precalculus

[rose locust]

Am I doing this correctly ?

[gilded prawn]

Hello can someone possibly explain how to get the answer on this question

I am a little confused

Thanks!

[limber bone]

sin(pi+theta) = -sin(theta)

an identity

true for all theta

@gilded prawn

[gilded prawn]

So the answer is -0.1?

@limber bone

[limber bone]

if sin(theta) = 0.1

[gilded prawn]

Or sorry

[limber bone]

then sin(theta+pi) = -0.1

ye

[gilded prawn]

Oh sweet

Thanks for the explanation

[limber bone]

sure

[light jewel]

Never mind

I got it

[graceful wing]

Express tan (−223°) in terms of a reference angle and the tangent function, is it tan ( 137° )?

[rose locust]

I just want to make sure. Are my options for sin theta (in the box) : sin= -1/2 and sin= -1/3?

or would it be 1/2 and 1/3

[limber bone]

1/2 and 1/3

2sin(theta)-1 =0

and 3sin(theta)-1 =0

[graceful wing]

do you know how to answer my question?

[sharp pagoda]

quick question , going into calcalus which units of precal should i have a strong grasp of ?

[hardy abyss]

functions, specifically polynomial, exponential, logarithmic, trigonometric

[short sorrel]

you should know your trig identities

or at least be able to review them when the time comes

make sure you're realllllly comfortable with your algebra.

calculus is just algebra with extra steps

(at least at an intro level)

[sharp pagoda]

i dont know why but not matter what i do i cant grasp algebra, i know i probably sound stupid but i never know where to start for any question

[upper flint]

the first ones

[stuck lark]

what did you try?

[upper flint]

honestly im confused with the whole page @stuck lark

i thought if x1 < x2 then the interval isnt increasing

but the y value changes so idk

[stuck lark]

i thought if x1 < x2 then the interval isnt increasing
i don't know what this means

[upper flint]

Its smtg my teacher said

idk if im interpeting it right

[stuck lark]

@upper flint hmm that doesn't sound right. what do YOU think makes a function increasing or not?

[upper flint]

@stuck lark if any value x or y increases?

[stuck lark]

it's not really about the x value

what you do is move along the x axis from left to right. as you do that, pay attention to the y value. if the y value increases... the function is increasing

[upper flint]

so from (-8,-4) to (-8,-2)

f increases

[stuck lark]

you mean from x = -8 to x = -2?

[upper flint]

wouldn't it be x=-4 to x=-2

so the answer to "is f increasing on the interval (-8,-2)" is yes?

[stuck lark]

when the question says "is f increasing on the interval (-8,-2)?" it's asking you to do this:

start at x = -8, then travel to the right along the x axis until x = -2. you pay attention to the corresponding y value as you go along. is it increasing?

[upper flint]

ooooh

i see

[stuck lark]

so is the function increasing on (-8, -2) or not?

[upper flint]

increasing @stuck lark

[viscid thistle]

Is there a channel here for stats

Cause I'm dying in this class

[short sorrel]

#probability-statistics

[stuck lark]

@upper flint 👍🏽

[late parrot]

This seems to be the correct place for discussion of intervals

[flint vale]

it is

if anything someone will redirect you

[late parrot]

Alright

I’m in a little bit of a dilemma

I’m unsure if it’s an issue with me, my teacher, or the textbook

So I’m covering intervals

[flint vale]

its pretty open ended here

draw your x and y axis

[late parrot]

The way we were taught to interpret interval notation (n, n) is (n < x < n)

[flint vale]

keep going, ill keep explaining

[willow bear]

(n,n) is (n < x < n)

[late parrot]

I’m having a really specific issue where the way we were taught to interpret it leaves a hole in the parabola

[willow bear]

no

you can't just denote two different numbers with the same letter

[late parrot]

(a, b) is (a < x < b)
I’ll use it that way

that’s how we were taught

[flint vale]

yes

[late parrot]

Here’s the issue I’m encountering

[flint vale]

thats it

not n, it makes no sense

[late parrot]

In the question, it states that it’s a quadratic
It says it decreases between -♾ and -2

Since it’s a parabola, it’s supposed

oops

formatting

[flint vale]

you know what decreasing is right?

draw your x and y axis

[late parrot]

Anyways, the way we were taught to interpret the notation meant that the y-values would be decreasing between -♾ and -2

[flint vale]

yes

[late parrot]

And it also states that y increases between 2 and ♾

Right?

[flint vale]

between x=-♾ and x=-2

[late parrot]

isn’t that when y is decreasing?

[flint vale]

and between x=2 and x=♾

ik, lag

[late parrot]

and that’s when y increases?

[flint vale]

yes thats what it says

[late parrot]

So if y stops decreasing after -2, and begins increasing after 2

What happens to y between -2 and 2?

