precalculus

[patent beacon]

@candid onyx
I don't know what these drunkards are saying, you're way off. If f(x) = x² + 1, what's f(x + h)?

The answer to that is (x + h)² + 1

Your difference quotient will involve x

[acoustic hazel]

Should be the numerator after simplification

[proud wraith]

Where do I start?

[uncut mulch]

which part are you struggling with?

[proud wraith]

I just don't really know where to begin

So I guess "a"?

[uncut mulch]

ok, so what is the value of the curve at x= -1?

[proud wraith]

1

[uncut mulch]

ok good

now for part b,
when is f(x) = -4?

[proud wraith]

(-1,2)?

[uncut mulch]

you didn't need help after all

[proud wraith]

Thanks. It helps when you have someone walking you through it.

[pine temple]

stuggling with understanding the domains

[mental rivet]

what do you not understand @pine temple ?

[pine galleon]

what is a domain maybe ?

[acoustic hazel]

A domain is the set of numbers you can input into a function and get a result

Like, for sqrt(x), the domain is x>=0, since and negative x won’t give a real result.

[pine temple]

i know what the domian is, i just dont understand how he got the conditions of it

bc when i solved for x>=0

[acoustic hazel]

So, is this your work?

[pine temple]

no its an answer key

want me to show work

[acoustic hazel]

What part do you get lost at

[mental rivet]

you shouldn't solve for x>=0 , you should solve for >=0 that is under the root

[pine temple]

i get lost at the number line

graphically it makes sense but our teacher doesnt give enough time to graph it

[mental rivet]

first make sure when the function exists, like in the previous example as @acoustic hazel said, everything under the root must be positive, and denominator shall not be 0.

now, find the roots of the eq, place the roots on the number line, now, the extreme right side of the roots is taken +ve and towards left change sign alternatively

[acoustic hazel]

Once you’ve factoried, you know the roots of the equation. Since the roots of an equation are where it changes between positive and negative, you mark the roots on the number line. Since you change between + and - at each root, you know it has to alternate between + and -, so you pick a random value between 2 roots, put it into the equation, and label the interval you took the value of x from as + or - depending on the output

Do this for a random number in each interval between roots

And you get the + and - for each interval

[pine temple]

ohh you have to pic numbers btwn said values

[acoustic hazel]

Yep

[pine temple]

okay that clearss it up because i was going off my factored equation

tyty

[acoustic hazel]

Preferably easy ones, like 1

Np

[pine temple]

👍

[mental rivet]

for the 4th question, the function exists only if the expression (x^3 - 2x^2 - 3x) is >=0 , now finding the roots for that gives --- {0,+3,-1} arrange them in ascending order on a number line.
- + - +
<--------(-1)------(0)-------(+3)-------->
initially our conditions were >=0 , so we need +,
which is why the answer is [-1,0] U [3, inf]

[torpid flax]

It seems to me like these width, length, and height values should be taken as "exact" as it doesn't say anything about those values being measured. Why does this not become the case?

Or should I assume numbers are inexact unless specified explicitly, or are countable in nature?

[slate wigeon]

What’s a.?

For the notes sorry I’m garbage at math

[mental rivet]

which a ?

[slate wigeon]

@mental rivet

[mental rivet]

set numerator = 0 then solve for x

[slate wigeon]

Ok

[viscid thistle]

,rotate 270

[viscid thistle]

[tepid flare]

@torpid flax if sig figs are mentioned, it implies inaccuracy is a possibility
most of the time

[sick seal]

can someone help me understand where 5/x+1 came from?

whats shown is all thats given.

[short sorrel]

g(x) isnt defined? at all?

[sick seal]

yeah, checking previous slides it dose not seem to define g(x)

[short sorrel]

huh

i guess they just forgot to mention that $g(x) = \frac{5}{x+1}$?

[obsidian monolith]

Namington:

[sick seal]

i would have to assume so.

ill email prof. just wanted to check in here first.

[short sorrel]

yeah, good idea.

[sick seal]

thanks man

[candid onyx]

the domain of f? do is set x^2+x to zero?

[uncut mulch]

what's the definition of the domain?

[candid onyx]

all possible x values

[uncut mulch]

and what would be the possible x values of this function?

[candid onyx]

all real numbers?

[viscid thistle]

(x+10)/(x-9) the x and y intercept for this is (-10,0) and (0,-10/9) but I am still getting marked up for this on my quiz zzzzz

I tried both ways and it doesn't even specify to put it as a point what am I doing wrong

[willow bear]

it doesn't look like you're doing anything wrong

[uncut mulch]

can you show the full question

[willow bear]

does it still mark you wrong if you try to put these down as points

ie (-10, 0) and (0, -10/9)

[viscid thistle]

yeah

[willow bear]

weird

[viscid thistle]

Yeah think it might be a bug on mymathlab, oh well.

[bleak lance]

Find an equation for the line with the given properties. Express your answer using either general form or the slope-intercept form of the equation of a line. Parallel to the line y=9x; containing the point (-5,7)

[willow bear]

what have you tried and where are you stuck?

[bleak lance]

The issue is that the problem is different and I always have issues solving problems that are of different kinds lol ;-;

[willow bear]

different than what?

[bleak lance]

Find the equation of the line parallel to 4x+y=-8 and through point (2,-3)

[willow bear]

how is this problem any different?

