it?

3pi/3 x= 1 I believe, so -2pi/3 is near -1 maybe

[past meadow]

its pi/6

for sin(a(x+c))+d the phase shift is given by c units to the left

[wide ocean]

Ahh I see

Thanks! @past meadow

and the period of pi/2 contributes to the point near -1

[wide sequoia]

@past meadow

got another question

or anyone?

also hod do you calculate the asymptotes of logarithmic functions

[uncut mulch]

are those supposed to be multiplication dots?

[wide sequoia]

no

decimals

@uncut mulch

[uncut mulch]

are you sure?

it makes more sense if they are

[wide sequoia]

pretty sure its decimals

[native sequoia]

(3.4)^x?

[wide sequoia]

yessir

[native sequoia]

LHS is always positive for real x then

[wide sequoia]

?

[uncut mulch]

missed that part

[native sequoia]

and I think it should be multiplication

[uncut mulch]

there won't be any real solutions either way

[wide sequoia]

How comre

[uncut mulch]

the exponential terms are positive for all real x

thus the lhs will always be positive for all real x

will never equal zero and thus has no real solutions

[wide sequoia]

ok

[supple yew]

@uncut mulch are you free right now

if you are could you prove the steps for sin(0.77x)=-1/2

provide*

[uncut mulch]

based on past experience with these types of questions with exponents

they usually deal with exponents with the same base and avoid powers of decimals (except for e)
certain places also use dots in that format for multiplication

find the general solution(s) for 0.77x and then isolate x

[opaque olive]

So i solved this equation via completing the square $\color{green} {(a + 2)x^2 − 2x − a = 0}$ and i got got x = +- 1. I subbed my solutions only to find that -1 doesnt make it equal to zero which was strange

[obsidian monolith]

Yes:

[opaque olive]

anyone got any idea as to why that happend?

[fading token]

Can you show us the working out?

[opaque olive]

sure gimme a sec

@fading token

[uncut mulch]

you applied the $\pm$ in the incorrect step

[obsidian monolith]

ramonov:

[opaque olive]

ugh

yep

thanks

[uncut mulch]

also some dodgy algebra

the "scribble" in the denominator was a little ambiguous

[opaque olive]

lol

did alot of the steps in my head u see

[viscid thistle]

i need some help with getting the formula for this

would k just be 0.1386 and i just ignore the 'double every 5 minutes' thing

[strong escarp]

How do you do this?

[tulip jay]

you know the y intercept

y=a(b)^x

solve for b

[strong escarp]

what would A be

[tulip jay]

50

[strong escarp]

oh

so what is y

[tulip jay]

(30,10)

10=y

[strong escarp]

thx

[tulip jay]

np

[strong escarp]

x is just 0 right

[viscid thistle]

'

help plz

[tulip jay]

30, 10

[strong escarp]

30

[tulip jay]

yes

[strong escarp]

thanks again

[tulip jay]

also you get (.2)^(1/30) and you can just write that in the equation as y=50(.2)^(x/30)

instead of 50((.2)^(1/30))^x

to make it less messy

[strong escarp]

idk how to do this

[viscid thistle]

Ok so

$a^{\frac{b}{c}}=\sqrt[c]{a^b}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@strong escarp

[strong escarp]

oh

4^x

?

[viscid thistle]

I haven't checked it wait a sec

[pale bison]

yes

[strong escarp]

thx

[viscid thistle]

Yes

Lol

Np

[strong escarp]

What's the formula to answer this?

[tulip jay]

or you could just do ((64)^(1/3))^x

4^x

[viscid thistle]

Late mate

[tulip jay]

I just got confused lol

[wide ocean]

This question got my head spinning

it can't be B right?

[acoustic harbor]

Try multiplying the top and bottom by 1-sin(theta)

[viscid thistle]

Do you have to choose one option or more than one?

^^

[wide ocean]

@acoustic harbor like this

oh wait wrong sign

[acoustic harbor]

like what

[wide ocean]

one sec sorry

ahh so it's C

[acoustic harbor]

what

[wide ocean]

thanks 😦 Got that question wrong

[acoustic harbor]

hold up

[wide ocean]

oh

[acoustic harbor]

what’s 1-sin^2(theta) equal to

[wide ocean]

ignore the 1+3sin(theta) btw

oh it's equal to cos^2(theta)

[acoustic harbor]

yes

and then you can get rid of the cos theta on top

[wide ocean]

cos theta / cos ^2 theta

=cos theta

[acoustic harbor]

so it would leave you with 1-sin(theta)/cos(theta)

answer is D

[wide ocean]

