#I need help here
86 messages · Page 1 of 1 (latest)
Hi @hushed fiber so, you have any ideas
a good place to start would be to think about what those vertical asymptotes mean, equation wise
Sorry, we won't give the answer
Well idk what to do or how to understand it!
It has a video!
And I can't understand the video!
SO WHAT THE HELL IS GOING ON!?!
😡
Can you show the video
Everything
Look at the graph?
I'm looking at the graph!
Does it touch the x-axis anywhere
positive 2?
Numerator?
negative 2?
Negative 2x?
No?
If x=2 results in 0, then it means....
x=2?
You're writing y as some rational function of x
if x=2, and you get 0, it means you're multiplying by 0
How do you multiply by 0, when x=2?
It's just 0 right?
THIS IS FUCKING STUPID!
This is why I hate working on this because the process is frustrating and annoying!
Sorry..
I don't see how it is, the video is 5 minutes long
I'll tell you the solution for it cutting the point at 2
Cause it's being solved by an expert!
It results in
[y = \frac{(x-2)^2}{\cdots}]
fishwhale
x-2=0 when x=2
We need an even power (strictly greater than 0), because it does not change sign as it goes towards 2 and after it hits 2
You basically keep using the information of it cutting axes or having vertical asymptotes to figure out how it looks like
also whats the bottom part?
x takes values from negative infinity to positive infinity
the bottom part is filler - I am not going to tell you the answer
when x takes the value 2, it just minuses off, and regardless of whatever other x components, it does not matter. Multiplying by 0 results in 0
So its (x-2) squared and 0?
no
It's (x-2) squared, multiplied by other parts
There's a constant to be multiplied
There's also the vertical-asymptotes to be multiplied
it's one big product of (x...) and a constant term
Which is what the video does
This is why it works
You are just using the fact of things being or not being 0 at times you want