precalculus

[willow bear]

x=-1 and x=1 are both kinda far from x=-7/2

[naive moss]

hmm any tips on beginning to approach this? should i try values that are close to -7/2?

[opal prairie]

Can you break the given function

[naive moss]

would i need to use the quotient rule here? if so, then u = |2x+7| and v = 2x+7

[digital cloud]

are -1 and 1 bigger or smaller than -7/2?

[naive moss]

both are bigger than -7/2

[digital cloud]

so you've checked the value when it's bigger than -7/2

why don't you test the other conditions

[naive moss]

right, they both resulted in 1. so i'll try something smaller than and equal to -7/2 then

it's undefined when x = -7/2, and -1 when x < -7/2

ok, so that's it...

why are those values called points of discontinuity?

[uncut hollow]

As you go past that point f(x) flips from 1 to - 1 (or -1 to 1)

[naive moss]

and it's undefined when x = -7/2, so in a graph it'd be the empty circle?

although, what i'm wondering now is: i'm only trying out -7/2 cos this is a multiple choice question -- what if i didn't have the clue from the multiple choices? i'll just have to do it by trial and error?

[willow bear]

?????what

[digital cloud]

this is what it looks like

[willow bear]

why would you need the quotient rule literally what are you talking about

[digital cloud]

you can see that it's not continuous

[willow bear]

we aren't taking any derivatives here...

[naive moss]

@willow bear maybe try to be a bit more constructive in your comments? hint: look at some of the other responses above

[uncut hollow]

Bruh

[digital cloud]

bruh

[willow bear]

oh sorry did i not sugarcoat it enough for you

[naive moss]

(like maybe keep quiet and not be hurtful if you have nothing better to say)

the world could use one less person like that

many thanks @digital cloud @uncut hollow

[willow bear]

so how should i have reacted to your suggestion of using the quotient rule in a problem that had nothing whatsoever to do with derivatives?

[naive moss]

it's not what you said but how you said it. if you're not happy or find a question too stupid for you, you don't have to respond

[willow bear]

...

[uncut hollow]

Anyway er

This basically comes down to understanding how |x|/x behaves

for future ref

[sick steppe]

<@&268886789983436800>

[sharp bridge]

Can someone help me with some college level precalc please?

[digital cloud]

don't ask if someone can help
just ask your question and if someone's free and sees it they'll help

[shut timber]

for this, why do you multiply by 1/(x^2)^1/3

specifically, why ^1/3

[viscid thistle]

@shut timber are you still wondering?

[shut timber]

yea

[viscid thistle]

okay

essentially, they multiplied numerator and denominator, by the same thing, x^(2/3)

the reason stands in the knowledge of the exp law: a^n * b^n=(a * b)^n

let me clarify it with another example.

[shut timber]

my quesiton is kinda like

go on

[viscid thistle]

oops.

let us have $\lim_{x\to \infty}\frac{x}{1+{\color{green}{x^2}}}$ for example. The usual step here is to multiply num and den by $\frac{1}{\color{green}{x^2}}$ so that: $$\lim_{x\to \infty}\frac{x}{1+{\color{green}{x^2}}}\cdot\frac{\frac{1}{\color{green}{x^2}}}{\frac{1}{\color{green}{x^2}}}$$ and then you can simplify to: $$\lim_{x\to \infty}\frac{\frac{1}{x}}{\frac{1}{x^2}+1}$$ and $\textit{then}$ you can evaluate the limit way easier than before. the logic applied here is precisely the same.

multiply num and den by the same thing, the highest degree term: $\lim_{x\to\infty}\frac{x+1}{({\color{green}{x^2}}+1)^{\frac13}}$ and the reason why they don't just multiply num and den by $\frac{1}{x^2}$ is because the denominator is inside the third root, and you can, loosely speaking, "let it inside" by adding a $\frac13$ in the exponent, since $a^n\cdot b^n=(a\cdot b)^n$: $$\lim_{x\to\infty}\frac{x+1}{({\color{green}{x^2}}+1)^{\frac13}}\cdot \frac{\frac{1}{x^{\frac23}}}{\frac{1}{({\color{green}{x^2}})^{\frac13}}}$$ so that it gets inside, $(x^2+1)^{\frac13}\cdot \frac{1}{(x^2)^{\frac13}}=((x^2+1)\frac{1}{x^2})^{\frac13}$ $$\lim_{x\to \infty}\frac{x^{\frac13}+\frac{1}{x^{\frac23}}}{(1+\frac{1}{x^2})^{\frac13}}$$

[obsidian monolith]

Al𝟛dium

[viscid thistle]

wow, first try

hopefully that clears it out.

@shut timber

[shut timber]

wow

thank yuou so much

one small question

wouldnt you add 1/3 only if its multiplication

like multiply (x^2)^1/3

[viscid thistle]

not sure what your question is right now.

are you saying that why we multiply by 1/(x^2)^(1/3) and not just (x^2)^(1/3)?

[shut timber]

like what do you mean by adding 1/3 into x^2

[viscid thistle]

oh i just mean like, instead of multiplying numerator and denominator by 1/x^2, we add a 1/3 to the exponent, as in 1/(x^2)^(1/3)

the term add here doesn't mean arithmetic, just put up a new thing, not add 2 things as in 2+2

[shut timber]

can you represent it as a square root

[viscid thistle]

as a third root.

but yes

the whole thing?

[shut timber]

how would you do it as a root

[viscid thistle]

the whole process?

[shut timber]

just the bottom part

[viscid thistle]

or how to express 1/(x^2)^(1/3) as a root

ah okay.

[shut timber]

yea

[viscid thistle]

by letting inside $\sqrt[3]{x^2+1}\sqrt[3]{\frac{1}{x^2}}=\sqrt[3]{(x^2+1)\frac{1}{x^2}}$: $$\lim_{x\to \infty}\frac{\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt[3]{1+\frac{1}{x^2}}}$$

[obsidian monolith]

Al𝟛dium

[viscid thistle]

hopefully it's this what you meant

this is just the bottom 2 lines of the big tex above

[shut timber]

but wouldnt you only take out the cuberoot(x^2) if its multiplication?

can you do that for arithmetic as well?

[viscid thistle]

where do you see a cuberoot(2)?

and you aren't being very clear with that last question

[shut timber]

cuberoot(x^2)**

[viscid thistle]

still unsure on what your question is

[shut timber]

how are you taking out the x^2

[viscid thistle]

try again expressing yourself, the better you explain yourself, the better i can help you.

[viscid thistle]

where am i taking a x^2?

[shut timber]

when you multiply cuberoot(x^2+1) by cuberoot(1/x^2), you are taking the x^2 with the cuberoot attached to it

in multiplication, like (x*b)^1/3, 1/3 is "attached" to both x and b, but how are you able to do it

[viscid thistle]

please replace the last "it" with what you mean, i can't be sure of what you are asking if you keep putting pronouns to hide meaning

[shut timber]

it just refers to this situation

[viscid thistle]

the upper line of this is what confuses you?

[shut timber]

yea

[viscid thistle]

if anything i'm bringing it "inside" not "outside", but this is just the same law mentioned earlier

that same law can be expressed with roots: $\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}$

[obsidian monolith]

Al𝟛dium

[shut timber]

ohh

ojk

ok,

[viscid thistle]

are you sure you are clear with doubts?

or do you still have any doubts? if you do, please do tell me.

[shut timber]

one smalll thing, y did you multiply by 1/cuberoot(x^2)

why do you do cuberoot

not about the multiplication

[viscid thistle]

still unsure on what your question is

are you saying why i chose cuberoot(1/x^2) instead of just 1/x^2?

[shut timber]

yea

[viscid thistle]

the whole point of this tex was just to answer this precise question.

it's the same question as with the "why ^(1/3)?".

remember that both things are equal, the same question is being answered: a^(1/3)=cuberoot(a).

[shut timber]

yes

[viscid thistle]

have you read the whole tex slowly and steadily without skipping anything?

usually that's what causes the lack of clarity

[shut timber]

yes, but it kinda just goes over the overview of why 1/3

[viscid thistle]

which is the same answer to why cuberoot here

i personally wouldn't call that overview, i expanded the explanation for an specific reason

[shut timber]

im confused with the concept of "letting it in"

[viscid thistle]

yes, so look here.