[bleak lance]

Hmmmmm

idk

[willow bear]

have you actually TRIED doing the first problem you posted here, instead of setting up a "this is different" mental block for yourself

[bleak lance]

Issue is that how can there be two y's?

[willow bear]

wdym by "two y's"?

[bleak lance]

idek

something like 2x+5x=7

where y=5x

[short sorrel]

i'm... confused

so x and y are constants?

[bleak lance]

I guess

[short sorrel]

i dont see how that relates to your original problem though

[bleak lance]

Idk either

[short sorrel]

...

ok so

we have a line "Parallel to the line y=9x; containing the point (-5,7)"

do you know what it means for lines to be "parallel"?

[bleak lance]

2 lines that dont intersect

[short sorrel]

yes, but what does that tell us about the lines?

[bleak lance]

They dont intersect. They simply dont touch.

[short sorrel]

that's true, but again, it's not what i'm looking for

lets say you have the equations of two lines

how do you tell whether they're parallel

without drawing them out?

[bleak lance]

The slopes have to be equal still

[short sorrel]

right, equal slopes

so lines are parallel if they have the same slope

[bleak lance]

And they need different y intercepts

[short sorrel]

therefore, if our line is "Parallel to the line y=9x; containing the point (-5,7)"

it has the same slope as y = 9x

so what's its slope?

[bleak lance]

Thats the tough part. The conufsing part to me is that y=9x.

[short sorrel]

y = mx + b

y = 9x + 0

so whats the slope?

[bleak lance]

9

[short sorrel]

right.

so the slope of our line is 9

and it has the point (-5, 7)

can you use that information to find its equation?

[bleak lance]

Yup

9x--5=9x(x-7)

[short sorrel]

no, this is a new line.

y no longer equals 9x

y = 9x was the equation of the other line

not this one

so you cant substitute that in

just leave it as y - (-5)

[bleak lance]

Why is that?

[short sorrel]

the equation y = 9x describes another line

we're not talking about that line

we're talking about a new line

a new line that happens to share a slope, but that's it

we cant bring equalities from the other line to this line

because theres no relation

because they're not the same line

Consider an example:

Tom makes $5 an hour, Jim makes $7 more than Tom an hour. How much does Jim make?

Well, the equation of Tom's revenue is r = 5h, while the equation of Jim's revenue is r = (5+7)h.

we can't just take Tom's revenue

and put it into Jim's revenue

because they're... different revenues

they're different lines.

and, in fact, if we tried that, we'd get:

r = (5+7)h
5h = (5+7)h
5h = 12h
0 = 7h

which means that 7 hours... equals 0? what? we're not even talking about revenue (r) anymore, we got rid of those... this is meaningless

---

if you REALLY prefer, you can think of it as different x and y values

like $y_A = 9x_A$, where the small $A$ means that it relates to the "first" line

[obsidian monolith]

Namington:

[short sorrel]

and then $y_B + 5 = 9x_B(x_B - 7)$

[obsidian monolith]

Namington:

[short sorrel]

but as you can see, doing this becomes cumbersome and a bit distracting

and it's generally pretty clear from context which line we're talking about

because there's never a reason to mix the lines, so there's never confusion.

so we don't bother to distinguish these variables.

[bleak lance]

I just hope that I can somehow remember all of this

Tbh my college scheduling is complete trash

That makes matters even worse

I sometimes spend 4 days overthinking about something

Ik that this is a waste of time to vent here but Ill move to #chill just to keep this on topic.

[viscid thistle]

What is the limit of gamma(k+n+1)/(gamma(k+2)*gamma(n)) as n goes to infinity?

where gamma(k) is bernoulli's gamma function thing

[frozen needle]

Bernoulli

the thing that does Γ(1+n)=n! for all n in N ?

@viscid thistle

[willow bear]

[upper flint]

determine if the sequence is arithmetic or geometric or neither. If arithmetic find the common difference. if geometric find the common ratio.

[next willow]

@upper flint what would the common ratio be? What the common difference?

[upper flint]

@next willow The ratio is the number multiplied by to get to the next term and the difference is just a number added or subtracted to get to the next term

[next willow]

Alright, now let's check which, if any, is constant along the sequences

[upper flint]

none

its +2,+3,+4

[next willow]

For the first right? Ok that excludes it being an arithmetic sequence

What about the ratios?

[upper flint]

nope

so its neither? @next willow

[next willow]

Yes

[upper flint]

find the sum

[next willow]

Ok, do you know the formula for the sum of finite terms in a geometric sequence?

$\sum_{k=0}^n r^k$

[obsidian monolith]

Mat:

[next willow]

What does this equate to?

[upper flint]

uhhh

im lost

[next willow]

About what?

[upper flint]

does this have smtg to do with a1 * r^n-1

[next willow]

It does

[upper flint]

o

[next willow]

Maybe you would have preferred it in the more general form

[upper flint]

do i plug smtg in or

[next willow]

$\sum_{k=0}^n a r^k$

[obsidian monolith]

Mat:

[upper flint]

ok

[next willow]

But we can take the factor a out of the summation as it is doesn't depend on k, right?

$a\sum_{k=0}^n r^k$

[obsidian monolith]

Mat:

[upper flint]

right

[next willow]

So we're back at the previous question lol

[upper flint]

o

[next willow]

Anyway, the value of the sum is

$\sum_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}$

[obsidian monolith]

Mat:

[upper flint]

so its 1-3^n-1 / 1-3

[next willow]

Mmh, not exactly, you're sum isn't precisely in this form

[upper flint]

o

[next willow]

To start, we factor the 4 out, don't you agree?