OMG yesss

wait how does it leave that on top

[acoustic harbor]

because you multiplied the top and bottom by 1-sin(theta)

[wide ocean]

omg oh yeah

[acoustic harbor]

it doesn’t magically just disappear

[wide ocean]

thank you!!!!!! 😄

This is helpful 👍 thank you so much

@acoustic harbor ty

[acoustic harbor]

no problem

[viscid thistle]

What's the d value for this

oh its 3 nvm

Is this right

[past meadow]

@wide sequoia sorry i didnt answer was asleep

[wide sequoia]

NO problem

[opaque olive]

last part

where i need to deduce that $1 < b - a \leq \frac{4}{3}$

[obsidian monolith]

Yes:

[opaque olive]

idk how to tackle it?

i guess we need to start with $p^2 > 0$ and $q^2 > 0 $

[obsidian monolith]

Yes:

[opaque olive]

(no spoilers pls)

[austere bramble]

have you gone through part a and b

[opaque olive]

yes i have

[austere bramble]

ok lemme take a look

[opaque olive]

done everything except the last line

[austere bramble]

have you tried plugging in p=b+a and q=b-a in?

usually once you go through and simplify you can work you way backwards to show what they ask

[opaque olive]

ive proven what i needed to prove for part 3

i just need to do the deducing

last line

$1 < b - a \leq \frac{4}{3}$

[obsidian monolith]

Yes:

[austere bramble]

ok so once you solve for p^2 in terms of q

what do you get

[opaque olive]

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

[austere bramble]

should be a q^2 in numerator

[opaque olive]

yep

[austere bramble]

and q in denom

[opaque olive]

lemme check one sec

[obsidian monolith]

Yes:

[opaque olive]

yeah that

typo lol

[austere bramble]

ye

ok

so now

try to find some restrictions on q

that this equation forces

[opaque olive]

q doesnt equal 1

[austere bramble]

in particular, consider what kinds of values the LHS can take

[opaque olive]

also

positive

[austere bramble]

well

[opaque olive]

so bigger than zero

[austere bramble]

non negative

[opaque olive]

yeah

0

[austere bramble]

so you have then $\frac{-q^2}{3-3q}\geq 0$

[obsidian monolith]

M8 of 48:

[opaque olive]

it cant eequal zero can it?

[austere bramble]

are there any other things we can get?

why not?

[opaque olive]

nvm it can

[austere bramble]

take a look at the parity between the numerator and denominator

[opaque olive]

variable on numerator one degree higher than variable than denom

i dunno

[austere bramble]

the parity meaning pos/neg

[opaque olive]

oh

ok

so

its negative

[austere bramble]

right

but the quotient must be positive right

which forces what

[opaque olive]

ah

bigger than 1

[austere bramble]

right

no

0

[opaque olive]

oohh

yes

0

[austere bramble]

so 3-3q>1

[opaque olive]

wait

[austere bramble]

oops 0 lmao

[opaque olive]

ok

lol

yep bigger than zero

[austere bramble]

that's where the LH inequality will come from

[opaque olive]

wdym by LH

[austere bramble]

1<b-a

oh

left hand

[opaque olive]

ok

[austere bramble]

for $q\leq 4/3$

[obsidian monolith]

M8 of 48:

[opaque olive]

ok one sec

ok

so i subbed $q = b -a$

[obsidian monolith]

Yes:

[opaque olive]

didnt get $q \leq \frac{4}{3}$

[obsidian monolith]

Yes:

[austere bramble]

ye

you need to try and deduce something else

from what we had earlier $\frac{-q^2}{3-3q}\geq 0$

[obsidian monolith]

M8 of 48:

[opaque olive]

ok

what next

from $3-3q>0$ i got $1>q$

[obsidian monolith]

Yes:

[opaque olive]

@austere bramble

[austere bramble]

yo did u figure it out

[opaque olive]

unfortunately not no

[austere bramble]

one sec

lemme look again

[opaque olive]

sure

[austere bramble]

hmm not sure actually

one min lemme think

[opaque olive]

Aw oh

[fervent girder]

how do i do this?

this is the answer^

[opaque olive]

If you figure anything out @ me please

[austere bramble]

go to #questions @fervent girder

[fervent girder]

Sorry

[austere bramble]

and @opaque olive my guess is part b might have some sort of hint for this last inequality

[opaque olive]

Ok

Lemme back track

[austere bramble]

wait

actually

[opaque olive]

I think ive deduced it to be bigger than 1

But not algebraicly

[austere bramble]

if you can show $3-3q\geq -1$

[obsidian monolith]