[shut timber]

wouldnt you only be able to do that when inside is multiplying

[viscid thistle]

i call letting it in since the ${\color{green}{\frac{1}{\sqrt[3]{x^2}}}}$ is outside of the other root here: $\sqrt[3]{x^2+1}\sqrt[3]{{\color{green}{\frac{1}{x^2}}}}$, and since $\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}$, you loosely speaking let it in by bringing it inside the other root: $\sqrt[3]{(x^2+1){\color{green}{\frac{1}{\sqrt[3]{x^2}}}}}$

wait a sec.

[obsidian monolith]

Al𝟛dium

[shut timber]

ok, so from 1/3, you bring that to x^2. and then since x^2+1 have same degrees, you just put the three on the outside

so cuberoot(x^2+1 * 1/x^2)

[viscid thistle]

and once again, you are being very reluctant to help with your vagueness and incompleteness of your questions, it's still unsure what you mean and what you are asking here. i'd suggeest to take a few breaths, think of what you want to say before writing it and then message me.

as in there are loads and loads of cases when "inside is multiplying", unless you clarify it, i can't simply help with it

[viscid thistle]

yes, again it's all the law a^n * b^n=(a * b)^n.

[viscid thistle]

be careful, parenthethising is important. cuberoot((x^2+1) * 1/x^2).

[shut timber]

yes, i think i understand it now

thanks

[mild pollen]

is the above statement true or false?

[past meadow]

no.

A function is continuous at a point when the limit exists and is equal to the value at that point

if the limit exists but is not equal to the value of the function

its not continuous

[viscid thistle]

can someone pls solve these before 6 pm

i really need them fast

with the steps

pls

this server doesn't work like this. we aren't going to give you answers.

read #info for more.

but pls cud u still help me out

pls?

someone?

pls?

im beggin u

no, what we can do is help you to get you to your answer, if you are willing to do so, let it know

[viscid thistle]

ok

just help me out

in the 2nd one

1st i got it

[ornate island]

not sure how to express a function in the term of another function

[civic sierra]

l(x) is just f(x) translated some units to the right and down

@ornate island

[ornate island]

o

[civic sierra]

You know how that works?

[ornate island]

ye

came up with l(x) = g(x-33)-11

[mild pollen]

thank you for this

[viscid thistle]

#linear-algebra @naive moss

[uncut hollow]

a non-square matrix maps vectors from R^n to R^m where n and m are different natural numbers, which makes it impossible to have an eigenvector

[cyan geyser]

ill pay someone for tutoring

hmu in dm

[proud raven]

more involved that just asking questions?

@cyan geyser

[hardy mango]

Can someone help me

[gritty raptor]

don't ask to ask sir

[hardy mango]

For example if I get a equation and it asks if a inverse exists how do I know

[patent beacon]

Answered in #help-1

[wary mulch]

can anyone say wat is the derivative for greatest integer function and fractional part function

[digital cloud]

what do you think it is?

[wary mulch]

i read the msg

[digital cloud]

👀

[wary mulch]

for gif undefined at integers

[digital cloud]

i hope it makes sense why that is

[wary mulch]

yea it is soo then for fractional part 0 at integers

[digital cloud]

what do you think?

i'll give you a hint, it's not 0 at integers

[wary mulch]

undefined at integers

[digital cloud]

yes

[wary mulch]

evrywere else it is 0

[digital cloud]

no

it's not

[digital cloud]

how would you write the function for the fractional part of number?

[wary mulch]

{x}=x-[x]

[digital cloud]

do differentiate that

whaddya get

[wary mulch]

1

[digital cloud]

yup

and discontinuities at the integers

[viscid thistle]

i need a good text book to start off with precal, does someone know of any? it can be anything, book, videol, course

[viscid thistle]

khan academy :P ?

try searching "precalculus book" in #book-recommendations channel.

[lofty prism]

How would you solve F(x)2-F(x)1/y2-y1 ?

[whole aurora]

x(x+\sqrt(3))(x+\sqrt(3))(x-1)(x-1)(x-\sqrt(5))(x-\sqrt((5))

can someone help me with math plz!!!!
find zeroes of (x^4-x^3+7x^2-9x-8) if 3i is a zero
obviously -3i is a zero but what else is

[sick steppe]

if 3i is a root, so is -3i by CCRT

so you know the function is divisible by (x-3i)(x+3i)=x^2+9

[whole aurora]

i got that far

[willow bear]

what is this supposed to mean?

[visual hinge]

there the same thing

its just a different way of writting it

[viscid thistle]

hey guys

i need help in hw

[gusty relic]

so I want to check if this function is continous: x+log(x)-e

how can I do it? I know theres a way to check if a function is continious in a point using limits but I dont know how to check it in general

[uncut hollow]

in general if f and g are continuous functions so is f + g

so it's enough to know that x and log x are continuous here

[gusty relic]

and if for example I wanted to check if log x is a continious functonis

like I know it is, but other than eyeballing whan can I do to prove it

[uncut hollow]

kinda depends on how you're defining log x i suppose here

[frigid holly]

if a function is continuous at x=a AND its derivative is continuous at x=a, does that mean it's differentiable at x=a ?

pardon if this is a dumb question, but I'm really tired atm 😄

I can understand that the function doesn't need to be differentiable just because it's continuous, but the textbook is a bit handwavey about differentiability, saying that if a function makes an abrupt jump or "makes a sharp turn", it's not differentiable, but how exactly is "making a sharp turn" defined mathematically? my intuition says that discontinuity in the derivative is what constitutes a "sharp turn", but I'm not entirely sure if I'm right or if I just haven't thought of any edge cases that might prove my intuition wrong

[uncut hollow]

you said its derivative is continuous at x = a which presupposes that it is differentiable there

(otherwise how would it have a derivative there)

[frigid holly]

now that you mention it, it does sound like I'm doing circular reasoning

[uncut hollow]

As for having a sharp turn, intuitively: you can think of differentiation as giving you a way to approximate a function around a point by a straight line, so if that's that's impossible (e.g. |x| at x = 0, as you have two lines with different gradients next to 0) then it can't be differentiable, if that makes sense

[frigid holly]

yeah that makes sense intuitively

but how do I prove that there is or isn't a sharp turn at some point?

[uncut hollow]

Also note you could have a non-continuous derivative somewhere, so that's not quite how i'd define 'sharp turn'

[uncut hollow]

in terms of proofs, actually show the function is differentiable (by taking the limit etc)

the 'sharp turn' is more of a vague idea for intuition

[frigid holly]

ah I see, the resources I consulted involved simply stating that "if sharp turn, then not differentiable", which implied that differentiability is proven/disproven by proving something called a sharp turn

but yeah, I'll just take a look at the actual proofs

thanks

[uncut hollow]

np

[frigid holly]

okay so, f(x) is differentiable at x=a simply when the derivative has a limit as x—>a? or should I say that the expression by which derivative is defined has a limit as x—>a? or are those two the same thing 😄

[viscid thistle]

Got a relatively simple average rate of change problem that has me stuck. Is my process until last step correct? Any help is appreciated!

[willow bear]

your use of 26 to mean 2*6 is questionable

or...

wait, those aren't sixes, are they

they're b's

@viscid thistle am i understanding correctly that your problem is this?

Find b such that the average rate of change of f(x) = 1/x over [2, b] is -1/10.

[viscid thistle]

yes sir

[willow bear]

...please don't call me sir.

[viscid thistle]

?

[willow bear]

i'm a woman lol

[viscid thistle]

its an expression but no problem yes maam

really just need math help so any is appreciated

[willow bear]

idc if it's an expression i don't like being called sir and i have strong feelings about it

anyway...

it's a little strange to me that you didn't simplify $\frac{\frac{2-b}{2b}}{b-2}$ into $-\frac{1}{2b}$ even though you could've done so

[obsidian monolith]

Ann

[willow bear]

not doing it isn't exactly fatal, but you will have b=2 appear as an extraneous solution of your equation

[viscid thistle]

is b-2 the same as 2-b

[viscid thistle]

thought that those were different, like 5-4 isnt same as 4-5

[uncut hollow]

neither of those, a function is differentiable at x = a when the derivative exists at a

[willow bear]

they're not the same but they are opposites of each other.

b-2 = -(2-b).