[flint vale]

up to God

doesnt matter

thats why its open ended

multiple answers work

[late parrot]

If I’m trying to draw it then it seems like it would matter since I’m trying to find an equation for it

I’d imagine it’d become a piecewise graph, but it’s not something we’ve covered and there’s only one equation in the end

[flint vale]

no it doesnt

no

it can be

but no one said anything

so its up to you

but you are missing one thing

the y-intercept

plot that

sketch something that decreases on the left of x=-2 and increases on the right x=2

but keep in mind its quadratic

[late parrot]

So do I just completely ignore the fact that there’s empty space between -2 and 2, because y can’t decrease after it passes x = -2

[flint vale]

doesnt have to be empty space you can make it increase, decrease whatever

[late parrot]

It says in the question that the interval of decrease can’t go past -2

At least from what I’m seeing in the notation

[native timber]

it doesn't say it doesn't decrease in that interval

[flint vale]

doesnt say cant

[native timber]

it only says that it decreases in (-\infty, -2)

[flint vale]

thats it

nothing else

dont assume

its telling you what it wants but anything outside of that doesnt matter to it

[late parrot]

but the notation translates to (-infinity < x < -2), right?
Isn’t that telling you where it stops decreasing, since they stop it at the -2?

Or is there just no reason to listen to the -2

[flint vale]

it stops at -2

[native timber]

it doesn't say that it stops decreasing at x = -2, it just says that for sure, the interval (-infinity, -2) is decreasing

[flint vale]

but that doesnt mean that anywhere not specified cant decrease

so that "empty" space can have a decreasing part

[late parrot]

but then what’s the point at stopping it at -2, for any other reason than to be vague lol

[flint vale]

and for this question, it will formally

dont ask me

ask the question

its vague

[late parrot]

is that essentially what it’s doing?

[flint vale]

to be vague yes

[late parrot]

jfc

alright

[flint vale]

anyways

you now have two "curves" and a point, curves look like a U-shape correct?

or V

[late parrot]

i can get the equation, i just wasn’t sure what happened between -2 and 2

that’s really all i needed

[native timber]

if it gave you two intervals (-inf, a) and (a, inf) then it's trivial to get the vertex

they probably just made the question this way to make it harder

[flint vale]

that didnt make it harder

[late parrot]

it’s our first day doing intervals

so to us, yes

it made it harder

[flint vale]

not even that, im looking at it as a way to open up more solutions

[late parrot]

either way i’m just relieved i’m not reading the notation totally wrong or whatever

[flint vale]

thats the point

good

[late parrot]

just wasn’t sure what happened between -2 and 2

so anything can happen outside of the stated intervals?

[flint vale]

inside, ye

[native timber]

no actually the fact that they gave you that leeway might've made the question easier since you can just say that the vertex is the y-intercept that they gave you

[flint vale]

just leave it lol

[late parrot]

except you can’t, because the answer is (x-2)^2

[flint vale]

lol no it can be a lot of other things

[late parrot]

as far as i’m aware the h value is the x value in the vertex

[flint vale]

yes

does that have to be vertex?

no

[native timber]

your vertex's x value can be anywhere from (-2, 2)

[flint vale]

right

wanna give a counterexample?

jira

[late parrot]

don’t see the point in having us do something with so much ambiguity on day one, when we’re still trying to understand the concept at a basic level, but whatever

thanks

[flint vale]

mp

[native timber]

(0, 4) is a vertex that still fits the constraints

[flint vale]

but i believe your instructor did not deliberately make it ambiguous

perhaps they are unsure of whether or not you can write an eq

with vertex given

so they gave you room to work with but 0,4 to make sketches close to home i.e. the origin

[neon drum]

????????

?????

[stuck lark]

??? what did you try ???

[neon drum]

plugging in 5 as the x for the first one

86 for g

and 2 for h

[stuck lark]

focus on part A first

you plugged x = 5 into what?

[neon drum]

sqrt(5+4)

[stuck lark]

why sqrt(5+4)?

[neon drum]

do i have to do g(5) then get that and plug it into f?

[stuck lark]

good idea

[neon drum]

I dont understand this.

[stuck lark]

same idea as what we talked about before

[neon drum]

i dont know how to plug in -1 to f

[stuck lark]

the graph of f will help you

[neon drum]

is f(x)=1/2x+1?

[short sorrel]

sure, but its not like you need the equation here

[stuck lark]

i don't care what the equation of f(x) is. just use the graph

[short sorrel]

^

what value does f(-1) have on the graph?

[neon drum]

ohh

0

[stuck lark]

good

[neon drum]

thanks

[stuck lark]

now plug THAT into g

[neon drum]

right, so g=0 because its at (0,0) right?

[stuck lark]

that's correct

[neon drum]

thank you

[stuck lark]

no problem

[novel ridge]

http://prntscr.com/p4m72m Please guys can someone help me

I just need to make sure this is correct

[proud sparrow]

@novel ridge looks right

[hollow bluff]

hello

Im not really sure i understand how to do piecewise functions

this is the problem and i do not know what the first step is

[flint vale]

well theres 3 questions

[hollow bluff]

is the the -5,-2.5

[flint vale]

the piecewise function essentially describes everything you need

[hollow bluff]

*-2,5

[flint vale]

no so the f(-5) means what is f(x) when x=-5

the piecewise function says something about when x does not equal -2 which works

[hollow bluff]

ok

So do i just plug in -5 for the x?

and is there a special reason as to why there is an equal sign with a slash?