M8 of 48:

[austere bramble]

then it'll follow

intuitively i think that'll be the solution

[opaque olive]

Strange

How do we know thats true

[austere bramble]

well

this is just my speculation

[opaque olive]

Ok

[austere bramble]

after looking for like 10 mins

as for the q>1 part

[opaque olive]

Yeah i think ive got it for q > 1

[austere bramble]

your argument is literally because since p^2 is non negative, and -q^2 is non positive, it forces the denominator to be non positive as well

[opaque olive]

Yeh

[austere bramble]

and it can't be 0

so it's strictly negative

so solving 3-3q<0 for q gives that part

im now very confident

you need to just come up with why $3-3q\geq -1$

[obsidian monolith]

M8 of 48:

[opaque olive]

Solving 3-3q<0 gives q<1

[austere bramble]

no

[opaque olive]

Wait

Nvm

[austere bramble]

xd

[opaque olive]

Lol

So demon has to be negative to make fraction overall positive is what you meant right?

[austere bramble]

yes

[opaque olive]

Ok makes sense

[austere bramble]

what class is this for btw

[opaque olive]

Its um

Not in the ciruculem

[austere bramble]

hm ok

[opaque olive]

Its competition math i guess

Kinda

[austere bramble]

ye

what i thought

im even more confident now that you just need to try and show $3-3q\geq -1$

[obsidian monolith]

M8 of 48:

[austere bramble]

since it's competition math

[opaque olive]

Ok

[austere bramble]

the reasoning might lie somewhere in part a/b

[opaque olive]

oh ok

ill give it a try

[austere bramble]

ye too tired to give it a more thorough look but just @ helper or something if you stuck gl

[opaque olive]

ok thanks

[austere bramble]

lemme know if u figure it out kinda curious

[opaque olive]

sure

[hardy abyss]

you have $\frac{p^2}{q^2} = \frac{1/3}{q - 1}$

[obsidian monolith]

Seoin:

[hardy abyss]

right?

[opaque olive]

from where

idk

[hardy abyss]

well you have $3p^2q - 3p^2 - q^2 = 0$

[obsidian monolith]

Seoin:

[opaque olive]

oh yes

yep

[hardy abyss]

so you have the thing I wrote first, yes?

[opaque olive]

yes

[hardy abyss]

but $p\geq q$ since $a$ and $b$ are nonnegative

[obsidian monolith]

Seoin:

[opaque olive]

yes

[hardy abyss]

so $p^2/q^2 \geq 1$

[obsidian monolith]

Seoin:

[austere bramble]

ah

[opaque olive]

yes

hm

[austere bramble]

how did i not see this lmao

[hardy abyss]

you threw away information when you used $p \geq 0$ instead of $p \geq q$

[obsidian monolith]

Seoin:

[austere bramble]

nah just rearranging into $\frac{p^2}{q^2} = \frac{1/3}{q - 1}$ was quite nice

[obsidian monolith]

M8 of 48:

[hardy abyss]

don't feel bad I didn't see it right away either 😂

neat problem I guess

[opaque olive]

so rearranging will get me the answer right

interesting

[austere bramble]

i just meant that once i saw it in that form, the answer was obvious

but the reasoning you need is that @hardy abyss said

[opaque olive]

yeah i see

[austere bramble]

what*

[opaque olive]

thanks for the help guys

[still quarry]

Plz help am I doing this right

For b), got derivative, plugged in 3 for t and ended up with 7

[rich flint]

@still quarry I got 7 as well for B

[still quarry]

Alright thx

And would you say that as in a growth rate of 7 bacteria per hour or every 3 hours or...

[rich flint]

@still quarry is rise/run also considered a slope?

We know the derivative evaluates the slope of a curved line

Output of the function/input of the function

[still quarry]

Well not in this case I dont think

But would you say that the growth rate is per 3 hours or per 1 hour?

[rich flint]

@still quarry slope of rise/ run is something associated with linear functions I’m saying the derivative is the slope of curves you can think of it in terms of y/x meaning your output function is the bacteria which is y and the x which is the denominator is the time.

@still quarry remember the average rate equation the only difference is the limit is added to it when discussing the derivative

@still quarry yes per hour

[still quarry]

So if x is 3 hours then there is an increase of 7 bacteria per hour?

[rich flint]

Yes at that time

Population of bacteria experiences a growth of what you just wrote

[still quarry]

Oh ok

So how would I properly write the rate?