[viscid thistle]

so then how would they cancel since there is no -(2-b)

[willow bear]

what do you mean?

you have $\frac{2-b}{2b(b-2)}$

[obsidian monolith]

Ann

[willow bear]

you can write this as $\frac{2-b}{-2b(2-b)}$

[obsidian monolith]

Ann

[willow bear]

or as $-\frac{b-2}{2b(b-2)}$ if you want, it doesn't matter

[obsidian monolith]

Ann

[viscid thistle]

that makes sense

[willow bear]

either way the cancellation can take place now

[viscid thistle]

yeah ok that was helpful thankyou for that wouldn't have got it without the help!

[viscid thistle]

idk why i forgot to mention the textbook says b=5 is there any chance this is a case with two different answers?

[willow bear]

hm?

i'm not sure what you mean

this problem has only one possible value for b

[viscid thistle]

the textbook says its 5 and after trying it, it makes sense using your steps

[viscid thistle]

I thought i had read u said b=2 at one point

[willow bear]

you will have b=2 appear as an extraneous solution

[viscid thistle]

extraneous means what in math? Heard before just can't remember

[willow bear]

it means that it'll be a solution of the equation you arrived at, but not for the problem you started with

[viscid thistle]

OH ok thank you for all the help. It is much appreciated truly

[frigid holly]

this is a very useful concept 😮 I hadn't heard it before!

[viscid thistle]

If i get a discount of 30% and then a discount of 15% on that price how do I find the percentage I saved? I know I have to use functions but I am truly stuck.

[willow bear]

"have to use functions" conveys about as much information as "have to use numbers"

@viscid thistle do you still need help with that problem?

[viscid thistle]

I got it. Thank you tho you were a godsend today!

[humble smelt]

I have a question, can you use the steepness of graphs to help see if the speed of a ball (for example) is increasing or decreasing? How can you use the slope to help find this trend?

let me know if I'm on the right channel

[proud raven]

in what context?

like if a ball was rolling on the graph? or if the graph is of the balls speed?

or in some other way

[humble smelt]

Like if the graph is of the ball's speed

[proud raven]

so if its going down it should be decreasing right

unless im misunderstanding

if the graph is of the balls speed and you want to know if the speed is going down, just consider the speed at two points

if the speed has gone down between the two points ...

maybe one way you could connect this to slope is to think of the slope of the line connecting some arbitrary two points together

if the speed has gone down during the time between the two points, the slope of that line connecting the two points will be negative

you can connect this into the limit definition of a derivative if youve seen that before

is that what youre asking? sorry i babbled a little bit @humble smelt

[humble smelt]

No problem, yes that's what was I asking! Yes I've seen the limit definition before, and this is very helpful.

[proud raven]

[humble smelt]

I see what you're saying, so looking at the difference between two points and seeing the slope is less means the speed is decreasing.

[proud raven]

well if the slope is less than 0

[humble smelt]

Right, so it would be negative

[proud raven]

yup

this is maybe leading you into $a = \dv{v}{t}$

[obsidian monolith]

jan Niku

[proud raven]

or idk if you are already given that

but thats basically all this is

[humble smelt]

Yes that's what I was given. Okay, sometimes I have a math thing going in my head but can't really sort it out until I talk about it, so this makes sense. Thank you so much!

[proud raven]

[viscid thistle]

can anyone help me with what the applications of adding, subtracting, dividing, multiplying functions such as f(x) +g(x)? I understand the applications of composite functions such as (f o g)(x), but I can't find a solid example of applications for other types of composite functions. Is it related to piece-wise functions?

[bleak lance]

Could someone please help me solve f(x)=4x^2-x-3?

[strong yew]

What's the point of using f(x) instead of y

?

[glacial lantern]

f(x) = 3x+8 x1 = 0 x2 = 3 find the average rate of change

of the function from x1 to x2

[vapid plaza]

Do you know what is rate of change?

[sullen zealot]

No

[rocky nacelle]

This is more a notation issue than anything. Firstly, just using y makes very important notions like inverse functions are composition of functions very difficult to impossible to define.

[vapid plaza]

f(x) represents a function of x, so when you see it you know that f(x) depends on x
y is just another axis that people use to graph functions primarily
For example, using f(x) and not y is good if you have multiple functions like g(x), h(x)

[bold temple]

points changed to (8,-6)

f(x) + 4

im guessing you do the same thing as before just with a slight twist.

f(-8) + 4

which would give me

(8, -4)

@viscid thistle

[viscid thistle]

Add y co-ordinate to 4 in 16th question

Multiply y co-ordinate with 2/3 and add 5 in 19th question

[bold temple]

so for 16 its 8, -4

[viscid thistle]

-6+4=-2

[bold temple]

or am i doing -6 + the positive 4 which is -2

alr

so 8,-2 is the answer? and all im doing down here is applying the Y co-ordinate to the number being added

[viscid thistle]

yeah

[bold temple]

quick question

why am I plugging in Y for the X coordinate?

wouldnt I be pluggin in the 8 where it was f(x) + 4

[viscid thistle]

f(x) is another way of saying y

so just replace f(x) everywhere here mentally with "y" and then solve

[bold temple]

@viscid thistle this looks right?

[viscid thistle]

Bruh I won't check all these answers. Feeling too drowsy. However, the one-two that I saw are right. Have self-confidence. Good night now. Too sleepy

[bold temple]

Ian want u to check all u good

I just wanted to see if they looked right not actully check one by one

prec8 the help doe fam

[restive hound]

Hello, I watched a video on YouTube on transforming a general form to a standard form of an ellipse. I’m confused on a part where he had to divide everything by 25. I understood why so that it would equal to 1, but the part when he can’t divide 16 and 25 is what I’m confused at. What he did is that he took the 16 and placed it into the denominator with 25 so it became (x-1)^2/25/16. Is it correct? Is that what you do when you can’t divide a number from the denominator?

[willow bear]

(x-1)^2/(25/16)

Is that what you do when you can't divide a number from the denominator?
as much as i hate to confirm imperative instructions like this, yes

but also it's not really any different than if it were, say, 16(x-1)^2/64 simplifying to (x-1)^2/(64/16)

(except that 64/16 would admit further simplification)

[restive hound]

Oh okay, thank you

Oh yep, is minor vertices just the same as covertices? And major vertices just the same as vertices?

[willow bear]

sure

[restive hound]

I have another question 😅 is the word “major vertices” just the same as “vertices” and “minor vertices” just the same as co vertices? Because when I watched organic chemistry tutor’s video, that’s what he said. Major and minor vertices

[willow bear]

uh

i guess so

[restive hound]

Ah ok, thank you

[viscid thistle]

how do you solve this equation for x? The correct answer is 1 but I can't seem to get it myself and some sites just straight up can't solve it

[viscid thistle]

f(x)=f^-1(x)

it is this form of question

[willow bear]

does that actually help in any way

[viscid thistle]

yes

[willow bear]

this equation reduces to a polynomial equation of degree 9

[viscid thistle]

No

wait

this one is easy

Ah there was a trick to solving this question

First root can be found easily

since these functions are inverse of each other

they intersect at the line y=x

So put x=3-2x^3
x=1

^one solution is this. I am sure

[viscid thistle]

solution^

[viscid thistle]

The questions with type f(x)=f^-1(x) no matter what the order of their polynomial is can be simplified to find at least one solution since they always intersect with the line y=x.
Therefore, we can instead solve a simpler equation by using this logic and just solve f(x)=x as shown above

[willow bear]

oh yes of course

[viscid thistle]

This makes sense, rather than solving them simultaneously, just solve one and y=x simultaneously becasue thats where they mirror

[fluid plank]

Hey guys i just want to know if this is correct

[harsh cobalt]

Can anyone explain this question?

[fluid plank]

<@&286206848099549185> can you help me with this

[restive hound]

I’m also learning about ellipses today, let me see what I can do. What number? Is it number 2?