[upper flint]

yes

[next willow]

$4\sum_{k=1}^{n} 3^{k-1}$

[obsidian monolith]

Mat:

[next willow]

Then we'd like to see k starting from 0 rather than 1

[upper flint]

ok

[next willow]

But if k starts from 1, k-1 starts from 0

And if k rises up to n, k-1 up to n-1

$4\sum_{k=0}^{n-1} 3^{k}$

[obsidian monolith]

Mat:

[upper flint]

ok

[next willow]

Then we apply the formula, but beware the n-1

[upper flint]

im confused on how u went from k=1 and n to k=0 and n-1

[next willow]

I see, that is just a reformulation of the notation for the sum

If you think about it, we're always summing the same terms

It's just an alternative way of writing it

$\sum_{k=1}^{3}( k-1) = 0+1+2 \
\sum_{k=0}^{2} k = 0+1+2$

[obsidian monolith]

Mat:

[next willow]

An example being this

[upper flint]

ok

so

now do i do a1/1-r

[next willow]

Apply this formula

To this

[proud sparrow]

^ how you can derive it is multiply the entire thing by r, and subtract to see which things cancel out

[next willow]

^^

[upper flint]

is is -1/2

[next willow]

The result should depend on n

In the formula you'd have r=3 and n-1 on top instead of n

[upper flint]

1^n+1 ?

[next willow]

Ok, wait a sec

$\sum_{k=0}^m r^k = \frac{1-r^{m+1}}{1-r}$

[obsidian monolith]

Mat:

[next willow]

This holds for any r and for any natural number m

Now you have

$4\sum_{k=0}^{n-1} 3^{k}$

[obsidian monolith]

Mat:

[next willow]

So substitute in it r=3 and m=n-1

And multiply by 4

[crude hemlock]

$\lim_{x\to 0} g(f(x))$

[obsidian monolith]

Kiako:

[crude hemlock]

How would one solve something like this

[mental rivet]

it depends on what f(x) and g(x) are generally

[crude hemlock]

Let’s just say like f(x) = (-2x)/(x) and g(x) = x + 3

[mental rivet]

what's g(f(x)) now?

[crude hemlock]

Wouldn’t it be 1?

[mental rivet]

yes it is

[crude hemlock]

Ah ok

And if f(x) didn’t have a limit at that point, would it have no limit for the whole thing?

[mental rivet]

yes

[crude hemlock]

Okay thanks

[ashen bay]

Hello, I am not sure how to solve | x -1 | + | x - 2 | >=4

[pale kettle]

What have you tried?

[ashen bay]

I tried the triangle

inquality theorem

with | x + y | <= | x| + | y |

then it becomes | x - 1 | + | x - 2 | >= | 2x - 3 |

but not sure if I can do that

[pale kettle]

Well you know that if |2x - 3| >= 4

[ashen bay]

yeah I can solve from there

[pale kettle]

Then |x - 1| + |x - 2| >= 4 right?

[ashen bay]

| x - 1 | + | x - 2 | >= | 2x - 3 | but can I do this

looks sketchy

[pale kettle]

Right

But would it be true that if |x - 1| + |x - 2| >= 4

Then |2x - 3| >= 4?

[ashen bay]

I don't think so

[pale kettle]

Exactly

So solving |2x - 3| >= 4 would give you some solutions

But maybe not all of them

[ashen bay]

So how should I approach this question

[pale kettle]

The best way is to solve the equation is three different regions

So look at when x >= 2

when 1 <= x < = 2

and when x <= 1

In each of these regions, you'll be able to get rid of the absolute value signs and then just solve the normal inequality

[ashen bay]

I can get rid of both abs value signs?

[pale kettle]

In each of those regions

You can

Because for |x| for example, we have that

|x| = x if x >= 0

[leaden ocean]

well if you know x >=2 then you know that |x| = x

[pale kettle]

and |x| = -x if x < 0

[ashen bay]

yes

[leaden ocean]

since x-1 and x-2 won't be <0

[ashen bay]

so to clarify

is x>=2

equation would be x - 1 + x -2 >=4

if x<=1 equation would be 1 - x + 2 -x >=4

if 1 <= x < = 2 equation would be 1 - x + x -2 >=4

[acoustic talon]

can someone explain to me basic limits

and when there is no limit for a function

[short sorrel]

the limit is what a function approaches as you get arbitrarily close to a specific value

a limit exists if the left-hand and right-hand limit both exist, and are both equal

[weak zealot]

Can anyone explain how you would go about finding f(x+1)? I get how to do the other one, but not that one

[viscid thistle]

Aight.

Wait are you trying to find the gray box?

Or are you talking about in general.

[weak zealot]

No, what i have is the key.

[viscid thistle]

Oh lol.

[weak zealot]

I just don't know how you would get those values

in the third column

[viscid thistle]

Idk how to do the last row but I can show you the first 3 rows.

[weak zealot]

The gray box is there because there's no value

i think

[viscid thistle]

Oh.

K nvm.

So if x = -2.

[weak zealot]

How do you do it??

[viscid thistle]

f(x+1) would equal to f(-2+1).

Which is f(-1)

And in the next row.

f(-1) = -4.

[weak zealot]

Ok so what if when you do that it equates to f(x) and that value isn't in the x column.