[rich flint]

Population of bacteria/ hour

[still quarry]

So 7/hour at the given time of 3 hours right

[rich flint]

Yes but your forgetting your rate 7 population/ hour not 7 divided by h

[still quarry]

Oh ok

So I shouldn't make it look like division

[rich flint]

Example is 7 miles/ hour is a rate

[still quarry]

I see. Thank you.

[rich flint]

When your calculating the derivative division occurs

[still quarry]

Yup

Do you think you could check out something else for me?

[still quarry]

Oh ya I remember

[rich flint]

I can try

[still quarry]

Trying to figure whether it's -39 or -18

[rich flint]

Are you guys using the limit equation still?

[still quarry]

For this the derivative is needed

[rich flint]

Yes I know but I’m asking you are you guys using the limit equation still cause this is precalculus or your in the channel of precalculus as opposed to calculus

I am trying to see how to go about showing you how to solve this

[still quarry]

I know how to solve it but not sure which derivative i should end up with

Basically the d/dx of the given equation and then plug in 2 for t

[rich flint]

Ok just take a picture and I will see what you did

I have no clue where you are in class I can’t explain a method that wasn’t taught

I know your taking a derivative but their are different ways to take the derivative

[still quarry]

Got these two just not sure which is right

[rich flint]

Ok your using the limit equation

[still quarry]

Pardon my ugly writing

[rich flint]

What happened to your negatives

For when you plugged in t and 2

Oh wait a minute you didn’t I see the ln 2 ok so your not using it

[still quarry]

Yeah I havent written the full answers yet cuz not sure which to use

[rich flint]

What did you do on the first page

Ok you can do quotient rule are you familiar with the derivative of a^x = a^x ln(x)

[still quarry]

Yes

[rich flint]

We can rewrite the equation

After you rewrite the equation you can then do the quotient rule

[still quarry]

So....meaning my second one is correct

[rich flint]

Whichever gives you -18 I suppose so

But notice that when you have a negative value rewrite it in terms that you can consume if it is too heavy

[still quarry]

Ohhhh ok makes sense

My first one gives -18

[rich flint]

That function is not an enjoyable function so try to find ways to make easier on yourself when you have a negative power in the numerator

[still quarry]

Thank you. You saved me a fair amount of grief sir.

[rich flint]

@still quarry I guess so but I literally have no clue what you did on one line it looks like you were trying to do the limit equation then on the next you weren’t but hopefully what I wrote made sense

@still quarry nevermind I see it good job

[still quarry]

Yep all good and got it in just in time thanks

[frail patrol]

How do I convert 4x^2-y^2+16x+12y=56 into standard hyperbola form I've tried taking the square but I end up taking to much and don't end up with a 1 I've gotten -4, -1 for the other side I'm just confused and I tried typing it into a calculator but they divided by 36 for the y and I've been stumped for a few hours now all I really need is to know why they divided by 36 I don't need an answer I need an explanation Ill also put the answer below this message

[viscid thistle]

what

that does seem like standard form

am i missing something?

[frail patrol]

how do you use the commands I could show it to you

[past meadow]

you can just type it up in plaintext

[viscid thistle]

?

[frail patrol]

(x+k)^2/a^2-(y-k)^2/b^2=1

[viscid thistle]

yes

that's literally what you have

[obsidian monolith]

Commander Vimes:

$4x^2-y^2+16x+12y=56$

[viscid thistle]

wouldn't that be expanded form

lol

pretty sure for hyperbolas you keep it factored

[frail patrol]

yes I'm just trying to simplify it

[viscid thistle]

for it to be 'standard'

you can't

it as simplified as possible

[harsh smelt]

no

[past meadow]

is what you posted in the first place the answer?

[viscid thistle]

what do you mean "no" @harsh smelt

[frail patrol]

yes that was the intended form

I'm just trying to figure out how to do one of the steps

[obsidian monolith]

Commander Vimes:

$4x^2-y^2+16x+12y=56$
can be reduced to standard form

[viscid thistle]

??

[frail patrol]

yes

[viscid thistle]

of course it can

[frail patrol]

it reduced to https://cdn.discordapp.com/attachments/363224154469826562/733576093352853514/unknown.png

[viscid thistle]

yes

so what's the issue?

you can't simplify any more than that

[harsh smelt]

you can't

[frail patrol]

yes I know

[past meadow]

polynomial thats the answer

that he just posted

[viscid thistle]

oh

[frail patrol]

I just need to know how to get to that form

[viscid thistle]

so he posted the answer

oh

now i get it

i thought you had that form

[willow bear]

complete the square

[viscid thistle]

anyway

yeah

[past meadow]

do you know how to complete the square?