[fluid plank]

Yes it's number 2 thanks

And the foci is actually (6√2,3) it got disrupted when i converted the file

[restive hound]

Okay, wait

[restive hound]

@fluid plank from what I can see, you solved it right. Your vertices are (9,0) and (-9,0) and your foci too is (6√2,3) and (-6√2,3). Your a^2 is indeed 81, but about your b^2, I’m not sure. Your a^2 is 81 and your c is 6√2. You simplified your c didn’t you? But before you simplified it, was it 72? Because √36 √2 ➡️ 6√2. When I solved for b^2, I got a negative answer which is -72.5. What I did was 6√2 - 81 = b^2. But when I tried using 72 instead of 6√2, I got -9. I’m not sure a negative number should be there. But all in all, for me, you’re right.

[wet pollen]

need help for nukber 5

number*

[restive hound]

I can help you!

Wait

[wet pollen]

sure

[fluid plank]

We actually used the same steps but i just divided both side by -1 to remove the negatives

[restive hound]

Ooooooo okay, is that what we should do if there is a negative?

Haven’t encountered one yet xD

[restive hound]

Okay so when we say “tangent to the x-axis” it means that it will touch at a point on the x-axis. I’ll draw to show you

[wet pollen]

okay

[restive hound]

First plot your center, so again, when you say “tangent to the x-axis” it means that if you were to draw a circle, the curve as you draw the circle will touch at a point on the x-axis. Same goes if we say “tangent to the y-axis”, it’s just that it will touch at a point on the y-axis. I did not draw a circle because it will be funny 😂 but you can know the answer by just counting! So from your question, the center was given to you and you have to find the equation in standard form. What your missing is your radius, that’s why in the problem it says “tangent to the x-axis” because that’s exactly how you’ll be able to find your radius. As you can see, I wrote 5 units. I counted 5 units starting from my center, down until I reach the x-axis. That’s your radius, so your r^2 is 25 and your r is 5. Remember this, if it says “tangent to the x-axis” and you have to find your radius, start counting vertically from your center until your reach the x-axis. If the problem says “tangent to the y-axis”, start counting horizontally from your center until your reach your y-axis. If you wonder which direction you will start counting, well it depends, usually when it’s tangent to the x-axis you either count downward or upward but if it’s tangent to the y-axis, you either count left or to the right. It depends on where your center is, in this case your center is on the first quadrant, on the right, so you count downward.

@wet pollen

Try doing this and if you have that thing that you rotate to make a perfect looking circle lol, see if it touches the x-axis on (3,0)

[wet pollen]

thank you so much

last

@restive hound

[restive hound]

You’re welcome, I’m glad to help

[wet pollen]

the answer here is (x+3)+(y-5)= 25 right?

[wet pollen]

am i right?

[restive hound]

Yes! You forgot to write the square though in (x+3) and (y-5), so it should be (x+3)^2 + (y-5)^2 = 25

[wet pollen]

oh shit i forgot

thanks by the way!

[restive hound]

Don’t forget that okay, it’s important!

[restive hound]

You’re welcome, happy to help

[restive hound]

Hello, what does “vertical major axis of length 8 and minor axis of length 3” mean? Do I have to multiply these two by 2? Because major axis is 2a and minor axis is 2b. Or do I have to square them because it’s a^2 and b^2

These are the given along with the center, I have to write them in standard form.

[viscid thistle]

Major axis = 2a
Minor axis = 2b

[viscid thistle]

ayo does this function have an inverse?

it looks like it should

[vagrant ingot]

if f(x) is a linear function with f(-1)=-4 and f(4)=11, find an equation for f(x) in the form f(x)=mx+b. Enter the values of m and b below

[warped thunder]

Sove

[vagrant ingot]

soe

[daring silo]

Stuck on these, can I get some help?

[untold jacinth]

posted the wrong question earlier, was stuck on this one, if anyone could help it would be greatly appreciated

[fringe raft]

What did you try

[jagged topaz]

the cost of goods and services in a urban area increased 1.5% last month. At this rate, what will be the annual increase rate?
answer is 19.6%
how come

[fringe raft]

This channel is busy try another one

[untold jacinth]

i tried plugging in the 3 for the x's but its just the breaking down part that gets to me

end up getting lost

[fringe raft]

You’re into calculus and trouble in squaring and adding ?

[untold jacinth]

im in a pre-cal class just generally stuck on this specific question

any other question on that subject was easy to me

maybe its because im tired but im out of it for this one

[fringe raft]

F(3)= -4(3)^2+4(3)-4
F(3)= -36+12-4

[fringe raft]

Try writing step wise that would help

[untold jacinth]

so in simpler terms its pretty much just f(3)=-4 right ?

[fringe raft]

Nope

[untold jacinth]

because this is how they see fit as to finalizing the answer

[fringe raft]

This is for f(1)

Your q asks for f(3)

[untold jacinth]

yeah i know, that was an example of how they were to answer that type of question

[fringe raft]

And f(1) is -4 not -14

[untold jacinth]

soooo?

[fringe raft]

This is a different function altogether

[untold jacinth]

yeah i know

i've been trying to tell you this is just a example

of that same type of question

[somber sigil]

If the qn is asking to simplify (I would assume so), factorise (2-x) out first and simplify the fraction, which should be also further simplified.

[inner sage]

im trying to come up with a arithmetic sequence with 5 terms, the last term is 100, and the first 2 terms can be anything you wish, but the common difference has to be the fibonacci sequence, so for example, 20, 20, 40, 60, 100 would be a valid answer. im looking for a formula or a pattern so i can find alternative answers

[willow bear]

so... you're not trying to come up with an arithmetic sequence at all

but one that follows the fibonacci recurrence and has 100 for its fifth term?

@inner sage

[inner sage]

yeah

@willow bear

[willow bear]

okay, so if your first two terms are x and y then the fifth will be 3x+2y

so for your case you would need 3x+2y=100

[marble briar]

What does this statement mean?

[patent beacon]

Give me some function

Any at all

[marble briar]

X²+1

[patent beacon]

f(x) = x² + 1
f(1/x) = (1/x)² + 1

LS:
f(x)f(1/x)
= (x² + 1)(1/x² + 1)
= 2 + x² + 1/x²

RS:
f(x) + f(1/x)
= (1 + x²) + (1/x² + 1)
= 2 + x² + 1/x²

So the function you gave me satisfies the relation they gave you

Which it had better, because they told you it would

[marble briar]

Excellent,thank you. I was trying to solve the same but did a silly mistake. Thanks😊

[patent beacon]

Np feel free to ask if you have anything else!

[marble briar]

Thanks!!

[marble briar]

How is it between [-1,1]?

[willow bear]

you mean either "between -1 and 1" or "in [-1,1]"

anyway this is just $|t|\leq 1$ being equivalent to $-1\leq t \leq 1$

[obsidian monolith]

Ann

[marble briar]

Thanks ,got it

[daring wolf]

Quick question what parent functions are even?

And Which are odd

I know Linear and Quadratic are even

What else am I missing

[trim hemlock]

linear functions are not even, and not all quadratic functions are even

even functions are functions such that for every x in the domain, f(x)=f(-x)

odd functions are functions such that for every x in the domain f(x)=-f(x)

and there are lots and lots of functions that can be even or odd

[daring wolf]

Absolute value is even

[trim hemlock]

again, not true

not all functions that include an absolute value sign are even

[daring wolf]

What

Exponential?

Cubic?

[trim hemlock]

you have to pay attention to what im saying
A function is even if it satisfies the following condition, f(x)=f(-x) ( or f(x) - f(-x) = 0) for every x in its domain

and not all functions have to be even or odd

[daring wolf]

Oh

[trim hemlock]

for example lets take the example of a quadratic function

a quadratic function is in the form ax^2 + bx + c

from what you have learnt ( supposed so) f(x)=x^2 is even

however f(x) = x^2 -3x + 1 is not

one thing you can also notice is that even functions are always symmetric along the y-axis

[daring wolf]

I just need to know the f(x)=

But yeah

[trim hemlock]

....

