SOLUTION: What would be the equations of the vertical asymptote(s) and horizontal asymptotes of these functions? 1. y = (x+4)/(x^2+1) 2. y = (x^2-x-6)/(x^3-x^2+x-6) 3. y = 2x^3 / (x^3-1

Algebra ->  Rational-functions -> SOLUTION: What would be the equations of the vertical asymptote(s) and horizontal asymptotes of these functions? 1. y = (x+4)/(x^2+1) 2. y = (x^2-x-6)/(x^3-x^2+x-6) 3. y = 2x^3 / (x^3-1      Log On


   

Question 639554: What would be the equations of the vertical asymptote(s) and horizontal asymptotes of these functions?
1. y = (x+4)/(x^2+1)
2. y = (x^2-x-6)/(x^3-x^2+x-6)
3. y = 2x^3 / (x^3-1)
4. y = root(x) / (2x^2 - 10)
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
What would be the equations of the vertical asymptote(s) and horizontal asymptotes of these functions?
HA=horizontal asymptote
VA=Vertical asymptote
..
1. y = (x+4)/(x^2+1)
HA:x-axis or y=0 (degree of top less than degree of bottom)
VA:none (bottom cannot become zero because of x^2)
..
2. y = (x^2-x-6)/(x^3-x^2+x-6)
HA:x-axis or y=0 (degree of top less than degree of bottom)
VA: Set denominator=0, then solve for x
x^3-x^2+x-6=0
use rational roots theorem to find zeros
...0...|.....1.....-1.......1.......-6
...1...|.....1......0........1.......-5
...2...|.....1......1........3........0 (2 is a zero)
VA: x=2
..
3. y = 2x^3 / (x^3-1)
HA: y=2 (degree of top= degree of bottom), (divide leading coefficient of top by leading coefficient of bottom=2/1=2)
VA: (x^3-1)=(x-1)(x^2+x+1) (difference of cubes)
VA: x=1
..
4. y = root(x) / (2x^2 - 10)
y=x^(1/2)/(2x^2-10)
HA:x-axis or y=0 (degree of top less than degree of bottom)
domain: x≥0
VA: 2x^2-10=0
x^2=10/2=5
x=-√5 (reject, not in domain)
or
x=√5
VA: x=√5