# Thread: Identify any vertical, horizontal, or slant asymptotes

1. ## Identify any vertical, horizontal, or slant asymptotes

Identify any vertical, horizontal, or slant asymptotes

b) f(x)=x³-4x²-6 / 2x²-2

b) Vertical: Take the denominator and simplify
2x^(2)-2 ---> 2x^(2)-2=0 ----> 2x^(2)=2 ---> x^(2)=1 ----> x=+/-1 (is this correct or should it be +/- square root of 1?)

The exponent 3>2 so there are no horizontal asymptotes

oblique is= x/2-2 ???? How do I get oblique?

2. Hello, lovetolearn!

Identify any vertical, horizontal, or slant asymptotes

. . (b) .f(x) .= .(x3 - 4x2 - 6)/(2x2 - 2)

Vertical: Take the denominator and simplify
. . 2x2 - 2 .= .0 . . . . 2x2 = 2 . . . . x2 = 1 . . . . x = +1

The exponent 3 > 2, so there are no horizontal asymptotes

oblique is = x/2 - 2 ? . . . How do I get oblique?

Long division.

(x3 - 4x2 - 6 ) divided by (2x2 - 2) .= .x/2 - 2 . . . remainder x - 10

We have:
. . . . . . . x3 - 4x2 - 6 . . . . .x . . . . . .. x - 10
f(x) . = . -------------- . = . -- .- .2 .+ .--------
. . . . . . . . . 2x2 - 2 . . . . . .2 . . .. . . . 2x2 - 2

As x → ∞, the second fraction → 0.

And the graph approaches: .y .= .x/2 - 2

3. Thank you so much! I understand the concept now!