[viscid thistle]

complete the square

[frail patrol]

Ive tried that

I simplified for x

[viscid thistle]

show your work

show your work

[frail patrol]

k

sorry for the messy handwriting in advance

[willow bear]

oh boy.

[frail patrol]

Here’s 3

[harsh smelt]

,rotate

[willow bear]

oh boy ok you overcomplicated it already

[frail patrol]

I attempted twice and realized I wasn’t going anywhere

[viscid thistle]

lol

ok

[willow bear]

right so like

[viscid thistle]

i'm just going to show you

how i would do it

[uncut mulch]

👀

[viscid thistle]

we start with $4x^2-y^2+16x+12y=56$

[obsidian monolith]

polynomial:

[willow bear]

.........ok poly wants to show off so i'm out

[viscid thistle]

then note that

$4(x^2 + 4x) - (y^2 - 12y) = 56$

[obsidian monolith]

polynomial:

[viscid thistle]

(fyi i'm showing more steps than you would do in practice, so you can skip some of these)

[frail patrol]

yeah Ik

[viscid thistle]

note that $(x+2)^2 = x^2 + 4x + 4$

[frail patrol]

I see how you got there

[obsidian monolith]

polynomial:

[viscid thistle]

so

$4(x+2)^2 - 16 - (y^2 - 12y) = 56$

[obsidian monolith]

polynomial:

[viscid thistle]

this is because we get the 4 when we expand

and then we multiply by another 4 at the start

[frail patrol]

ahh

[viscid thistle]

so that's completing the square

now

note that

[obsidian monolith]

Commander Vimes:

[viscid thistle]

$(y-6)^2 = y^2 - 12y + 36$

[obsidian monolith]

polynomial:

[viscid thistle]

so now, we have an extra 36

but in our problem, we have a - infront of the parentheses

[frail patrol]

so you add

[viscid thistle]

so it's effectively going to get multiplied by -1

yes

so we have

$4(x+2)^2 - 16 - (y - 6)^2 + 36 = 56$

[obsidian monolith]

polynomial:

[viscid thistle]

we can simplify further to get

$4(x+2)^2 - (y-6)^2 = 36$

[obsidian monolith]

polynomial:

[viscid thistle]

by moving all the constants to the right

now we divide both by 36

[frail patrol]

ohhhhhhh

[viscid thistle]

$\frac{(x + 2)^2}{9} - \frac{(y-6)^2}{36} = 1$

[obsidian monolith]

polynomial:

[frail patrol]

thank you that makes sense

I was just missing the part of moving the constants and diving by the constant

thank you once again

[copper vigil]

https://cdn.discordapp.com/attachments/369860765957488642/733655096910544938/20200717_080257.jpg

anyone mind telling me what exactly i did wrong here

i mean for the original inequality to be true, x cannot be greater than both the zeros of the parabola

i understand what it's supposed to be but i don't get why it's not working out algebraically

[viscid thistle]

there is no mistake

[copper vigil]

but it's clearly incorrect

[viscid thistle]

no it's not

[copper vigil]

if x is greater than -2 then the parabola can be less than 0

[viscid thistle]

yup

[copper vigil]

specifically in the interval (-2,-1)

[viscid thistle]

but you have two restrictions

so you take the one which has the greatest 'impact'

[copper vigil]

huh

[viscid thistle]

i.e. restricts it further

ok

suppose i tell you that

a > 5

and a > 7

Its incorrect.

what this means is that a > 7, since that one restricts it more than the other

if both have to be true

@viscid thistle no it's not

nothing about that work is incorrect.

[copper vigil]

oh i see what you're saying and that makes sense

[viscid thistle]

yes

[copper vigil]

but why doesn't that account for the left side of the parabola

[viscid thistle]

because you haven't done those cases

both factors can be negative

[viscid thistle]

and that will also satisfy the inequality

[harsh smelt]

you separated w/o pointing cases

[viscid thistle]

@harsh smelt that matters not

[copper vigil]

so you're saying that i have to split it into 4 different cases, allowing each factor to be both positive and negative

[viscid thistle]

yup

that is one way

[harsh smelt]

product of two factors is positive if and only if all factors agree in sign

[viscid thistle]

@copper vigil so how can you do the left side of the parabola, as you described it?