[flint vale]

equal sign with slash means not equal to

[hollow bluff]

yeah i get that now

[flint vale]

you dont plug in -5, its asking it

so -5 isnt -2 right?

[hollow bluff]

yea

[flint vale]

so what does your piecewise say

[hollow bluff]

x not equal to -2

[flint vale]

right

what does that correspond to

theres one more piece of info that it gives

[hollow bluff]

x = -2

[flint vale]

no

when x does not equal -2

what is f(x)

it gives it you you

[hollow bluff]

f(5)

[flint vale]

no its 1

[hollow bluff]

how did you get 1

[flint vale]

the piecewise says f(x) is equal to 1 if x is not equal to -2

the piecewise also says that f(x) is equal to 2 if x = -2

i guess its just understanding the notation here

[hollow bluff]

wait

that doesnt mean case 1 and case 2?

[flint vale]

you can think of it as two different cases

[hollow bluff]

i thought it was like 1) x not equal to -2
2) x equals -2

[flint vale]

no

lol

[hollow bluff]

man im stupid xD

[flint vale]

its okay, its new to you

the curly bracket just puts it all together

[hollow bluff]

ah

[flint vale]

f(x) is equal to 1 if ...

[hollow bluff]

so would it be f(5) = 1

[flint vale]

f(x) is equal to 2 if ...

yes

what about f(-5)

[hollow bluff]

1

[flint vale]

and then f(-2)

[hollow bluff]

and the last one would be 2

[flint vale]

good

[hollow bluff]

man this isnt that hard

thanks!

[flint vale]

its not anymore

thats a good thing

mp

[hollow bluff]

I just got one with an interval

are there different rules>

That is what i put

piecewise is easy

thank you so much Lejoro

[flint vale]

That looks good to me

It’s like reading a table

[twilit shadow]

Did I do this right? My answers are on the left, and the right is my really messy work. I’m trying to find the relative minimum, relative maximum, and point of inflection of the equation S(w)=w^3-w^2+3

[spice urchin]

i have this problem

im supposed to find the equation of the parabola

with those 2 x - intercepts

i need hepl finding the line of symmetry

[viscid thistle]

Find the slope

[twilit shadow]

Nevermind, I’ve realized my mistake

[rose locust]

why is arcsin √1 = 90 and 270?

[kindred grove]

How would you work that hoe out

c)

Need a hint not an answer

Thanks in advance

[viscid thistle]

Is this all equal to zero?

[uncut mulch]

it is in the form of a quadradic
you can use a substitution to see it more clearly

[kindred grove]

I have to factor it

[viscid thistle]

Yes

You can factor

Hmm whats a hint

Ah, multiplying exponents add each other

And the middle term is one less than the last term.

[kindred grove]

I know I have to factor

[viscid thistle]

(a+2)(a+1)

[kindred grove]

Stop typing you are literally

Giving up the answer

[viscid thistle]

🤔

Thats the answer?

Then you flunk already

[kindred grove]

Close to it

[viscid thistle]

🤔

Ok

[uncut mulch]

the hints i provided above should be sufficient

[kindred grove]

I know thanks mate I think I remember now

[rose locust]

why is arcsin 0.5 = 90 and 270?

[patent beacon]

It isn't!

,calc arcsin(90 deg)

[obsidian monolith]

The following error occured while calculating:
Error: (intermediate value)(intermediate value)(intermediate value) is not a function

[patent beacon]

Good then

[hollow bluff]

So im learning a new topic and i want to make sure i got a question right

[short sorrel]

not quite

the piecewise function is defined in two parts:

one part for x <= 2

another part for x > 2

but it seems like there's a "gaping chasm" in the interval from 1 to 4

[rose locust]

@patent beacon i know that arcsin 0.5 is 30 and 150, but the answer key says 90 and 270

[hollow bluff]

Can you explain how to graph piecewise defined functions to me?

[rose locust]

or... maybe i forgot to solve for cos theta...

[hollow bluff]

So the two parts you said there is a chasm

So was i right about having two different lines?

[patent beacon]

It's two linear functions on the same graph

[short sorrel]

well, you drew a chasm

but there shouldnt be one

[hollow bluff]

oh

[short sorrel]

so we look at each "condition"

and draw the line it defines

[patent beacon]

They look like the right functions though

[short sorrel]

first off, we look at the "condition" for x <= 2

now, this applies to ALL values of x

that are less than or equal to 2

your line somehow starts at -1 and ends at 1

but it should apply to EVERY x value less than (or equal to) -2

then, we look at the next case, which has "condition" x > 2

[hollow bluff]

Wait

for the first case

[short sorrel]

again, this new line should apply to EVERY value greater than 2

[hollow bluff]

I cant add the next point because it goes off the graph

[patent beacon]

Your lines are in the right positions, they just stop for no reason

[short sorrel]

and yeah, the lines are correct

[hollow bluff]

I cant fit the next dot on the graph

[short sorrel]

but the intervals they're defined on arent

the software you're using doesn't have the ability to add arrows to lines?