[viscid thistle]

That's why there's a gray box there.

No one knows what f(2) is.

[weak zealot]

Ok, no need to be a d-head.

Thanks.

[viscid thistle]

Lol np.

[tidal rain]

how do i solve this?

[uncut mulch]

what have you tried?

[viscid thistle]

@tidal rain Start by squaring both sides.

I suggest moving the 2 first though.

Squaring also might introduce new answers, so just check your answer at the end.

Then you got it.

[languid dust]

how to do 1 and 2

[torn swift]

You have to select all intervals

You’ve only checked one and none in the other

Your zeros basically tell you what intervals to choose

I’m assuming you just did part b of each one using your calculator first

So that means you aren’t understanding the Intermediate Value Theorem?

[brisk forge]

anyone here good at physics?

that can help me a bit

its baby physics

[torn swift]

[brisk forge]

❤

im on 2b

well 2a

[torn swift]

Actually, physics server might be the better fit for this one lol

b and c, I can do that

a, I’m not sure how the rules of units within trig functions

I believe that the input must be unitless

[brisk forge]

hmm

can u help me with b and c then

and do you have a link to a physics servcer

server

[torn swift]

#old-network

Assuming that he input it the trig function must be unitless, you can see what units B and A have

t has units of seconds

So then units of B * seconds = 1

Which is to say, the units must cancel out

Then you have to get meters somehow to that means A must have units of ?

Though check me on that reasoning with the unitless input for trig functions

I’m thinking perhaps rad/s for B

As for b and c

Velocity is the first derivative, and acceleration is the second derivative.

Those two facts come from how the units work out

[brisk forge]

okkkkk

ok that clears some stuff up

ill be back lol im gonna try it

[brisk forge]

elo

anyone up?

does this make sense to anyone

[stuck lark]

yes

i assume b is some constant

if v represents an object's velocity and t represents time, then for any t, v is equal to t multiplied by some constants

[willow bear]

@brisk forge context

[brisk forge]

oh sorry

[willow bear]

yeah so d/dt (a + bt^2) = 2bt

[brisk forge]

can u explain to me the 2bt part

does the exponent move in front of b?

if so why

[willow bear]

...

have you like

ever done any calculus

[brisk forge]

nope

barely

[stuck lark]

then why are you doing this?

is this... ap physics hw?

[brisk forge]

its physics hw. and i have my calc class tomorrow

lol im learning both at same time

[stuck lark]

jeez, usually one would learn some calc BEFORE taking calc-based physics, but alright

[brisk forge]

yeah my prof said if i dont know why its fine

but i still wanna know

what happened that made bt^2 into 2bt?

[willow bear]

the derivative of t^2 is 2t

power rule

[brisk forge]

ohhhhhh

is that rule learned in precal?

[willow bear]

idk

[stuck lark]

it's usually taught after you learn the definition of the derivative

there are a bunch of rules out there for taking derivatives of functions so that you don't have to keep applying the definition of the derivative every single time, like the power rule

[brisk forge]

fuck thatd be handy to know lmao

ty guys

[brisk forge]

eloo

im back lol

on 2c. acceleration would be a= cos(Bt)- B_0/ t right?

final velocity minus initial divided by time?

[willow bear]

no

acceleration is the derivative of velocity

[compact rivet]

Hey everyone. I have a question y = -3x^2 + 6x + 3. I have to take out a common factor, complete the square, find the turning point and find the intercepts. How do I do that?

[charred hull]

look at every coefficient present

so for the form ax^2 + bx + c, what is a common factor among a, b, and c

[compact rivet]

3

[charred hull]

yep

wait

oh wow same default pfp

don't just give answers please

[compact rivet]

Sorry?

Oh wait, I changed my name lol

[charred hull]

oh lol

[compact rivet]

Haha

[charred hull]

okay, you can factor that out into the form

0 = 3(-x^2 + 2x + 1)

[compact rivet]

Yep, thats as far as i got

Then i got stuck

[charred hull]

consider instead

0 = -3(x^2 - 2x -1)

that might help a little bit

[compact rivet]

Would that be easier?

[uncut mulch]

oh, don't replace y with 0

[charred hull]

it's the same problem, i'm just trying to make it easier to visualize

[compact rivet]

Ah ok

[charred hull]

oh wait that says y lmao, gp ramonov

[compact rivet]

Ok i wrote that down

[charred hull]

now since you want to complete the square

think of the form x^2 -2x + 1

[compact rivet]

Ok

We divide teh 2 by 2 then square it right?

[charred hull]

nope

[compact rivet]

Oh

[charred hull]

so x^2 - 2x + 1 = (x-1)^2

how can you get the form x^2 - 2x -1 to x^2 - 2x + 1

[compact rivet]

Hmmm, I'm not sure.

[charred hull]

do you know what completing the square is?

[compact rivet]

I've done it briefly. When I have dont it, you divide "b" by 2 then square it.

done*

[charred hull]

that's

not what it is

[compact rivet]

Oh rip me

[charred hull]

you can rewrite a number, such as 3, into the form 4-1

then you can write that as 2^2 - 1

[compact rivet]

That makes sense

[charred hull]

now, how can you rewrite x^2 - 2x - 1

[compact rivet]

Do I change the 2 or the 1?

[charred hull]

well

you want to change it into a form that's the square of something

which i've demonstrated would be x^2 - 2x + 1

how can you get that

[compact rivet]

Could you write -4+3?

