Question

How do you find the asymptotes for `(2x^6 + 6x^3 )/( 4x^6 + 3x^3)`?

Answer 1

HA at `y=1/2`

VA at `x=0` and `x=root3(-3/4)`

Explanation:

When we want to find the horizontal asymptotes, we want to analyze the highest degrees on the top and bottom.

`(2x^6)/(4x^6)`

Since we have the same degree on the top and bottom, the horizontal asymptote just simplifies to the coefficients. We're left with

`2/4`, or `y=1/2`

For our vertical asymptote, we want to think about what value(s) make our function defined.

We can set the denominator equal to zero. We get

`4x^6+3x^3=0`

We can factor our an `x^3` to get

`x^3(4x^3+3)=0`

Setting both factors equal to zero, we get

`color(steelblue)(x=0)` and

`4x^3=-3`

`x^3=-3/4`

`color(steelblue)(x=root3(-3/4))`

These are our vertical asymptotes.

Hope this helps!