[trim hemlock]

this is what you need to know

its very wrong of you to put a category of a family of functions into the category of it being even or odd

[daring wolf]

Yeah I get the jist of it

[viscid thistle]

f(x) = square root of x-1

got it from a textbook for reviewing, it said to specify the domain

"The radicand must be nonnegative, so x is bigger or equal to 0 and thus x is bigger or equal to 1. Therefore, the
domain (D) is D = {x|x is bigger than 1}"

having a hard time understand how the logic worked

[sick steppe]

x-1>=0 since you can't sqrt a negative

[viscid thistle]

got the logic, thanks tho

[viscid thistle]

use product rule
$$\dv{x} f(x)g(x)= f(x)g'(x)+g(x)f'(x)$$

[obsidian monolith]

Spica

[viscid thistle]

so its
$$2\qty(t \dv{t} e^t + e^t \dv{t} t)$$

[obsidian monolith]

Spica

[tough jolt]

How do you go between these two steps

[uncut mulch]

difference of two squares

[novel echo]

Good afternoon, I am new and I need to solve a factorial problem but I do not understand much. Could you please help me?

Please is the only exercise for my homework

help please :C

can you help me?

😖

[digital cloud]

what are you confused by? and what have you done so far?

[cosmic crown]

already answered in #prealg-and-algebra

[novel echo]

no he hecho nada, lo borre :c

[digital cloud]

i have no idea what that means

[night cave]

hey hey wrong language

[nova scroll]

what the question

[kindred panther]

g(x) would be at the bottom box and f(x) would be at the top box but what would I put as the denominator for the fraction?

[cold oar]

I'm having problems understanding this.

[plucky basalt]

whos good with yr 10 maths
i need help lol

[cold oar]

S h o w.

[sick steppe]

dont ask to ask.

[plucky basalt]

?

[sick steppe]

You cant get much help w/o asking an actual question. You asked for help, which is a pointless question in a help server

[pearl stump]

how would I simplify 15 a?

[copper sparrow]

What I would do is divide first. So when you have a fraction divided by a fraction, take the denominator, find the reciprocal of it, and multiply that by the numerator

So you'd get -14x^3 + 14x^2/16x^2-16x

Then try factoring it to see if anything can cancel out

[low blade]

Hey, I will try and explain it simpler.

So we have:

$\frac{\frac{-x+1}{8x}}{\frac{2x-2}{14x^{2}}}$

[obsidian monolith]

TJ89899889

[low blade]

Which is equivalent to:

$\frac{-x+1}{8x}\cdot\frac{14x^{2}}{2x-2}$

[obsidian monolith]

TJ89899889

[low blade]

After multiplying, we get:

$\frac{-14x^{3}+14x^{2}}{16x^{2}-16x}$

[obsidian monolith]

TJ89899889

[low blade]

Which can be factored to:

$\frac{-14x\left(x^{2}-x\right)}{16x\left(x-1\right)}$

[obsidian monolith]

TJ89899889

[low blade]

Which can then be factored to:

$\frac{-7x\left(x^{2}-x\right)}{8x\left(x-1\right)}$

[obsidian monolith]

TJ89899889

[nova scroll]

$f(x)=1/x or x^-1 and g(x)=x^2-5x+3$

[obsidian monolith]

mustapha56

[viscid thistle]

What should I google to find an explanation of how to do this? In other words what is the term or subject of this question?

[uncut mulch]

algebra

[novel echo]

xd?

[somber sigil]

Move to algebra but a tip for part b maybe
let (1/2) - {1/[1+(x-1)²]} > 0

Then I think u should be able to do part a?

[viscid thistle]

Well... No, No I don't get it. But I'll move my question to algebra. I'm missing or forgot the education as to what 'let function > 0' should imply.

[somber sigil]

Function more than 0 or function is positive

[safe cobalt]

my pre calc education was kinda shaky and im now pretty worried about my calc class,,, anyone have any pre-calc tips or video refrences?? it would be much appreciated (keep in mind its AP (american)highschool calc so pretty testing/ memmory focused)

[maiden spear]

Hey guys. I’m currently in Pre-calculus but I’m out with covid and there’s no notes online. Is there any place good to start? (My school is doing basics unit circle and advanced currently)

[versed cape]

Hey I need some help

So here is the questions
The graph of f passes through (-1,5) and is perpendicular to the line whose equation is x=6

[late mica]

So what exactly are you trying to find?

[versed cape]

I already got help it's fine

[restive hound]

Hello I’m confused on this part. I have to find the standard form of the ellipse, but the given are only the vertices which are (-2,-8) and (-2,2) and the foci (-2,-7) and (-2,1). The center was known by using the midpoint formula. So the center is (-2,-3). From the example, “a” was solved like this: 2a = 2-(-8). Why was it like this? I don’t understand how it became “2-(-8)” Why did it have to be subtracted? And from the example too, “c” was found by doing this: k+c = 1, just substitute the the value of k which is -3. I don’t understand 😅

I reckon I could just find the value of a by graphing the given vertices and counting the distance between them and getting the half of the value. But with c, I’m not sure 😅

[restive hound]

Hello, you can check Khan Academy, I know they have Pre-calculus lessons there. I wish you a good recovery

[meager vessel]

I didn't know 😶

Asked for the soln from my teacher

And then got to know

[crimson geyser]

I need help 😭

Someone ping me If you can help :)))

[restive hound]

@crimson geyser what topic? I can help if it’s about circles and ellipses

[viscid thistle]

@ayushithesushi#9381 do you still need help?

oh uh.

[willow bear]

suppose not.

[autumn dirge]

why do we use g in calculus?
for eg like g(x)

instead of f(x)

[sick steppe]

cause you cant name every function f...?

same way we use variables other than x

[autumn dirge]

oh so g(x) isnt just for calculus

[sick steppe]

no...

[autumn dirge]

its just a general variable

[sick steppe]

function

[autumn dirge]

oh i just saw it commonly thats why

right

[sick steppe]

g(x) isnt a variable

[autumn dirge]

g is

thought?

though

right?

[sick steppe]

depends on the context

[autumn dirge]

and together g(x) is function?

wait what do u call the x inside the brackets

[sick steppe]

yes, g(x) is a function named g of variable x

x

[autumn dirge]

parameter lol?

oh juts x

is there like a special name for the place

[willow bear]

g is just a name just like f, by the way.

[sick steppe]

well yeah.. it's just the variable(s) that are inputs for g

[willow bear]

the only reason f is commonly used as a name is because it's the first letter of "function"

and g is just the next letter after f

nothing else to it

[autumn dirge]

ahh big brain

hahah they really think it like that

thank youse

[stable depot]

(2x^(5/2) + 10x^(3/2) - 18x) / Sqrt(x)

can someone simplify and show the the steps

[uncut mulch]

$\frac{2x^{(5/2)} + 10x^{(3/2)} - 18x}{\sqrt{x}}$?

[obsidian monolith]

ℝamonov

[daring wolf]

Someone help me with 2 and 4

I don't know what to do with the -8 in 2

And the √ in 4 and the 2 x's in 4

[stable depot]

yes. i dont see how to simplify

[uncut mulch]

consider first splitting that into 3 fractions

and then apply your exponent laws

[brisk gyro]

If a function is even, it is symmetrical with respect to the y axis. If it’s odd, it’s symmetrical with respect to the x axis and flipped/reflected with respect to the y axis. If it’s translated, like in your case, it’s symmetrical and/or flipped/reflected with respect to the translation(s) (in your case it’s symmetrical w.r.t. y=-1 and flipped/reflected w.r.t. x=2). If a function is neither even nor odd, then you can’t tell if it is symmetrical to some axis or not.

It's symmetrical with respect to y = -1 and reflected around x = 2

Your function looks like this

@peak crystal So it's not reflected on neither the x nor the y axis since it's translated

[viscid thistle]

I'm thinking of doing calc and was wondering how difficult is it to understand

[copper sparrow]

If you are good at algebra, then it's like learning any other thing

[viscid thistle]

Ah ok id say I'm like 80% good at algebra but I think some extra practice could help out

[faint prairie]

how hard is pre calculus compared to an algebra ii/trig honors class

[copper sparrow]

At my school, precalc was easy af compared to algebra II

[faint prairie]

is it a wise decision to take precal over the summer or nah

[copper sparrow]

depends on why you are doing it

it never hurts

[wet void]

I know this is really simple and I’m probably not thinking rn, but if someone could help me go through 33 that would be great

[magic latch]

im s lost can anyone help?

[digital heron]

The second one is x=2 because 2^(2+1) = 2^3=8.

The first one is x=4

[magic latch]

i dont understand

wount it be 2^(x+1)=8

[digital heron]

8 = 2^3

2^x+1 = 2^3

What is the x?

