Question

Find the horizontal, vertical, and oblique asymptotes, if any, for the given rational function G(x)=(x^4-1)/(3x^2-3x)

Answer 1

\[
g(x)=\frac{x^4-1}{3x^2-3x}\\
g(x)=\frac{(x^2+1)(x^2-1)}{3x(x-1)}\\
g(x)=\frac{(x^2+1)(x+1)\cancel{(x-1)}}{3x\cancel{(x-1)}}
\]

Answer 2

we can cancel only if \(x\ne1\)

1) vertical asymptote:
the value of "x" for which the function becomes undefined.
this is a rational function. When the denominator becomes "0", we cannot solve it,
what value of "x" makes it 0?

2) horizontal asymptote:
rearrange the equation interms of "y"
and see if you will get a rational function or "x" becomes undefined for some value of "y"

Answer 3

So the vertical is x=0 and the horizontal is y = (1/3)(x^2 + x + 1) ?

Answer 4

3) oblique asymptote:
perform the polynomial division:
\[g(x)=\frac{(x^2+1)(x+1)}{3x}\]

Answer 5

how did you get the horizontal asymptote?
horizontal asymptote must be a straight line with slope = 0

Answer 6

What would the horizontal look like?

Answer 7

(1/3)(x^2 + x + 1) + 0 , which is a paraloba

Answer 8

@electrokid

Answer 9

horizontal means Flat. no curve

Answer 10

so, it should simply be:
y = a number

Answer 11

you still did not tell me how you found
\[{1\over 3}(x^2+x+1)\]

Answer 12

when x --> ∞
1/3x --> 0
so we left with
(1/3)(x^2 + x + 1) + 0 , which is a paraloba
by simplifying

Answer 13

so is the horizontal 1/3?

Answer 14

@electrokid

Answer 15

horizontal is y=0

Answer 16

that parabola is an oblique asymptote. you have to extract two lines from it.
and there is no horizontal asymptote

Answer 17

(1/3)(x^2 + x + 1) + 0 so this is the oblique?

Answer 18

yes

Answer 19

Oh ok

Answer 20

not this y = (1/3)(x^2 + x + 1) ?