[hollow bluff]

Im using aleks

[short sorrel]

or arbitrarily extend them?

[hollow bluff]

I dont think there are arrows

Im using aleks

I dont think there are arrows

[short sorrel]

so there's no way to draw a line that continues infinitely?

that seems like a... weird limitation

[hollow bluff]

So this program does give an explanation but I just dont understand

[flint vale]

You need to be able to draw arrows lol, there has to be some way

Let your instructor know

[hollow bluff]

alright

Well this time they gave me a new problem

[rose locust]

How do I get 90° and 270° from this equation ?

[hollow bluff]

one with the arrows now

[flint vale]

Divide by cos

And then you get 90

[hollow bluff]

so would there be 1 dot on -4 and -2?

[flint vale]

Circle at x=

Open circle btw

And then a flat line

Bro it’s like y=

[hollow bluff]

oh ok

[flint vale]

Right because as you go on the y value will not change

Like the y=

[hollow bluff]

Ok i got it

[flint vale]

270 works bc cos is on both sides

[rose locust]

The only way i can get 90° is if I do arcsin 1 but i don’t think that’s right

[flint vale]

So 0=0

[rose locust]

am i just stupid

[flint vale]

Same idea with 90

Cos 90 = 0

So 0=0

It’s on both sides so you can get 0=0

[hollow bluff]

Ba bam

[flint vale]

Nice

[willow bear]

@flint vale no, you can't divide by cos(θ) without accounting for its zeroes!

[flint vale]

i did

after

it was segmented

[harsh cipher]

Hello

[last linden]

hi

[harsh cipher]

says solve graphically. I type it in ti83. says error

😦

[last linden]

just read it from the graph they gave you

[harsh cipher]

I looked over my lesson it doesn't show something similar...however I can see that x= -1 thats where vertical asymptote is

[last linden]

oh why did you choose 0

[harsh cipher]

clock was ticking

[last linden]

that is simply saying y = (whatever that function is)

and you want y to be 0

[harsh cipher]

I understand

[last linden]

at which x coordinate is y = 0

and u see from graph at x=-1, ok u got it good

[harsh cipher]

but why am I not able to put this in ti83

[last linden]

it should solve

[harsh cipher]

it says error..sort of frustrating not being able to do so

[last linden]

for x

[harsh cipher]

(8x^3+8)/(-9x^3+6x^2)

[last linden]

wait does it have a solve function on there?

[harsh cipher]

= error

[last linden]

or you trying to graph it

[harsh cipher]

I'm trying to graph it

[last linden]

ah ok

do you have to put like f(x)= in front of it

or like y=

[harsh cipher]

y =

[last linden]

it should graph it fine

its not a complicated function

[harsh cipher]

no it doesn't graph I'm not kidding 😛

[last linden]

can u show calculator xD

[harsh cipher]

no I can't

but it doesn't do it. I set window ranges to standard as well

[last linden]

well graph some other function like (x-5)(x+5)/x(x-6)

[harsh cipher]

you know the best thing I can do right now is move on to next problem

[last linden]

see if it graphs other functions

[harsh cipher]

it graphs fine

maybe it's the x^3 it's having problems or I have to set window range a lot higher

[rigid sun]

uhhh

whats your error

@harsh cipher

[harsh cipher]

error is the calculator would not graph.

[rigid sun]

no

what does it tell

you

it should tell you the type of error

it won't just say error

[harsh cipher]

It says Error: Syntax

1. Quit

2. Goto

[rigid sun]

that means you typed it wrong

[harsh cipher]

Ok I need to try it again then

[rigid sun]

clear it and try again

ill whip out mine

[harsh cipher]

okay thanks!

I'm typing it it now

Okay..

[rigid sun]

i just typed it and it worked

[harsh cipher]

I typed it without the brackets

and it worked

[rigid sun]

wtf

[harsh cipher]

brackets it doesn't work I have no idea why

[rigid sun]

don't use brackets

use parenthesis

brackets=bad

[harsh cipher]

got it, that was easy

[rigid sun]

yup

[harsh cipher]

thanks

[last linden]

you used brackets

bruh no wonder didnt work xD

[hoary valley]

Guys in finding the slope < we will always start with the y right ?

[serene heath]

wdym

[hoary valley]

Slope = y-y1/x-x1 ?

[willow bear]

no, the slope of a line going through $(x,y)$ and $(x_1,y_1)$ is not $y - \frac{y_1}{x} - x_1$.

[obsidian monolith]

Ann:

[hoary valley]

We always subtract the big number from the smaller one for example if x was 5 and x_1 was 10 we must subtract 10-5 right?