@magic latch

You have the same bases

So its simple equation

x+1=3

x=3-1 => x=2

[magic latch]

ohhh

okay i get it now lol thanks

[forest jewel]

never worked with this before... what if the second one is 2^(x+1) + 9 = 16?

Oh I get it now, tried it via photomath XD

[restive hound]

Hey guys I have a question about parabola, if we are dealing with a parabola wherein their orientation is upward and downward, then that means their axis of symmetry is vertical and always y=0? Even though the Vertex(h,k) is not at (0,0)? And if the orientation is to the left and to the right, that means their axis of symmetry is horizontal and always x=0? Even though if the vertex is not at (0,0)??

[willow bear]

y=0 is not a vertical line

if you meant x=0, then no, the axis of symmetry of a parabola is not always the line x=0.

[restive hound]

Ah wait I think I get it now

So if the parabola opens upward/downward then that means it is x=0, but it is not always y=0 because it depends on where your vertex is because the axis of symmetry always passes through the vertex. But if the parabola opens left/right then it is x=0?

Am I right? 😅 @willow bear

[willow bear]

it doesn't need to be x=0 specifically

[restive hound]

yeah

but from my example, the graph opens upward then the vertex is (0,0) but the axis of symmetry is x=0. Is it right?

shouldn't it be y=0?

[willow bear]

the graph opens upward then the vertex is (0,0)
just because the graph opens upward does not mean the vertex has to be at (0,0)

the vertex could be anywhere independently of which direction the graph opens

[restive hound]

yeah but from the example it is x^2 so the vertex is (0,0)

I'm confused, why is it x=0? why x when the axis of symmetry should be vertical

[willow bear]

only now did you say you meant specifically the parabola is y = x^2.

x=0 is a vertical line.

x=0 is the equation of the y-axis.

[restive hound]

oh so this whole time i've been thinking it's x-axis

ah okay I understand now lol

Thank you very much @willow bear

[restive hound]

Oh wait y=x^2 just the same as x^2= 12y (this is the example)?

[willow bear]

no these aren't the same lol

[restive hound]

Ah oh lol, but yeah that’s the example x^2= 12y so it is still right, that its axis of symmetry is on the y axis? x=0?

[willow bear]

yes

it's just y = (1/12)x^2

[restive hound]

Ah okay then if it is something like y^2= -8x then it’s axis of symmetry is y=0?

[restive hound]

Ohh okie I see

[willow bear]

yes sure

[restive hound]

Thank you very much! :DD

[willow bear]

@restive hound don't ask to ask, just ask

[restive hound]

Ok sorry

[rocky finch]

I'm having a tough time fundamentally understanding the difference between a function and an equation. So I have the following reasoning in my head but don't know if I'm right, a function tells us how something changes with respect to something, for example v(t) = t^2 tells us that velocity changes as the square of its time. But an equation like for example, f = ma only tells us how two things are related, but doesn't tell us how something changes with respect to something since there is no independent and dependent variable. Is this last part correct or utterly wrong, thanks for the help 🙂

[patent beacon]

A function is anything that has one input/output.

For example, consider this rule:
f(1) = 1
f(2) = -1
f(3) = 2
f(4) = -2
f(5) = 3
f(6) = -3
...

It should be clear what the pattern is, and how this function can input any natural, and output any positive/negative integer. However, it would be difficult to write it as a single equation

An equation is any statement with an equal sign.@rocky finch

[rocky finch]

First of all thank you for the help, but I'm still a bit confused because for example x^2 + y^2 = 1 is a circle, it's not a function but it has a clear input and relationship (btw don't mind using complex terms I've had courses in calculus, it's just that I suddenly went down a philosophical thought pattern and got stuck in it)

@patent beacon

[patent beacon]

It has a clear input and output? What are they?

[rocky finch]

Ooh no yes I see what you mean now, but like y + 2x = 1 can be manipulated to have a clear input and output but it's considered an equation and not a function

[patent beacon]

y + 2x = 1 is clearly an equation cause =

It can also be considered a function, by letting one of the variables be dependent/independent, then solving for it.

So in that case, it is pretty easily seen as both.

Something like x² + y² = 1 is much more difficult to see as a function

Oh sorry, you said x² + y² = 1 is not a function. I'm just waking up sry haha

[rocky finch]

hmm, yes the one input one output correspondence

[rocky finch]

No worries buddy, thank you for helping me anyways that's already super nice of you

So would you say a function is an equation who only has one output per input

[patent beacon]

A function is any rule that satisfies one-input to one-output.

As above, I posted a function that isn't an equation!

Likewise, you showed F = ma, which is clearly an equation. It can also be seen as a function if we chose to hold one of the values constant, like so:
F(a) = ma

[rocky finch]

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOH WAIT so you have equations like y + 2x = 1 and sometimes these equations imply a function like y = -2x + 1

[patent beacon]

But we won't usually choose to look at F = ma this way

[rocky finch]

@patent beacon Alrighty, thank you for the help 🙂

[patent beacon]

Np! Feel free to ask if you need anything else.

[rocky finch]

@patent beacon Sorry to bother you again, this is the last time I swear, but I've been thinking more about it and are functions and equations just 2 different things but sometimes equations can imply functions or is this incorrect?

[patent beacon]

I think the big thing is interpretation.

Sometimes an equation can't be interpreted as a function.

Sometimes a function can't be interpreted as an equation.

Sometimes you have something that can be interpreted as both.

2y + x = 1 is clearly an equation.
If you want to assign x as "input" and y as "output", then you can interpret 2y + x = 1 as a function as well. y = (1 - x)/2

You don't necessarily have to do this, of course. Depends if you benefit or not from seeing things this way

[rocky finch]

Oh yes, I see. I think I'm just confusing my self by trying to understand it and overthinking it

[patent beacon]

Haha sorry if it isn't the treasure trove you thought it might be. Sometimes it can get really interesting though. That function I posted above:
f(1) = 1
f(2) = -1
f(3) = 2
f(4) = -2
f(5) = 3
f(6) = -3
...
Throws a lot of people off

If you try to write that as an equation, it gets messy quick, and isn't really helpful. This is best as a function without an equation.

[rocky finch]

Ooooh yes I see what you're talking about

So it's a matter of convertibility in general sometimes you can write a function as an equation sometimes you can't

[rocky finch]

No worries hahaha, It's just that I enrolled into a mathematics program at uni for next semester and I'm revising everyhing I've seen in HS to make sure I'm ready but now I just feel dumb and too dumb to pass as a Math major

[patent beacon]

I'm pulling these examples from years of experience. I'm not trying to make you feel dumb haha. I'm sure many students applying with you have similar questions, but don't know how to ask. You're doing well.

[rocky finch]

Well thanks for the reassuring words, and thanks for all the help you've given me so far, Have a nice day 🙂

[patent beacon]

Np, feel free to ask if you have anything else!

[rocky finch]

Okidoki, thank you!!!

[haughty wraith]

Heh what unit is this

[restive hound]

How can I get the standard form of a parabola if I’m only given the focus which is (2,5) and and the directrix that says that it is 6 units to the right of the vertex? I can’t plot the directrix because I need to know the vertex first, but I know that the distance of the focus and the directrix to the vertex is just the same but I can’t figure it out

[willow bear]

@restive hound the directrix is 6 units to the right of the vertex, so the vertex is 6 units to the right of the focus.

[restive hound]

Oh okay, if the vertex is 6 units to the right of the focus. Do I get the half of 6? The answer to this question is (y-5)^2=-12(x-5) and if I get the focus it would be 3. Btw, I was stuck at this question so I took a look at the back of the book and tried to figure it out lol

[willow bear]

...

that doesn't look right

[restive hound]

The one where I asked if I should get the half of 6 or the answer?

I think the book is wrong 🤷‍♂️

[willow bear]

can you show screenshots/pictures of the book

[restive hound]

Okay

The question is on number 21

Even the question on number 23 I can’t figure out

[willow bear]

hold on

let me just try graphing this

[restive hound]

Okay, thank you very much for the help, I appreciate it. I’m slowly giving up but I really don’t want to leave this unanswered. I have no teacher to ask because I’m learning this again by myself

[willow bear]

what i'm getting is this

which places the directrix 6 units right of the focus but only 3 units right of the vertex

this blue parabola is the answer they give

blue = the answer they give
red = the true answer

[restive hound]

Ahh okay I see, so if the red is the true answer then the answer on the book is wrong?

