Question

G(x)=(x^4-1)/(x^2-x)

vertical asymptote is/are x=______
OR
There is none

Horizontal asymptote is/are y=_________
OR
There is none

The oblique asymptote is/are y=_____
OR
There is none

Answer 1

have you factored the numerator and denominator yet?

Answer 2

Yes I did

Answer 3

So, then we hopefully saw that something canceled out. I would note that as a hole in the graph (but not as an asymptote).

Answer 4

So what do you have left in the denom now?

Answer 5

would x is left in the denom

Answer 6

First factor the numerator and denominator for G(x) and cancel. When you factor you get:

G(x) = (x - 1)(x + 1)(x^2 + 1)
--------------------
x(x - 1)

Here the 'x - 1' cancel out so our G(x) becomes:

G(x) = (x + 1)(x^2 + 1) = x^3 + x^2 + x + 1
--------------- -----------------
x x

Note, the domain is always on the ORIGINAL function. The asymptotes, intercepts, and other things are based on the second G(x) once you have cancelled and simplified, but the domain is ALWAYS based on the ORIGINAL function, NOT the new simplified function.

Domain: x not= 0, 1 (Based on original function)

EVERYTHING BELOW IS ALWAYS BASED ON THE NEW SIMPLIFIED FUNCTION
--------------------------------------------------------------

Vertical asymptotes are where the denominator is 0.

x = 0 --> V.A: x = 0

Horizontal asymptotes occur when the degree of the numerator and denominator are the same or the degree of the denominator is greater than the numerator. Here the degree of the numerator is greater so we will have an oblique/parabolic asymptote. So no there is NO Horizontal asymptote.

To get the oblique asymptote, we divide the Numerator by the Denominator using long division.

x^3 + x^2 + x + 1
----------------- = x^2 + x + 1, Remainder = 0. Ignore the remainder.
x

So the oblique asymptote, which is actually a parabolic asymptote, has the equation:
y = x^2 + x + 1

@flutterflies

Answer 7

that is what I got too. So lets think about this for a minute. What happens to the graph as x gets REALLY close to zero? We know it can't BE zero because that would make it undefined, right?

Answer 8

Sorry I made a typo, in the line that says: "x^2 + x + 1, Remainder = 0. Ignore the remainder.", the Remainder = 1, not 0. Sorry about that.

@flutterflies