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

1 Answer
Jun 30, 2018

Answer:

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!