[willow bear]

yes that's what i've been saying all along

the focus is (2,5), the vertex is (8,5), the directrix is x=14.

[restive hound]

Ah okay, what did you app/website did you use to graph that? Because I’m trying to figure out question seven too, the directrix on the book is different from my answer and I want to see if I’m right. Cause I know I did the solution right

[cosmic crown]

desmos

[restive hound]

Ah! Okie, thank you

Thank you very much Ann for the help 😄

[willow bear]

yeah it's desmos

[restive hound]

Oh what if the directrix is x=5 and it says that the vertex is 3 units to the left of the focus. In order for me to know the focus, should I start counting 3 units from the directrix?

Or 6 units from the directrix?

[willow bear]

3

er

no

it's always focus --- vertex --- directrix

so yes 6

[restive hound]

Okay so what I did is I counted 6 units to the right of the vertex and I landed on x=11 then since it says that the vertex is 3 units to the left of the focus, I count 3 units to the left and I landed on x=8. And so that’s my vertex but I don’t have the k value yet, but the distances between the directrix and from x=11 is 3. So x=11 is like my focus? Because the answer is is (y-3)^2= 12(x-8) so the h value is 8 but how do I find k=3?

I mean the focus is 3 when solved but if I didn’t know the answer yet, x=11 would be like my focus?

[trim hemlock]

Given the function y=x^3-8x^2+8x, whose graph is (C), and the function y=x^2+(8-a)x - b, whose graph is (P). Given that (C) intersects (P) at 3 points with x-coordinates in the interval [-1,5]. When a is minimum, what is the value of the product ab?

So i have to admit, i had to use desmos to visualise this a bit even tho this is a multiple choice question which implies that i shouldnt need any visualisation here

tho so far, what i have come up with is that for (C) to intersects (P) three times, a has to be larger than 15 (and with an appropriate b value that i havent found the relation from a to b yet)

which is weird, because if a is larger than 15, then there would be no minimum for a?

(when a=15 and b=25, the two graphs intersects each other at x=-1 and x=5)

(also, when a=15 and b=25, (C) and (P) shares a common tangent at x=5 which has slope 3)

and thats all i can conclude so far about the problem. I think i will need some guide for this because it seems like i might have missed something or i could have went down on a wrong path

https://www.desmos.com/calculator/yx5bvcdccq oh and heres the graph for the visualisation if it helps add any context as to how i have found the a=15 information

[boreal marlin]

how do i find a domain?

<@&286206848099549185>

or like whats a domain?

[restive hound]

I know that domain is x-values

[boreal marlin]

idk how to do this

[warm forum]

if $f(x)=\pm\sqrt{g(x)}\
\text{The domain is the interval of x which satisfies }g(x)\geq0\
\
\text{if }f(x)=\frac{k}{g(x)},k\in\mathbb{R}\
\text{The domain is the interval of x which satisfies }g(x)\neq0$

[obsidian monolith]

Muzan Jackson

[boreal marlin]

huhhhh??

[sick steppe]

what?

[warm forum]

wha-

did I do something wrong

[sick steppe]

no I was asking Normie.. cause their question was pointless

[sick steppe]

.

[warm forum]

lol okay

[boreal marlin]

no coz i legit dont understand

[warm forum]

this problem seems fun to solve

[trim hemlock]

it sure is, been pulling my hairs off over it for 4 hours

[sick steppe]

so you ask meaningful questions....

not "huhhhh?"

[boreal marlin]

that usually mean "wdym?"

[sick steppe]

I mean if you're not going to freely convey what you're struggling with, hard to help.

Domain was explained, if you dont understand something in the explanation you ask something. You wouldn't in a classroom setting go "huhhhh?" and have the educator know automatically what exactly you're struggling with

[boreal marlin]

ok so im struggling with understanding complicated symbols

symbols im not familiar with...

[sick steppe]

what in specific

cause literally everything here is symbols, the words you're typing are made of symbols

[boreal marlin]

like where did the g come from?

[sick steppe]

it's just an example function

[boreal marlin]

ok so im focusing on the top one because it seems simpler than the one below it right

[sick steppe]

they're equally complex but ok

1st one just says you cant take an even root of a negative, 2nd says you cant divide by 0

[boreal marlin]

that makes it easier to understand...

can you give me an example of a domain?

or is it too much to ask?

[sick steppe]

sure.... what function?

[sick steppe]

ok well you cant divide by 0, so x!=0 and x-7!=0

so all real numbers except 0 and 7 work

$D_h={x\in\mathbb{R}|x\neq 0,7}$

[obsidian monolith]

Mosh

[boreal marlin]

how do you say that in english

[sick steppe]

.

[boreal marlin]

ohhhhh

wait so whats a domain

[sick steppe]

Set of all allowed inputs

[boreal marlin]

meaning?

[sick steppe]

... meaning the stuff you can input

cant really phrase it any other way then "allowed inputs"

[boreal marlin]

which means anything besides 0 and 7 right?

[sick steppe]

any real number except for 0 and 7, yes

[boreal marlin]

so if i ask myself what the domain of this function is...

my answer should be 3 and 2 right?

[sick steppe]

...

[boreal marlin]

or am i still missing something

[sick steppe]

have you like, read anything I've written?

[boreal marlin]

i read them

[sick steppe]

you know... the part where I said what the domain was

which was nowhere close to {2,3}

[boreal marlin]

sorry.. let me review real quick

wait isnt 3 and 2 not 0 or 7?

[sick steppe]

...

yes, 2 and 3 are in the domain

but there are infintely many more values in the domain

it's just R where you cant equal 0 or 7

2 works, 3 works, pi works, sqrt(78767) works

0 doesnt work

[boreal marlin]

whats R?

[sick steppe]

Real numbers

[boreal marlin]

does that mean R should not be equal to 0 and 7?

[sick steppe]

......

R is the set of real numbers

ie the number line

D_h is the set of x in the real numbers such that x isnt 0 or 7

[boreal marlin]

can you say the domain of this function and explain after? because in all honesty im not getting most of this...

[sick steppe]

...

I did

many times

[wise matrix]

I don't think they understand notation

[sick steppe]

I explained the notation.

[boreal marlin]

are you saying 0,7 is the domain?

[sick steppe]

no

[wise matrix]

No

Sry I will stop shaodwing you

Proceed

[sick steppe]

Im saying everything EXCEPT 0 and 7 are in the domain

and I genuinely cannot explain it any other way

[boreal marlin]

so meaning anything EXCEPT 0 and 7 is the answer?

[sick steppe]

yes.

[boreal marlin]

12345689?

[sick steppe]

Like I just said, and explained prior

that isnt 0 or 7, so yes..

[boreal marlin]

so x does not equal 0 or 7?

[sick steppe]

yes

which is why I wrote such

[boreal marlin]

so meaning X could be anything besides 0 and 7?

[sick steppe]

yes........

[boreal marlin]

so if i say x=2? am i not understanding anything?

[sick steppe]

x=2 is in the domain.

[wise matrix]

Explaining limits for range is gonna be a bit of a challenge lmao

[sick steppe]

cause 2 isnt 0 or 7

[boreal marlin]

you said a domain is a set of..?

inputs

[sick steppe]

yes..

I did in fact say the definition of domain

[boreal marlin]

if i say h(2)=3 over 2 + 2 over 2- 7 .... what would that make me?

so if change x to 2...

that would make me wrong

[sick steppe]

h(2)=11/10

since 2 is in the domain.. you get an output..

[boreal marlin]

so what are you saying the domain is?

[sick steppe]

.....

Im not answering that

[boreal marlin]

wait how did you get 11/10

[sick steppe]

I have answered that sufficiently

Im also not going through explaining adding fractions

[boreal marlin]

oh ok that makes more sense

x=2

?

[sick steppe]

Im unfortunately done trying to help, as I have explained domain many times and it feels like you're just not listening.

[viscid thistle]

could someone help with this here:
Find the range of values of k such that the equation a2 + 2a + k = 0 has real solutions.

