maths

[frozen peak]

i need help with “b”.
i know how to find asymptotes when the function is rational, but how do i find when the function is like this?
how do i find the vertical asymptote for instance (if there are any)?

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[sweet barn]

What value can u put into the function to make it go to infinity or negative infinity (for vertical asymptone)

@frozen peak

Or u can try to plug in infinity or negative infinity for y and solve for x

Same thing

[dusk ledge]

remember that when $x\leq 0$, that $\ln(x)$ is not defined

[craggy bay]

;( | 追放された興奮

[frozen peak]

x —> 2?

[frozen peak]

yesss

[sweet barn]

Yep

Try to solve for it by plugging positive and negative infinity for g(x) as well

It's good for learning experience

[frozen peak]

but this is something i can do to find the horizontal asymptote, right?

[sweet barn]

For horizontal put x to be infinity or negative infinity

[frozen peak]

but negative infinity doesn’t work

right

[sweet barn]

That means it doesn't exist

[frozen peak]

because ln is undefined when x is less than zero

[frozen peak]

ahh ok

[sweet barn]

No solution means no asymptote

[frozen peak]

okok

[sweet barn]

@frozen peak U need to do some interpretation as well

Like try putting positive infinity for y

It gives u x = infinity

That would not be a asymptote in this case

Since it interprets as as x goes to infinity y is infinity

.

As a general rule if both values seem to be infinity, it's not an asymptote

[frozen peak]

why for y?

[sweet barn]

Imagine a function u know which has asymptotes

[frozen peak]

wait which values?

[sweet barn]

For example 1/x

[sweet barn]

x and y

[frozen peak]

ok

aha ok

[sweet barn]

Vertical asymptote at x=0

Y is going to infinity and negative infinity

So putting y as infinity gives x which is vertical asymptote

.
Similar logic for horizontal one. Try to think about it

[frozen peak]

i seee

thank uu

[sweet barn]

Np you're welcome

[frozen peak]

but when i want to find the vertical asymptote for -->2+, should i then write x as 2.01?

cause i can not get zero

[frozen peak]

@glossy quarry

[sweet barn]

I think u have confusion on what a asymptote is

A asymptote exists when some gets close to something else but never meets it.
So by definition of the concept it occurs only when the thing approaching the other thing meet at infinity

In case of 1/x the curve meets the y axis (x = 0) only at infinity
It doesn't meet it finite distance but only gets close

The curve meets x axis (y=0) when x is infinity
It gets closer but just shy of meeting it at finite distance

[sweet barn]

•

U need to equate it with limit methods and it will evaluate to -infinity when x approaches 2+
U can't plug in value directly in this case.

[glossy quarry]

I have been summoned

[glossy quarry]

,rotate

[craggy bay]

Couldn't find an attached image in the last 10 messages.

Oct 20, 2024, 9:50 AM

[glossy quarry]

🗿

,rotate

[glossy quarry]

Lovely

Which point do you need help with @frozen peak ?

[sweet barn]

He was asking about how to find the vertical asymptote

[glossy quarry]

Ah

Alright

I am in the middle of something now, if you still need help in a few hours, ping me

[eternal raft]

@frozen peak

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[frozen peak]

+close

[eternal raft]

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