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

Question

1552 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-30T19:49:06

HA at $y=1/2$

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

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!

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