At the moment I understand for there to be a real solution it needs to be greater or equal to 0

[sick steppe]

"it" you mean discriminant

[viscid thistle]

i think 1 >= k, but don't think this is right

[viscid thistle]

yes

[boreal marlin]

i wouldnt say you failed to help me learn tho... i will keep in mind everything you just said and try processing it...imma get a tutor for this thanks tho!

[viscid thistle]

I have done this:

since b^2 - 4ac is discriminant

a = 1
b = 2
c = k

b^2 - 4ac >= 0
2^2 - 4(1)(k) >= 0

4 - 4k >= 0 for real solutions
but not sure what to do afer this

[sick steppe]

yeah just solve the inequality

$4-4k\geq 0 \implies 4\geq 4k \implies 1\geq k$

[obsidian monolith]

Mosh

[viscid thistle]

ok ty 🙂

[viscid thistle]

Could someone explain to me what I'm supposed to do here or what the goal is? I'm new to calculus and this exercise just makes no sense to me. Perhaps it's the phrasing... What would be a possible answer for example (not the actual one but a possible one)

[sick steppe]

I mean, you can note that x=0+(1-0)x, and that this question aims to prove the generalization

however just do the cases

[viscid thistle]

I got that far pretty much, but I don't understand what b>a means or a < b means concretely. I literally have 0 examples of this in the course literature so it's basically rocket science to me until I have some kind of reference

[sick steppe]

you've never seen an inequality before..?

[viscid thistle]

I have but I don't understand what they want from me here. I'm basically a math idiot untill if seen 1 example then I can reproduce

I don't get the question

[sick steppe]

well when a=b, you have y=a or b

cause b-a=0

[viscid thistle]

so how do I attach that to the (0,1) coórds

[sick steppe]

it's an interval.. not a point

[viscid thistle]

or is that not the question

[sick steppe]

$b>a \ (b-a)>0 \ (b-a)>(b-a)x>0 \ (b-a)+a>a+(b-a)x>a \ b>y>a$

[obsidian monolith]

Mosh

[sick steppe]

so ive shown y is between b and a if b>a

the 3rd line being $0<x<1 \implies 0<(b-a)x<b-a$

[obsidian monolith]

Mosh

[viscid thistle]

if b > a then b - a has to be > 0. roger that. But on the second line shouldn't you put the a + part in as well?

or what do you do there

[sick steppe]

ok so we assume b>a => b-a>0

now we know $x\in (0,1)\implies 0<x<1$

[viscid thistle]

well you can do that cuz its only real numbers right

[obsidian monolith]

Mosh

[sick steppe]

yes.. obviously we're working in a field with order like R

[viscid thistle]

hm hm. how is $x\in (0,1)\implies 0<x<1$

[obsidian monolith]

Dio

[sick steppe]

definition...

$x\in(c,d)\iff c<x<d$ is the definition of an interval of this form

[obsidian monolith]

Mosh

[viscid thistle]

why is that

[sick steppe]

.......

Not answering stupid questions, anyway

that's just how it's defined, google interval notation or check your notes if you dont know it

[viscid thistle]

okay then we roll with it

[sick steppe]

You literally cannot prove a definition

anyway $0<x<1\implies 0<(b-a)x<b-a$ since $b-a>0$

[obsidian monolith]

Mosh

[viscid thistle]

that feels weird

(b - a)x is always < b -a?

[sick steppe]

yes..

cause 0<x<1

[viscid thistle]

cause definition I suppose

[sick steppe]

no

cause of how inequalities work

b-a is pos so you can multiply through by it and maintain the direction

[viscid thistle]

okay

[sick steppe]

$0<(b-a)x<b-a\implies a<a+(b-a)x<b$

[obsidian monolith]

Mosh

[viscid thistle]

how do you know that a is on the left and b is on the right? is that part of the definition then

or is that cause b -a

[sick steppe]

...

I added a through

[viscid thistle]

ah right

okay understood

didnt know you could do that here.

[sick steppe]

You should review your inequality rules then..

[viscid thistle]

so what happens next after $a<a+(b-a)x<b$

[obsidian monolith]

Dio

[sick steppe]

...

you're done

a<y<b

you've shown $y\in (a,b)$

[obsidian monolith]

Mosh

[viscid thistle]

does that answer the question "what values does y run through"? solving the function and having

$y\in (a,b)$ as your answer

[obsidian monolith]

Dio

[sick steppe]

yes.....

y is always between a and b

[viscid thistle]

alright cool. like i said i literally never did this so what do I know 🙂 but okay thats a nice example then to try and solve the others myself

thanks.

[viscid thistle]

Okay so I'm trying a>b instead of b>a that you went through. So far so good:
a>b implies (a-b)>0
(a-b)>(b-a)x>0
but in the next step last time you added a through.
is that still logical to do? or did I already take a wrong turn

[bleak lance]

Why is that a “+6”?

And not a “-6”?

[viscid thistle]

If I’m looking at a polynomial function how can I determine the degree of the graph if it’s not given on the diagram

I’m working with 1

[sick steppe]

the number of turns made by the polynomial is at most 1 less than the degree

[viscid thistle]

Always?

[fervent stag]

no of times it changes direction...no of zero slopes + 1...thats the min degree possible

[viscid thistle]

Sorry if I sound stupid I just learned this today

So for example the first one would have a degree of 4?

[sick steppe]

yes

[viscid thistle]

For 3, I) I don’t get how you find the end behaviour based on the equation given

I think I need to dine out what quadrant but idk what that is from the equation

[restive hound]

Hello, how can I get the perfect square trinomial for the y^2 if there is no other term with y?

[kindred grove]

Hello this is revolving about the x axis

But my radius is “y”

How can I make it x?

I guess my r is just 3 since it’s revolving about the x axis. Am I correct?

[copper vigil]

what is the equation relating y and x

[torn acorn]

you would have to get two terms that have y in them that when combined, make 0y

[ashen otter]

Hello, need actual help here.

[ashen otter]

Let point P be (a, -a+1)
Let the points of tangency to y=-x^2 be (t, -t^2)
So the tangent is y = -2t(x-t)-t^2 = -2tx + t^2
Substitute point P to the tangent
-a+1 = -2at+t^2
t^2-2at+a-1=0

[ashen otter]

I manage to get this far, not sure what to do next...

[mild swan]

I'd start by drawing a graph to see where point P is

And see what the tangent line is like

[viscid thistle]

Am I correct in saying that the range is [-inf, 2.41, inf] here? (And the domain is Real numbers)

[mild swan]

You just need the interval where there are no domain strictions

This means what's under the √ must be nonegative:

7 - x^2 ≥ 0

Solving this for one or more interval(s) will give you your domain

Given your domain, what are the possible outputs for the square root? After that, how does your interval for the range change when every output is increased by 1?

@viscid thistle

[viscid thistle]

I feel like i did the kindergarten solution by just looking what 2,3,4 and -2, -3 and -4 would bring and make a conclusive answer out of that, when the answer would probably be something closer to 2.xx or 3.xx, but how do I determine this? only thing i can come up with judging by your tips is x = square root of 7 then

@mild swan

[mild swan]

Your domain will just be the solution set to 7 - x^2 ≥ 0, not necessarily equal to

[viscid thistle]

so what are you saying, 7 - x^2 ≥ 0 is the full range?

[mild swan]

domain

[viscid thistle]

oh fck

im a noob here. my bad, but could you explain what the domain then does for the range?

[mild swan]

Once you figure out what possible inputs you can put into the function, you can then consider how the interval of inputs can affect what outputs there are

[viscid thistle]

okay but like, there are still millions of numbers to put there then. i just went with 1, 2, 3,4 because it would be close to 7 if squared.

how do I retrieve what input I want to try out other then just what I did above

[mild swan]

Just solve for the domain first. Don't worry about the range before the domain.

[viscid thistle]

okay I thought the domain was just in general a real number here. So the domain is the solution set to 7 - x^2 ≥ 0. English is not my first language, what am I supposed to retrieve if a solution is set to a formula

in other words what does set to mean here

[restive hound]

Thank you 😃

[mild swan]

set isn't a special word in this case - it just means a group of numbers that satisfy a certain condition

In this case, the condition is that 7 - x^2 ≥ 0