In slope

[willow bear]

no

you can do either $\frac{y - y_1}{x - x_1}$ or $\frac{y_1 - y}{x_1 - x}$; these two are equal. but you have to have the points in the same order in the num and the denom

[obsidian monolith]

Ann:

[hoary valley]

@willow bear Thanks for answering, but which one should i focus on when i solve problems, you'd say?

I don't want to confuse myself

[willow bear]

doesn't matter

just write out clearly what your points are

[proud sparrow]

Just remember it's a symmetric expression @hoary valley, like the distance formula $\sqrt{x^2+y^2}$, you can swap $x$ and $y$ here by flipping across $y=x$.

[obsidian monolith]

Element118:

[crude hemlock]

Can a function have a limit of infinity? For example with the equation

$\lim_{x\to 0^{+}} \frac{1}{x}$

[obsidian monolith]

Kiako:

[lucid crow]

?

if you evaluate this limit you get infinity

but if you approach it from the left you get -infinity

[willow bear]

yes, this limit is $+\infty$

[lucid crow]

so the two sided limit does not exist

[obsidian monolith]

Ann:

[crude hemlock]

Okay

I was just wondering if a limit can output infinity

[willow bear]

yes of course it can

[crude hemlock]

Okay, thanks

[hoary valley]

Is $
1/-2= -1/2$
?

[obsidian monolith]

♥ Joshwa ♥:

[willow bear]

yes, $\frac{1}{-2} = \frac{-1}{2}$

[obsidian monolith]

Ann:

[novel ridge]

http://prntscr.com/p5bxqi

[woven cairn]

Bruh

Das ez

Pz

Lemon

Squeezy

[willow bear]

alright lemme just type this out for people who don't want to squint at the badly-proportioned screenshot crop

[novel ridge]

http://prntscr.com/p5c0g6

[willow bear]

or that

thank you

so

you selected option A

[novel ridge]

no i didn't mean to select it, i cant unselect it though

I'm thinking about 4

[willow bear]

4?

there's no 4

[novel ridge]

D

sorry

[willow bear]

He must first multiply 360 by 12 then divide it by 8.5.

[novel ridge]

wait

So

the hour goes around it 2 times a day

[willow bear]

yes, that's true

[novel ridge]

lemme reevalualate my options

I cant find anything suitable? 😔

so what does the 8.5 have to do?

oh thats the hours he spends there

@willow bear what does the question mean by "degree measure"?

the angle between the hour and minute hand?

[wind igloo]

In chill, Kyrix has said this is a timed quiz.

And in the rules, asking for help during tests is bannable.

So.

Bye!

[novel ridge]

cya

[gilded rune]

https://prnt.sc/p5ckke

[gilded rune]

<@&286206848099549185>

[slate vessel]

Hello

Have anyone ever here understood what integral calculus is or we just put the value in formula and get the answer

[sick seal]

do i jsut FOIL the (x-5-i)(x-3i) and call it a day

[uncut mulch]

not quite

[serene heath]

2 problems with that

[sick seal]

i got to add the degrees and coefficient...

sheii

[serene heath]

not degree 4

[sick seal]

no?

[uncut mulch]

what's special about complex roots in a polynomial with real coef

[sick seal]

im still unsure

[slate vessel]

how is that calculus

[sick seal]

its a pre calc class man idk

[slate vessel]

uh

[sick seal]

it feels like algebra plus

[slate vessel]

what grade are you in

[sick seal]

freshman in college dude

[slate vessel]

oh fuck lol

I am in 11th grade from India

[uncut mulch]

are you familiar with conjugate roots?

[slate vessel]

yeah somewhat

Also you asked

what's special about complex roots in a polynomial with real coef

Since P(x) has real coefficients, complex roots occur in conjugate pairs

[uncut mulch]

was directed at ramen

[slate vessel]

oh shit okie

[sick seal]

lol

and no im not

[slate vessel]

I'll try solve it

Brb 5 minutes

[sick seal]

you smart people -_-

[uncut mulch]

review that as you should have been taught that before being given this question

[sick seal]

aight

[slate vessel]

oh no I am not smart lol

[sick seal]

ok

im just dumb

@uncut mulch the conjugates are also zeros.

meaning p(x) = 7-i =p(x) 7+i

is that what you mean by special because that seems wonky

[uncut mulch]

the conjugates are also zeroes
but what do you mean with those p(x)?

[slate vessel]

@uncut mulch can you transform this into a qurandtic equation

(x-3)^2-i^2

how to open this

or do anything

with it

[uncut mulch]

how did it end up like that?

[slate vessel]

if you know how to open (x-3)^2-i^2

then we can answer the question of ramen

[uncut mulch]

the expansions are completely wrong

[slate vessel]

lol

then I can't do anything

I can't answer sorry

hm

wait what

how

@uncut mulch which step is wrong

[uncut mulch]

splitting 3i into -3+i etc
multiplication of (X - (5-i))(X - (5+i))

[proud sparrow]

you changed + to times and times to +...