Idk im kinda dumb tbh

[charred hull]

no, don't blame yourself, i think i didn't explain it well in my example

[uncut mulch]

ignore the constant for now
how would you turn
x^2 - 2x
into a square

[compact rivet]

you could make the 2 a 4?

[charred hull]

so if you want to get from 3 to something that can be square rooted, you would write in the from 3 + 1 - 1

which goes to 4 - 1

[compact rivet]

Yeah

[charred hull]

actually

look at what ramonov said

[compact rivet]

Oh, I was gonna say 4-5

[charred hull]

just think of x^2 - 2x

[compact rivet]

So we need to make the -2x something like a 4?

[uncut mulch]

no, you need to add a constant to make the whole thing a perfect square

you sort of touched on the process earlier

you would get the terms x^2 - 2x from the expansion of
(x - 2/2)^2 or ( x - 1)^2

[charred hull]

think of this

you have the term x

you can write that as x + 1 - 1

try that with x^2 - 2x

[compact rivet]

Just as a quick question so i understand something first, is what i did here completely wrong?

[uncut mulch]

thats comepletely fine

[compact rivet]

OK good

What about x^2 - 2x +4 -4?

[uncut mulch]

whats the b value in x^2 - 2x?

[compact rivet]

would it be -2?

[uncut mulch]

and half of that?

[compact rivet]

1

[uncut mulch]

and the square of that?

[compact rivet]

1

Oh ok, i did that in my work earlier

Ill send a screenshot of what i actually did

[uncut mulch]

all that working out is messed up

cross it out and redo it

[compact rivet]

Ok

[uncut mulch]

missing x values and incorrect signs

[compact rivet]

Oh

I can see one of the things i did wrong

It should be a negative 1

[uncut mulch]

from the 3rd line, your x term just disappeared

[compact rivet]

Yeah

The -2x right?

[uncut mulch]

yeh

[compact rivet]

Mmhmm

[uncut mulch]

so yeh, just redo the whole thing

[compact rivet]

Ok

Still don't klnow if im on the right track but im here now

[uncut mulch]

your last line is wrong

[charred hull]

you're doing fine kinda

[uncut mulch]

everything before that is fine

[compact rivet]

Yeah, I didn't know what to do for that line

[charred hull]

(x^2 -2x + 1) is not (x - sqrt(2))^2

[uncut mulch]

well what squared gives you
x^2 -2x + 1

[compact rivet]

Maybe (x -(idk) +1_^2

[uncut mulch]

huh?

[compact rivet]

(x-(idk)+1)^2

[uncut mulch]

whats with the idk

[compact rivet]

I can't figure out what to put for -2

[charred hull]

think of it like this

wait

just remove the idk

you're so close lol

[compact rivet]

(x+1)^2?

[uncut mulch]

nope

[compact rivet]

(x-1)^2

[charred hull]

yep

[uncut mulch]

thats better

[charred hull]

now, first

i want to make sure you know how that works

try to expand (x-1)^2

[compact rivet]

(x-1)(x+2)?

[uncut mulch]

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

[compact rivet]

Oh, close

I accidentally put that 2 in

[charred hull]

well, if it's a square

then that means it's itself multiplied by itself

[compact rivet]

Yeah

[charred hull]

x^2 = x * x

[compact rivet]

Yep

[charred hull]

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

[compact rivet]

Ah ok

[charred hull]

@uncut mulch hm, am i being a bother? do you mind me interrupting or is it fine

[uncut mulch]

nah its fine

[charred hull]

alright so

you've managed to simplify (x^2 - 2x + 1) to (x-1)^2

try writing this change

[compact rivet]

Yes, but where does the -2x go?

If tyhat isnt a dumb question

[charred hull]

(x-1)(x-1)

try using foil on that

[compact rivet]

Oh wait im dumb, i forgot you can add like terms

[charred hull]

use foil on (x-1)(x-1)

you'll see where the -2x comes from and where it goes

[compact rivet]

Yeah, i forgot you can add like terms

-1x-1x = -2x

[uncut mulch]

that has nothing to do with it

sort of

[charred hull]

i think i understand what red means

well, so write it all out now

[compact rivet]

x^2-1x-1x+1 goes to x^2 -2x +1?

[charred hull]

yes

so the reverse is true as well

which means you can get (x-1)^2 from that

[compact rivet]

Yes, I understand that now 🙂

[charred hull]

ok, can you rewrite the last line please?

[compact rivet]

As in y= -3(x-1)^2?

-2

[uncut mulch]

the -2 is still within the parentheses

[compact rivet]

Oh yeah it is

So are we writing it like: -3((x-1)^2 + 1 -2)

Wait one sec

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

the one is in there already6

[uncut mulch]

and then redistribute the -3

[compact rivet]

Do we divide by -3?

[uncut mulch]

multiply it back in

[compact rivet]

So x +3 so it would go back into the brackets?

multiply not x

[uncut mulch]

-3

[compact rivet]

So would the -2 become a -5?

or a -6?

[uncut mulch]

(-3)*(-2) = ?

[compact rivet]

-6 right?

+6

[uncut mulch]

which one

[compact rivet]

How do you do a faceplam

So would the equation be ((x-1)^2 +6)?

[uncut mulch]

missing the -3 in front of your square

[compact rivet]

Oh so it stays there and goes inside?

