Question

Asymptotes, removable discontinuities, and intercepts
f(x)=2x^3-3x+1/x^3-5x+7
would someone mind helping me and walking me through how to solve this and checking answers and stuff?

I know that to find the removable discontinuities, I have to factor the top, but I'm not sure how to do that with the 2 in front of the x...... I don't think yhu can even factor the top...
for the asymptote, I would do long division with the denominator, and then set it to 0 I think, right? (I don't know how to find the horizontal asymptote)
The intercepts are kind of easy.

Answer 1

\[f(x)=\frac{ 2x^{3}-3x+1 }{ x^{3}-5x+7 }\]

Answer 2

you could try and plug it in to this http://web2.0calc.com/ i use this a lot its very helpul

Answer 3

i havent done math in the longest time so i honestly wouldnt know how to do this

Answer 4

I'm not 100% sure how to work this out tho. Like, the steps haha.
I'll keep that link cuz I'm sure it'll come in handy in the future. c: Thank yhu for that.
I'm just missing a few steps and I'm not totally sure on how to work it out.
But thank yhu very much for that reference and at least looking at this c:

Answer 5

thanks now i feel useful xD

Answer 6

xD No problem~

Answer 7

to factor the top separate the numerator into x-1(2x^2+2x-1)

Answer 8

Can yhu explain how yhu got it like that?

Answer 9

oh trial and error, sometimes one root out of three is obvious in a cubic equation, (here x=1), and as you know you can write any cubic equation as
(x-rootno1)(x-rootno2(x-rootno3)
so if you know one root you can factorise a cubic equation

Answer 10

So then... I don't have any removable discontinuities do I?

Answer 11

actually, discontinuity, the domain, and asymptotes, are calculated before factoring and simplifying the function.

Answer 12

Can yhu help me work this out. ;-;