[slate vessel]

lmao somehow i am coming close to answer

I put the value of iota square into that equation

so it transformed into a qurandtic equation

Answer is

4x⁴-12x³+140x²-28x+1000

@uncut mulch @proud sparrow @sick seal have a look pls

[uncut mulch]

I told you the steps you had were very wrong. Why did you continue from there?

[slate vessel]

aaaaaaaaaa

I went to eat dinner

then I came back and thought lets try and continue

ping me if anyone posts a solution

[flint vale]

Try and continue something that’s wrong?

[viscid thistle]

how do i find the zeros of 2(x+1)^2+15(x+1)+25

[flint vale]

Right now I see that you can set that expression equal to 0

And solve for x

[viscid thistle]

then move everything to the other side

[flint vale]

What’s everything?

[viscid thistle]

everything except the ^2

2(x+1)^@

2

[flint vale]

No

[uncut mulch]

do you notice what type of equation you have?

[flint vale]

Pls expand everything

[viscid thistle]

quadratic

[uncut mulch]

don't expand

[flint vale]

Then answer what he wrote

[viscid thistle]

alright

what do i do 2 in 2(x+1)^2

[flint vale]

No for the question

Not to get it right away

[uncut mulch]

what methods are there to solve a quadratic?
it may help to use a substitution to see things more clearly

[viscid thistle]

completing the squre

sqr

factoring

quadratic form

[flint vale]

Try that with substitution

[viscid thistle]

which

[flint vale]

Quadratic

[uncut mulch]

what substitution do you think would be appropriate here?

[flint vale]

And sub out the binomial

[viscid thistle]

no idea

[flint vale]

I was gonna delete it

[viscid thistle]

wydm by substitution

[flint vale]

But I guess you don’t get what I said

And what he said

[viscid thistle]

nope lol

like substitution method to find x and y

[uncut mulch]

you noticed that it was a quadratic in (x+1) but it may be harder to see roots and factors in that form

[flint vale]

So why not switch it out for now

And come back later

[uncut mulch]

so you can let u = x+1 to see things more clearly

[viscid thistle]

ahh

i got 2(x+1)^2 + 15x + 40

idk how to find roots when theres a 2(x+1)^2

[uncut mulch]

oh you didn't understand what i said

[flint vale]

Yeah you didn’t

Replace the (x+1) terms with u

[uncut mulch]

instead of expanding (15(x+1) + 25 to
15x + 15 + 25 = 15x + 40,
replace the x+1 terms with another variable like u

[flint vale]

Or any letter

[viscid thistle]

ahh

[flint vale]

Go for it

Pls come back

[uncut mulch]

and you should reach something you can clearly see the factors of

[viscid thistle]

2(u)^2 + 15(u) + 25

aight

i got (2u + 5) (u + 5)

just plug in (x + t ) for u?

[uncut mulch]

*(x+1) not (x+t)

[viscid thistle]

si

Lol

ooHhH

Alright thanks COol people

[flint vale]

Ur not done yet lol

Come back when you are done.

So you don’t mess up in the end

[viscid thistle]

Lol i used a zeros calculator online

its -6

and -7/2

i checked my answer

[flint vale]

Ok nice

Though this method is longer, you can also brute force it

ie expand the binomials and solve for x

[viscid thistle]

Oh true

thanks

[clear glade]

how the hell do find the horizontal stretch

[rigid sun]

stretch of what

@clear glade

there is none

[clear glade]

where it says horizontal stretch or compression

oh really, not sure why I keep getting it wrong then

[rigid sun]

its 0

the parent function for exponential isn't a function you can just graph

kb^x

[clear glade]

look

I set vertical shift to 1 and everything else I left the same

[rigid sun]

ok

im still not quite sure what you're trying to do

let me see the whole page

[clear glade]

ok

[rigid sun]

wtf

it doesn't say you have to find stretch

[clear glade]

let me show you

[flint vale]

No

He’s saying that from the eq given

Use a fundamental function that you know

[clear glade]

or actually just check the first pic, when I select the line there's the stretches and shifts I have to put

[flint vale]

To alter it

[rigid sun]

ohhh

lol

what's your original function

that's what we need to find out

[clear glade]

wat

[rigid sun]

so

the computer is telling you to change the graph to match the function right?

[clear glade]

f(x) = 5^x + 1

[rigid sun]

that's the function you need to graph?

[clear glade]

well I have to plot a point and change the shifts/stretches if needed

[rigid sun]

ok

you never told me your target function

what function do we need to graph?

[clear glade]

sorry

well now it's f(x) = 6^x +4

[rigid sun]

ok

so

is the graph already preset, then you have to modify it?

[clear glade]

yeah, i just have to add a point and kinda change the shifts

[rigid sun]

ok

post a picture of the default graph

don't make any changes yet

[clear glade]

ok

[rigid sun]

[rigid sun]

ok

so your graph is completely blank

we have to graph from scratch?

[clear glade]

yes

if u wanna see full page

[rigid sun]

ok

so

first

we need to fix the function to the y axis

our first component will be the y intercept

[clear glade]

alright

[rigid sun]

which is?