[uncut mulch]

well you only distrbuted the -3 to the -2

[compact rivet]

So it would be -3(x-1)^2 +6?

[uncut mulch]

yeh

y=

[compact rivet]

So that gives turning point or not yet?

[uncut mulch]

that will be the vertex form for your function

[compact rivet]

So the vertex would be (-1, 6)?

[uncut mulch]

not quite

[compact rivet]

Do we need to remove the -3 in some way?

Omg im actually so dunb

(1, 6) right?

[uncut mulch]

yeh

[compact rivet]

Excuse my low mental capacity

[uncut mulch]

are you able to find the intercepts?

[compact rivet]

Almost. I know how to get the intercepts but not when there is a -3 at the front

do i just divide by -3?

and get -2?

[uncut mulch]

just follow the standard rules.

for the y-int what are you substituting?

[compact rivet]

make x = 0

[uncut mulch]

and what happens to you equation

[compact rivet]

(0-1)^2?

[uncut mulch]

what happened to the other terms?

[compact rivet]

I didn't change them. Was I meant to?

[uncut mulch]

well you didn't include them

[compact rivet]

I got my y int as 3 but idk if that was right

I did, -3*1+6

[uncut mulch]

yeh. now for the x int

[compact rivet]

Make the whole equation = 0?

So is the x int just 1?

[uncut mulch]

you set y=0 so
-3(x-1)^2 +6 = 0

and then continue to solve for x

[compact rivet]

Is this by removing all the other terms?

moving them to the other side?

[uncut mulch]

just isolate x. you should've done this plenty of times before this

[compact rivet]

Yeah, I have. But not at this level.

So far I'm like this:

Pretty sure its wrong

[charred hull]

look at the -2

in the -3

[compact rivet]

Yeah?

[charred hull]

lets say you have -3(4 - 2)

try writing that as -3(4) +x

and find x

[compact rivet]

x is 6 right?

[charred hull]

oh wait. hm

lemme check something

[uncut mulch]

error in the last line

[charred hull]

did my brain just mega crippl

[uncut mulch]

yes it did

[charred hull]

oops

ignore that lol

[compact rivet]

Mine has been crippled the whole time lol

So should i not square root it? or leave it as squareroot 2?

[uncut mulch]

sqrt( n^2 ) isn't just n

preferably leave it in exact form

[compact rivet]

Ok

[uncut mulch]

the main issue was you're missing a solution

by leaving something out

[compact rivet]

Am i trying to make the brackets = 0?

[uncut mulch]

first lets address this part
sqrt( n^2 ) isn't just n, it would be |n|

[compact rivet]

I'm sorry I have no idea what |n| means

[willow bear]

absolute value of n

[uncut mulch]

absolute value

[compact rivet]

Ah ok

I think I found the answer but i have no idea how i got it. I think the x ints are -1 and 3

[uncut mulch]

they shouldn't be integers

[compact rivet]

Well i guess i didnt then

[uncut mulch]

remember that it is a sqrt(2)

[compact rivet]

Oh yeah

[uncut mulch]

in this case
| x - 1 | = sqrt(2)

[compact rivet]

yeah

how can i get the x out of the brackets?

[uncut mulch]

depending on the sign of (x-1), either
x - 1 = sqrt(2) or
-(x-1) = sqrt(2)

[compact rivet]

Ok, so x = 1+ sqrt(2)

Is the other x intercept x = 1 - sqrt(2)?

or is it -sqrt(2) + 1 = x? or neither?

[uncut mulch]

x = 1 - sqrt(2) is fine

[compact rivet]

Awesome! So would this be correct?

Would that parabola be right?

[uncut mulch]

not AND

[compact rivet]

Oh, but are they the right intercepts?

[uncut mulch]

use a U for union/or

[compact rivet]

Ah right. So x = sqrt(2) + 1 or x = -sqrt(2) +1?

[uncut mulch]

for the 3rd line from the bottom, you should write a +- in front of the square root

[compact rivet]

Oh yes I forgot about that.

[uncut mulch]

you can even go there directly from
2 = (x-1)^2

[compact rivet]

Ah yep. Thank you so much for the help! And thank you too @charred hull !

[charred hull]

[long crown]

To break down numbers or expressions into its different parts is something I found useful taking calc, for example when solving intergrals: This video might take you one step closer to that mindset: https://youtu.be/OJ6w-f6zpdk 😊

[tidal rain]

Can someone explain the work my professor did here please?

[shrewd urchin]

🤔

@tidal rain

[obsidian monolith]

Krishna:

Krishna:

Krishna:

Krishna:

[sick seal]

what is the diffrence between ) and ]

[stuck lark]

) = exclusive, ] = inclusive

[lucid crow]

) excludes values, ] includes

for infinity you always want ()

[sick seal]

oh, so i basicaly told the hoe that 8 is a good number

[lucid crow]

the reason they used () for 8 is because if you include 8, t hen the denominator will be 8-8

so you divide by 0

and thats not allowed

[sick seal]

https://tenor.com/view/alonzo-lerone-oh-iget-it-gif-10735355

[lucid crow]

so u dont wanna include 8 there

[sick seal]

^ my favorite gif for this disc

[spice urchin]

im not sure how to do this

can someone help

[proud sparrow]

argh

$f(x)=x^2-2x$, find $\frac{f(x+h)-f(x)}{h}$?