[copper vigil]

let each of the factors be negative so that x+2<0 and x+1<0, so x<-2 and x<-1

and then the -2 is the stricter condition

oh i see that makes a lot of sense thanks

[viscid thistle]

so usually, there is one more thing you have to check. not for this particular polynomial, but in general

which is the in between case

for this one, it doesn't matter

but for more complicated examples, it can matter

$$x²+3x+2>0$$ factoring $$(x+2)(x+1)>0$$ divide the cases $$\begin{cases} x+2>0 \ x+1>0\end{cases} \ \begin{cases} x+1<0 \ x+2<0\end{cases}$$ now find the intersection of both cases and then find the intersection again to reach that $$x\in (-\infty, -2) \cup (-1, \infty)$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

but casework will always work

[copper vigil]

so it's not a contradiction to say that x>-2 AND x>-1. kind of stupid in hindsight but glad you told me that

[viscid thistle]

it isn't, they're both true. it's just that they both have to be true the same time, so one ends up overpowering the other, simply put

[harsh smelt]

@copper vigil in most of the cases in quadratics which have two different roots it is enought btw to check middle one interval

[copper vigil]

what is the other thing you'd check for @viscid thistle

[viscid thistle]

well

suppose you have something like

$\frac{(x-5)(x+2)^2}{(x+3)} \ge 0$

[obsidian monolith]

polynomial:

[copper vigil]

oh so just discontinuities then

i know that

[viscid thistle]

not particularly

in fact, they don't even matter here

at all

[copper vigil]

what is it then

[viscid thistle]

just an inequality

[copper vigil]

but what would you have to check for

[viscid thistle]

well, you have these things right

so try plugging in some large negative numbers

like

minus a million

that is roughly

$\frac{(-1,000,000)(-1,000,000)^2}{-1,000,000} \ge 0$

do you agree

[obsidian monolith]

polynomial:

[harsh smelt]

in case polynomial gives it is enough to check signs of (x-5)/(x+3)

[viscid thistle]

@harsh smelt can you go to literally any other channel

[copper vigil]

roughly i guess

[viscid thistle]

yup

so that immediately tells us that

when x is -infinity, then our inequality is satisfied

since the square is going to be positive

and we have 2 negative numbers

so the negatives cancel out

and we're left with a positive number, roughly 1million^2

now

look at the original expression

$\frac{(x-5)(x+2)^2}{(x+3)} \ge 0$

[obsidian monolith]

polynomial:

[viscid thistle]

when do you think we no longer have two negatives in our fraction?

so say we begin at $x = -1,000,000$

[obsidian monolith]

polynomial:

[viscid thistle]

and we climb towards being positive

at what point do you intuitively think we will first lose a factor that is negative

[copper vigil]

-3

[viscid thistle]

yes, after we hit -3 (since we can't actually hit -3 as it is an asymptote)

[copper vigil]

so that's just a vertical asymptote

[viscid thistle]

but like -2.999999999

at -3.00000001 we're still negative

[copper vigil]

so it goes to +- infinity yeah i know that

[viscid thistle]

but as soon as x = -2.9999999

that factor becomes negative

and our inequality is no longer satisfied

because the denominator is negative

but one of the factors of the numerator is positive

right

so currently

we know that

$(-\infty,-3)$ satisfies our inequality

[obsidian monolith]

polynomial:

[copper vigil]

get to the point lol

[viscid thistle]

lol

really man

[copper vigil]

i know what an asymptote is

[viscid thistle]

so

at what point do you think we start to satisfy our inequality once more

[copper vigil]

5

[viscid thistle]

yup

as soon as we hit 5

(since 5 isn't going to make anything undefined)

we satisfy our inequality

so the point is, we don't have to write down anything on a piece of paper

most inequalities you can solve in your head

for example

i could do the one i posted above in 2 seconds

i plug in -1 million

i see that it's satisfied

i plug in 0, i see that it is not satisfied

i plug in + 1 million, i see that it is satisfied

and i'm given the bounds at which these changes happens

i just have to list them in order

and it's done

but if you ever get lost, you can always do casework manually

[still quarry]

Plz help with #2

[elder junco]

Positive slope

A derivative...?

[still quarry]

Hmm yes that proves that it does but as to why is the question.....

[past meadow]

think about the definition of the derivative

[viscid thistle]

hm mean value theorem is the best way to explain #2

@still quarry

[wide ocean]

quarter year meaning 4 times a year?

does anyone understand?

<@&286206848099549185>

[viscid thistle]

@wide ocean yes.

[still quarry]

Plz help

[wide ocean]

thanks!

