Question

g(x)= x^3-2x^2+x/x^2-1
Horizontal asymptote, oblique asymptote, hole, domain, range is what I need to find I am having trouble. Will fan and medal.

Answer 1

u can try typing the function in the Wolfram web site...

Answer 2

it should be like this right:

g(x)= (x^3-2x^2+x)/(x^2-1)

Answer 3

There it is

Answer 4

hahahaaa u trying 2 confused me? @Maddy1251 :P

Answer 5

No :p

Answer 6

Do you know how to find oblique, horizontal and vertical asymptotes?

Answer 7

factor and cancel as a first step

Answer 8

For the vertical asymptotes, set the denominator to 0 (x²-1). For the oblique asymptote, effectuate long division until you find the equation of a line.

Answer 9

\[\frac{x(x-1)(x-1)}{(x+1)(x-1)}\]
then cancel the common factor of \(x-1\) top and bottom

Answer 10

since you are cancelling the \(x-1\) term there is "hole" at \(x=1\)

Answer 11

there is no vertical asymptote however

Answer 12

once you get to \[\frac{x(x-1)}{x+1}\] you find the "oblique" asymptote by division
you get a quotient and a remainder when you divide
ignore the remainder, the quotient is the "oblique" asymptote

Answer 13

there is also a vertical asymptote
set \(x+1=0\) get \(x=-1\)
the vertical line \(x=-1\) is the vertical asymptote

Answer 14

@satellite73, rewriting it wouldn't work only for calculating limits? Are you sure there is not an asymptote at x=-1 and an oblique asymptote y=x-2?

Answer 15

ok i lied, there is a vertical asymptote at \(x=-1\)
there is no horizontal one

Answer 16

@satellite73 I thought the hole was at -1?

Answer 17

@ChillOut I thought the same

Answer 18

@Maddy1251, rewriting it like that only works for calculating limits. x=-1 is a vertical asymptote.

Answer 19

There are no holes whatsoever in this function. But there is an oblique asymptote (which in turn makes horizontal asymptotes "impossible" to exist).

Just remember that whenever the numerator is one degree higher than the denominator there will always be a slant asymptote!

Answer 20

@ChillOut So, the hole was found how exactly?

Answer 21

@ChillOut the cubed x makes it slant because the x squared on the bottom, yes?

Answer 22

And how isn't there a hole for the graph?

Answer 23

Generally speaking, holes exist when you get 0/0.

Answer 24

@Maddy1251, yes. You have a cubic function on top of a squared one. So, yeah, you get a slant asymptote.

Answer 25

x=-1 would produce that.

Answer 26

Gotcha, that makes sense.

Answer 27

Remember when you rewrite a function to calculate a limit and that limit exists? That's how you find a hole.

Answer 28

Let's say, f(x)=\(\begin{cases} x+1\:\: if\:\: x≠ \: 5 \\ 10\:\: if\:\:x=5 \end{cases}\) There is a hole at x=5

Answer 29

@ChillOut I just did the math and you are right, my bad terribly. Lol. 1 cubed = -1, then its -1-2=-3 then -3+(-1)=-2/ then the bottom would equal 0, soooo -2/0 is undefined, so yeah you're right. That's how I checked. The limited function I kinda know.

I see.

Answer 30

Any questions?

Answer 31

So the horizontal does exist because of the oblique?

Answer 32

Doesn't*.

Answer 33

Thats what I meant. I can only type with one hand, I was in a riding accident lol, but okay. I think I may have this. Satelite factoring really helped actually, but you are very smart in this. Your help was really really useful

Answer 34

No problem, :P. Glad to help!

Answer 35

Sorry I couldnt medal, @superdavesuper got you I think. I fanned @ChillOut

Answer 36

I'm not here for medals. Do not worry about that :).

Answer 37

You're awesome!! Thank youu :)