[obsidian monolith]

Element118:

[proud sparrow]

I might have read it as $x^\alpha$

[obsidian monolith]

Element118:

[spice urchin]

kk

sorry

[proud sparrow]

Okay, so what happens when you substitute in?

@spice urchin

[spice urchin]

so i get

$\frac{x^2-2x+h)-(x^2-2x}{h}$

[obsidian monolith]

xWolf:

[spice urchin]

then i get stuck after

[proud sparrow]

@spice urchin Are you sure you substituted it correctly?

You need to replace all x in the function with x+h when you compute f(x+h)

[spice urchin]

so would it be $f(x)={(x+h)}^2-2{(x+h)}$

[obsidian monolith]

xWolf:

[spice urchin]

or without the f(x)

[proud sparrow]

not like that, but

$f(x+h)={(x+h)}^2-2{(x+h)}$

[obsidian monolith]

Element118:

[spice urchin]

and then you substitute that into the original equation

[proud sparrow]

yes

[spice urchin]

so it would be $\frac{(x^2+2xh+h^2)-(2x+2h)-(x^2-2x)}{h}$

[obsidian monolith]

xWolf:

[spice urchin]

@proud sparrow ?

then te answer is ${(2x+h-2)}$

[obsidian monolith]

xWolf:

[sullen sorrel]

How do you determine if this is a 1:1 function algebraically

[lucid oracle]

Is there a trick to remembering the rules for the ambiguous case?

[sullen sorrel]

no idea

looking at a graph i see how it's 1:1 but when i try to solve algebraically i get lost

[lucid oracle]

basically if you plug in two different values that yield the same result the function is 1:1

this function isnt 1:1 since its roots is only at 2

[pale kettle]

No that's backwards

if you plug in two different values that yield the same result

Then the function is NOT 1:1

if plugging in two different values always gives you two different results

then the function is 1:1

[lucid oracle]

oh I mixed it up then ig

[sullen sorrel]

alright so i get y=(+or-x +2)

wouldn't that make it not 1:1

[pale kettle]

I'm not sure exactly what you're trying to say

[obsidian monolith]

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

[sullen sorrel]

wouldnt that plus/minus make it not a 1:1 function

[pale kettle]

writing it as y = \pm x + 2 is not very correct

It's not like when x = 1

we have that both y = 1 and y = 3

[sullen sorrel]

you're right

i was looking at the inverse function i did on my paper

but when there is a \pm in general i dont see how it could be 1:1

[pale kettle]

I mean if you have y = \pm x

This isn't even a function so

[sullen sorrel]

right but theres also a ${x}\leq{2}$

nvm

i see

[viscid thistle]

how do you factor something like this

fraction exponents

[stuck lark]

you still look for the greatest common factor

[viscid thistle]

would it be ^-2/3

[uncut mulch]

argh tired

[river kindle]

@viscid thistle For those, the trick is to look for the lowest fraction exponnents, so -2/3 would be correct for 2x-1.

[viscid thistle]

and other side will be -4/3

[river kindle]

Seems that way, yes.

[viscid thistle]

ok

thanks

[river kindle]

It's intimidating at first, but if you slowly do the manual process of it, it makes sense why you pick the lowest fraction

[viscid thistle]

ye

[tidal rain]

Not sure how to factor this

[brisk forge]

hey reeeaaaaaaaal quick

whats the 3^- mean

and 3^+ mean

[uncut mulch]

lim x-> 3- :approaches 3 from the left
lim x-> 3+ :approaches 3 from the right

[brisk forge]

ohhhh ok

sweet thanks

ok i know i said real quick lol

but i fucked up on my hw

could someone help me understand how to do a few of these

[uncut mulch]

follow the curve,
what y-value does it approach at 3 from the left?

(its not asking for the value of that point)

[brisk forge]

1

[uncut mulch]

what about it approach at 3 from the right?

[brisk forge]

4

[uncut mulch]

do you know how to determine whether a limit exists?

[brisk forge]

no?

[viscid thistle]

Limit doesn't exist if there's a restriction on x. But I think you should be working with continuous functions.

Or whatever.

[brisk forge]

what do you mean by restriction on x?

[viscid thistle]

Have no clue tbh.

What it means by 3- and 3+ is the way you approach it.

So when x is 2.999999999999 (3-)and when x is 3.0000000000000000...01 (3+).

[brisk forge]

yeahh

so is that the restriction?

[viscid thistle]

I think.

[brisk forge]

just that you can get to 3 exactly?

oo

makes sense

[viscid thistle]

It means when it is infin.

Or something.

But not sure.

Like maybe for this.

There is no exact lim to 3

But only 3+ and 3-.

[brisk forge]

oooohhh lmaooo

this is fun

[viscid thistle]

Lol ye.

[brisk forge]

ok coool

[viscid thistle]

Your pfp looks like a bootyhole.

[brisk forge]

LOL

[viscid thistle]

If anyone pointed it out to you before.

[brisk forge]

It is

[viscid thistle]

LOL.

BRUH.

bruh

[brisk forge]

LMAO

[viscid thistle]

lol

LMAOOOO.

im last in my class.

:(

i forgot everything over summer

second day of school

I have low IQ.

already failing

:|

Sophomore year = doodoo.

[brisk forge]

@uncut mulch do you know if there is a specific way to determine how a limit exists?

[uncut mulch]

limit from the left and right need to be equal to the same value

[viscid thistle]

900000 IQ.

In general cases sure.