[clear glade]

4

[rigid sun]

gg

[clear glade]

rip

[rigid sun]

ok so make any exponential graph, but make sure it's y intercept is 4

we'll make edits on the stretch

[clear glade]

ok

[rigid sun]

wait nvm

the y intercept is 5

remember cus 6^0 +4=5

[clear glade]

oh right

[rigid sun]

lol don't forget that like i did

$6^x +4$

[obsidian monolith]

maleb1964:

[rigid sun]

ok

tell me when you've graphed a random exponential function with y intercept 5

[clear glade]

ok

so i put it on 0,5 and this was made automatically

[rigid sun]

ok

well

the first step is to change that vertical shift

we want to make the y intercept 5

have you got it?

[clear glade]

alright

I changed it to 4

Good?

[rigid sun]

yes

ok send a picture of your graph now

lets see what it looks like

[clear glade]

alright

[rigid sun]

[clear glade]

lol that pepe tho

[rigid sun]

lol

ok

[clear glade]

[rigid sun]

so you know how exponents work

6^-1= 1/6

[clear glade]

yea

[rigid sun]

yeah

we've accounted for the y intercept in

$6^x+4$

[obsidian monolith]

maleb1964:

[rigid sun]

now all we have to worry about is that 6 part

[clear glade]

I see

[rigid sun]

so, in order to improve the shape of our graph, we need to add more points to it that work

we know that 0,5 is one of the points on the target graph

what is a relatively close x value you that would be easy to find the y value for?

[clear glade]

1*

[rigid sun]

yup

[clear glade]

(1, 10)

[rigid sun]

and when you plug in 1, you get 10

yeah

[clear glade]

noice

quick maffs

[rigid sun]

now all you gotta do is change you graph so it runs through the point 1,10

[clear glade]

so add a point at 1, 10?

[rigid sun]

yeah

so we'll have two points

0,5

and 1,10

[clear glade]

hmm alright

[rigid sun]

we'll need to change something in the controls to get it there

you know which control?

[clear glade]

I think when I submitted my earlier attempts there was only one line tho

what do you mean?

[rigid sun]

wait nvm

wtf

[clear glade]

lol

[rigid sun]

you got lucky

the graph defaulted to 6

[clear glade]

lol

[rigid sun]

ur so lucky

[clear glade]

lol

so do I have to change the horizontal shift or anything

[rigid sun]

no

lucky

basically, graphing exp is a two step approach

[clear glade]

😮 i got it

[rigid sun]

set the y intercept

then, find a close x value, usually x=1

[clear glade]

Oh I see

[rigid sun]

find the corresponding y value, then use the horizontal shift to make the graph intersect that point

make sure it's horizontal shift

vertical shift will move your y intercept, which is bad

only horizontal shift changes the rest of the graph without change the y intercept

[clear glade]

I understand, will keep it in mind

[rigid sun]

ok

did you get it right?

[clear glade]

yea

[rigid sun]

gg no re

[clear glade]

the rest of the part was to find the domain and range and asymptotes which I know how to do

Thanks so much man

[rigid sun]

mhm

[clear glade]

I was stuck on that for like an hour

I appreciate it man

[compact raft]

Hi i need help with this specific question we're doing imaginary numbers in my class
(2+\sqrt(2)i)^(-1)

[uncut mulch]

$(2 + i \sqrt{2})^{-1}$

[compact raft]

yeah

thats the one

[obsidian monolith]

ramonov:

[uncut mulch]

what have you tried?

[compact raft]

so since its raised to the negative power i just made it a denominator of 1

the multiplied it by the congregate

[uncut mulch]

denominator of 1?

[compact raft]

1/$(2 + i \sqrt{2})^{-1}$

[obsidian monolith]

I AM YEETZUS:

[uncut mulch]

pretty bad way to word it,
also ditch the negative 1

(after turning it into a fraction)

[compact raft]

$1/(2 + i \sqrt{2}) * (2 - i \sqrt{2})$

[obsidian monolith]

I AM YEETZUS:

[uncut mulch]

you'll need to multiply both the numerator and the denominator by the conjugate

[compact raft]

ok i got $2 - \sqrt{2}i / 4 -2i$

[obsidian monolith]

I AM YEETZUS:

[uncut mulch]

somethings wrong,
multiply a complex number by its conjugate results in a whole number

(which is why you multiply by the conjugate to rationalise)

[compact raft]

hmm lemme try it again and get back to u

[uncut mulch]

use \frac{}{}
for fractions

otherwise it isn't any better than writing normally without parentheses

[compact raft]

\frac{2- \sqrt{2}i}{4-2i}

ops

[uncut mulch]

that was the denominator you had before which was wrong

[compact raft]

yeah i multiplied it again and it gave me the same answer

[uncut mulch]

do you know the identity for
(a + bi)(a - bi)?

[compact raft]

a^2 + bi^2

[uncut mulch]

nope

[heady jewel]

no

multiply properly

[uncut mulch]

expand it out step by step

[heady jewel]

step by step

i hope you didnt memorize the result

[compact raft]

a^2 - abi + abi -bi^2 ?