@still quarry (0,1) is y intercept

plug in 3 for x, and solve for c since it should equal 12

and same for b and c

[uncut mulch]

also consider properties of the vertex

[still meadow]

^

[still quarry]

I just got it all anyways but thanks

[still meadow]

Sometimes it's easier to convert from another form to the required rather than finding the function in the required form right away.

[still quarry]

Yes exactly what I was thinking

[odd helm]

how do i solve this

[elder junco]

Good question

[still quarry]

Jeez never seen one of those

[elder junco]

Polar equations...

[still quarry]

Oh....well they are certainly neat when graphed

[copper vigil]

what is cos(3t)

[acoustic harbor]

what do you mean by that?

[upbeat bone]

You can use $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$

[obsidian monolith]

Wilston Lynx:

[upbeat bone]

@copper vigil

[viscid thistle]

For exponential growth and decay I can use any base except for 1, right?

[copper vigil]

any positive base besides 1 or 0

not really

give context

[willow bear]

as stated no

is there a problem you're doing rn

thank you for not answering my question

as if that needs any proof beyond just writing out the array...

[willow bear]

what's "it"

oh so negative numbers don't exist now?

m=-1 and n=19 would fit your restriction

m=-100 and n=118 would still fit the restriction of m+n=18

:/

then m and n can only take values between 1 and 17, excluding 9.

[obsidian monolith]

Xoc:

Xoc:

[wide ocean]

I can't seem to get t=20 (answer) after

converting to logs

[kindred garnet]

Can some help me?

[hexed ermine]

Think about it, the lowest it can go is 1 meter

When it's at its top it's at 10+1 which is 11 meters

So from this theres a vertical translation of 1

The min is 1 which is at time = 0

Assuming the wheel turns constantly, at t=4min it's at its max

So you know it is something to do with a sinusoudal equation

[tulip jay]

1st step should be to draw a picture

to try to understand the problem

min is not at t=0 tho

[hexed ermine]

How is it not

You are loading at t=0 which is your min

[tulip jay]

at 0 you are at 6 o clock

which is 11pi/6

wait

[past meadow]

how is 6'oclock 11pi/6

[tulip jay]

idk how clocks work

lol

[hexed ermine]

Bruh

6 o clock is south

[tulip jay]

remember how I said to draw a diagram? I didn't do that

smart me

[hexed ermine]

Anyways yeah, draw it out, you'll notice that the min is at t=0, so that makes you think that it has to do with cosine because cosine has an extrema at t=0, so you can manipulate it to give you what you want

Just need to scale it by some factor and stretch out the wave so that it reaches its extrema at t=8,

[opaque olive]

How would i go about sketching this graph in the plane of x - y? $\color{red}{|x| + |y| = 1}$

[obsidian monolith]

Yes:

[past meadow]

consider cases

[opaque olive]

yeah

when x > 0 and y > 0

and x < 0 and y < 0

[past meadow]

what happens when both y and x positive, what happens when both negative, what happens when one negative etc

[opaque olive]

the only 2 cases?

hm

[past meadow]

x>0 y<0, and X<0, y>0 you missed

[opaque olive]

why do i need to consider those?

x can only hold 1 value at a time?

[past meadow]

wym

[opaque olive]

i mean

if x > 0

y is bigger than zero

until 1

so

[past meadow]

why is that true?

what about x=0.5, y=-0.5

that satisfies |x|+|y|=1, but has x>0 and y<0

[opaque olive]

oh

so y is a value that changes too?

[past meadow]

yeah theyre both variable

just like with when you're graphing anything else

[opaque olive]

usually

theres only 1 variable?

like

f(x) = y

y value depends on x

[past meadow]

i see 2 variables there

sure, and y's value depends on x here too

but

y can take multiple values for the same value of x

[opaque olive]

okay

so

when x > 0

[past meadow]

like x=1/4 gives y=3/4 OR -3/4

[opaque olive]

oh i see...

[past meadow]

so, now consider what the graph looks like for each of those four cases

[opaque olive]

yeah a cube

[past meadow]

cube in 2 dimensions?

[opaque olive]

lol

[past meadow]

do you mean square?

[opaque olive]

yep xD

[past meadow]

if so, yes it is shaped like a square

can you draw exactly what it looks like?

[opaque olive]

yeah

square

[past meadow]

like with axes and y and x intercepts?