[brisk forge]

ohhhhhhh

shit

ok so what exactly happens in thatformula

when it jumps like that

and skips 3

[viscid thistle]

There's a break in the middle of the graph.

So in some cases what you said is correct.

However in a lot of cases what you stated is incorrect.

[brisk forge]

😮

[viscid thistle]

Lim for 3 can be infin and negative infin (whether + or -)

For that graph I just showed you.

Or 2.5.

[brisk forge]

ohhh shit ok

[viscid thistle]

I can't tell.

[brisk forge]

how far in math have u gone

[viscid thistle]

I'm in PCH rn.

Since I can't take any higher classes.

Since I'm sophoomore.

[brisk forge]

precalc?

[viscid thistle]

Ye.

[brisk forge]

whats the h for lol

[viscid thistle]

Honrs.

[brisk forge]

oooo

[dull breach]

what power of x is associated with a0 ??

[viscid thistle]

In a polynomial?

If it is in a polynomial then it is 0.

You have

$a_{n}x^n + a_{n-1}x^{n-1} +.....+ a_{n-n}x^{(n-n)}$

[obsidian monolith]

leviosa:

[viscid thistle]

And the last term is equivalent to.

$a_{0}x^0 = a_0$

[obsidian monolith]

leviosa:

[viscid thistle]

@dull breach

[dull breach]

what is a ?

the cooefficient?

[viscid thistle]

Yes.

[dull breach]

what is a and n?

what bout n

[viscid thistle]

What?

[dull breach]

n is the power yea?

[viscid thistle]

Yes.

n is the highest power in your polynomial.

[dull breach]

why is n below the a

what is the point of that

[viscid thistle]

Not really a point of that.

I can replace it with a b c d etc.

[dull breach]

i dont understand why its like that

[viscid thistle]

It's trying to match coefficients to the power of x.

[dull breach]

but why cant it just be ax^n+ax^n-1...

[viscid thistle]

Because not all "a"s are the same.

[dull breach]

ok

so

u could just say

ax^n+bx^n-1....

[viscid thistle]

Yes but it's not prefered.

[dull breach]

why ?

[viscid thistle]

Because with using an and a(n-1).

You know what power the x of it has.

So if I ask you "what is the term for n-1"

You can say.

$a_{n-1}x^{n-1}$

[obsidian monolith]

leviosa:

[viscid thistle]

Which includes the coefficient that correlates with x^(n-1)

[dull breach]

ok

what power of x is associated with a0 ??

i dont understand this tho

[viscid thistle]

x^0

[dull breach]

isnt x^0 a exponential?

[viscid thistle]

Yes.

But it also equals to 1.

That's why the last term is always a constant.

(Or 0 if there isn't one)

[dull breach]

huh

so will the last term always be 1?

[viscid thistle]

No.

The last term will always be a0.

Or 0.

[dull breach]

oh ok

[viscid thistle]

👍

[dull breach]

but the power to x will be 0?

[viscid thistle]

Yes.

For the last term.

[dull breach]

got it

[viscid thistle]

If you can't see it then it's 0.

[dull breach]

thx bruv

[viscid thistle]

Mhm.

[dull breach]

what grade u in?

[viscid thistle]

10th.

[brisk forge]

brain

can u help me

im getting my ass ate on this math problem

it would be x^3+h^3+3x^2h+3xh^2 -x^3 /h

[uncut mulch]

-3? did you mean -x^3
also parentheses

[brisk forge]

uhh

oh yeah soz

-x^3

ramo u wanna be my tutor lol

[uncut mulch]

after cancelling the x^3 and dividing by h, what do you have

[brisk forge]

uhh

3x^2+3x

[uncut mulch]

you're making a mistake with this term 3xh^2

and the h^2 should still be there,
i didn't ask you to take the limit yet

[brisk forge]

OHHHH

ok sec

wait xh cant be divided by h?

[viscid thistle]

No.

[uncut mulch]

it can, but 3xh^2/h isn't 3x

[viscid thistle]

If you think ahead, h is essentially 1/inf.

You can't divide inf over inf.

[uncut mulch]

h isn't 0

[viscid thistle]

No.

But 1/inf isn't even a defined number.

Not even a real number.

[brisk forge]

ur confusing me lol

[uncut mulch]

ignore him

[brisk forge]

ok back to this

x^3+h^3+3x^2h+3xh^2 -x^3 /h

[viscid thistle]

:c

[brisk forge]

soz lol

[uncut mulch]

$\frac{x^3+h^3+3x^2h+3xh^2 -x^3 }{h}$

[obsidian monolith]

ramonov:

[brisk forge]

yeah ok lol

i was typing that

urs is better

[viscid thistle]

Oh wait I'm wrong.

Nvm.

[uncut mulch]

what is $\frac{3xh^2}{h}$

[obsidian monolith]

ramonov:

[brisk forge]

3x?

wait

3x^2?

[uncut mulch]

what is $\frac{h^2}{h}$

[obsidian monolith]

ramonov:

[brisk forge]

2

[uncut mulch]

huh

[brisk forge]

wut

lmao

[viscid thistle]

h^2 = h x h

Right?

[brisk forge]

yeah

[viscid thistle]

$frac{h*h}{h}$

[short sorrel]

\

[viscid thistle]

$\frac{h*h}{h}$

[obsidian monolith]

leviosa:

[viscid thistle]

Yeah I put it in the wrong place lmao.

What does that equal to.

[brisk forge]

1?