[uncut mulch]

now simplify

wait

not quite, missing the ^2 on the b

[compact raft]

oh

i thought the b was acting like a whole number

a^2 - b^2 i^2

[uncut mulch]

and simplify that

[compact raft]

i^2 = -1

a^2 + b^2

[uncut mulch]

yep

[compact raft]

so its going to be $2 - \sqrt{2}i/{6}$

[obsidian monolith]

I AM YEETZUS:

[uncut mulch]

$\frac{2 - \sqrt{2}i}{6}$

[obsidian monolith]

ramonov:

[compact raft]

Damn ! i thinks thats right

that is correct right?

[uncut mulch]

yeh

[compact raft]

thanks man this really helped a bunch was lumping in the {i} with the sqrt 2

thats what was messing me up

[uncut mulch]

use a hook to indicate where the sqr ends or write the i before it

to make it less confusing

[clear glade]

need help on this one

[viscid thistle]

That's the wrong graph.

If you think about it, the smaller x goes, the larger y becomes.

[flint vale]

review power function before you extrapolate

and then plot trivial points

[clear glade]

ok

how about for this

or is there something I should change for the stretches?

[flint vale]

the trivial point in incorrect

figure that out with stretch

[clear glade]

horizontal stretch?

and also should i change the base to 2 (from e), forgot about that

[rigid sun]

@clear glade you still need help?

[hexed bolt]

Hello

What is part B asking

[willow bear]

factor f(x) completely

[clear glade]

Yea, sorry ; ( @rigid sun

[hexed bolt]

it gives me (x+1)(x^2 + 6x + 73)

Now what?

[willow bear]

uh

[short sorrel]

it does?

[hexed bolt]

Oh wait

I did that wrong

[short sorrel]

you have the roots right there

[hexed bolt]

Anyway

What do i do once I get that

[rigid sun]

@clear glade

is it the same problem up above

[short sorrel]

"product of linear factors" means "of the form (x-a)(x-b)(x-c)..."

[clear glade]

yep @rigid sun

[short sorrel]

each factor (x-a) is linear (no exponent on the x)

so, product of linear factors.

[clear glade]

yea same one

[short sorrel]

it just wants you to factor fully

luckily for you, you already know the roots

[rigid sun]

uhhh

[short sorrel]

so you shouldnt have to do all the work again

[hexed bolt]

Yeah

[rigid sun]

the quality is bad for some reason

is that a negative 2?

[clear glade]

yeah lol

one sec

[rigid sun]

ok

well you know the proeperties of exponents

[hexed bolt]

How do I get it without the exponents @short sorrel

[rigid sun]

(1/4)^x=?

[hexed bolt]

I got (x^2+16)(x^2+25)

[short sorrel]

you. already. know. the. roots.

[rigid sun]

4^(-x)

[short sorrel]

use. those. roots. to. find. your. factors.

[hexed bolt]

ok

[clear glade]

Oh

[rigid sun]

@clear glade remember what i told you earlier?

fix the y intercept first

[clear glade]

right

[rigid sun]

the y intercept is?

[clear glade]

2

[rigid sun]

its 3

[clear glade]

really

ok

[rigid sun]

remember, anything to the power of 0, with the exception of 0 (in this case), =1

so your y intercept will always be 1 + c where your equation is in the form b^x +c

[clear glade]

but won't it be (1/4)^0 = 1 * 2 = 2

[rigid sun]

is that a multiplication sign

or addition

[clear glade]

multiplication

[rigid sun]

ooof

well then you're right

[clear glade]

; )

[rigid sun]

my quality cucked me

[clear glade]

rip

[rigid sun]

lol

[clear glade]

my bad if it's not good quality, I tried resending

[rigid sun]

so the equation is 2(1/4)^x?

[clear glade]

yep

[rigid sun]

k

ok

so for this, you'll need to know how vertical shift influences graphs

[hexed bolt]

I can't seem to get this one either

[rigid sun]

like you pointed out to me, the two multiplies the 1 from b^0 to equal 2

so whenever you have a multiplying factor, you'll need to adjust

[clear glade]

Alright

[rigid sun]

ok

so first, we're gonna do the stretch

[clear glade]

ok

[rigid sun]

stretch is determined by the x coefficient before the main operation ie exponent

[clear glade]

so 2

[rigid sun]

not in this one

[clear glade]

oh

[rigid sun]

you remember from earlier, we rearranged the function in 4^(-x)

[clear glade]

Oh yea

[rigid sun]

now here's the simple trick

-x, indicates a stretch of -1

which means that the graph will be backwards across the y axis

[clear glade]

oh so it will reflect across y axis

[rigid sun]

yes!

so all you have to do is graph 4^x, except backwards

[clear glade]

since it's -x, i get it

alright

[rigid sun]

once you do that, make sure you do a vertical stretch by a factor of 2

since it was being multiplied by 2

[clear glade]

right, makes sense