[opaque olive]

it was just a typo xD

yeah

0,1 to 0,1 -1,0 0,-1

[past meadow]

i assume you mean 1,0 for the first one there

and yeah

[opaque olive]

yes

its late here

lol

sorry for mistakes

[past meadow]

nw

[opaque olive]

yeah im gonna give the rest of these questions a go

[past meadow]

gl

[opaque olive]

thanks

[viscid thistle]

can someone

help and give me the steps to this

https://media.discordapp.net/attachments/731060280431476766/734212701307404328/image0.png

[pale bison]

what have you tried?

you should review some examples first

[gloomy path]

pretty straight forward
you need to learn how to multiply matrices

[pale bison]

no shit

[gloomy path]

Ik but like
in this case
there is no way around it

either get a tutorial

this dumb intuition of mine also helps

[opaque olive]

quick question, whats the newline function

in Latex

[upbeat bone]

\

[opaque olive]

ok thanks

[upbeat bone]

double \

[opaque olive]

ok

For this equation, $|x-1| + |y-1| \leq 1$
I need to consider these equations right?
$\y=-x+3
\
y=x-1
\
y = x +1\y=-x+1$

[uncut mulch]

could. not need.

[obsidian monolith]

Yes:

[opaque olive]

i want to sketch it

[uncut mulch]

|x-1| + |y-1| = 1
would have a horizontal and vertical shift of 1 from
|x| + |y| = 1

[opaque olive]

yeah i saw that

i could have just translated it by the vector 1 1

[fleet yew]

,w plot abs(x)+abs(y)=1

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by AMD#4434

[viscid thistle]

Daaamn

[fleet yew]

Oh my god its a square thats so cool

If you consider that abs(x) = sqrt(x^2)

You can actually see this as somewhat analogous to the circle equation

I wonder if you could parametrize this

Like maybe a square sine and a square cosine

[barren sapphire]

@fleet yew You can. Here's what it looks like if you parametrize by angle (in analogy to regular sine and cosine) rather than area (in analogy to both regular and hyperbolic sine and cosine): https://www.desmos.com/calculator/mtprksk0af

[wide ocean]

I tried 5^2x+6 / 5^4x-16 = 5^6x+21

do all 5's cancel out? Or the left side only

I can't seem to get the answer either way "1/8"

[barren sapphire]

It was good to convert everything to powers of 5s. Now you need to think, what happens when you divide two powers of 5? For example, (5^3)/(5^2). What does that become (as a power of 5)?

@wide ocean

[uncut mulch]

apply the proper exponent laws

[fleet yew]

@barren sapphire for some reason i was expecting it to look like a straight sawtooth wave

[barren sapphire]

You get one if you go by area instead

Like, sin θ is the y-coordinate on the unit circle where the area is θ/2. So if you make a "square sin" that's the y-coordinate on that square where the area (counterclockwise from the positive x-axis) is half of the input, you do indeed get a straight sawtooth

https://www.desmos.com/calculator/bijucp1jqy

[barren sapphire]

@fleet yew you can see the area and the angle in this animation: https://www.desmos.com/calculator/vpwwvfqgjg

[fleet yew]

@barren sapphire wow

That looks really nice

I still dont get how exactly you parameterized it though. Those are some really complicated formulas youve got there

i'm gonna try doing the math myself because this is really interesting

[still quarry]

Plz help with a communication question

[spring relic]

if d2y/dx2 >0 min pt, if d2y/dx2<0 max pt

[opaque olive]

https://gyazo.com/13b8f1987c6d8c7aadc01992c35fd2f1

part 2

(ive done part 1)

all the x values are squared i see?

do i just square root my x intercepts?

all my stationary y values should be the same

i could just sketch it from scratch but i assume the question didnt intend for it to be done like this

[hushed sorrel]

from part i's sketch, you can again find how many real values of x satisfy the equation for each k in part ii (remember the plus-minus when square rooting, and that square rooting a negative is not real!) and also notice f(x^2) is an even function so it exhibits symmetry about the y axis. Finally it is obvious as x tends to infinity, y tends to infinity, so you should be able to piece this all together @opaque olive

they wont expect you to label specific points I dont think, just to get the general shape right

is this STEP :p I remember taking it thisyear and step 2 was ass

lmao

[opaque olive]

yeah STEP lol

[hushed sorrel]

its quite hard to start off with

[opaque olive]

im pretty sure i gotta label intersections right?

[hushed sorrel]

u UK or international?

[opaque olive]

UK

wbu

[hushed sorrel]

same

camb maths?

[opaque olive]

yeah its late here

nah

just for fun

[hushed sorrel]

I think bc its a sketch q then maybe label them?

[opaque olive]

yeah

you camb math?

[hushed sorrel]

but in general you dont have to, and in recent years, the papers have been getting longer and longer qs

[opaque olive]

oh ok

[hushed sorrel]

well hopefully lol