Find the asymptotes of f(x)= x/square root(4x-1)

1. May 19, 2005

m0286

Hello,
Well Im 1 review question away from completing my calculus independent learning course WOOO! But I am stuck... can some one PLEASE HELP

Find the asymptotes of f(x)= x/square root(4x-1)
THANKS SO MUCH YOU GUYS HAVE BEEN AWESOME!

2. May 19, 2005

arildno

What is an asymptote?
(I know; but show that you have thought a bit upon the problem)

3. May 19, 2005

m0286

sorry, i showed my work now.

Sorry I know what an asymptote is, i just didnt want to type it all out sorry I guess i should have..
I know a vertical asymptote is a vertical line on the graph that the function comes closer and closer to but never touched, and horizontal is the same thing but a horizontal line.

I know for vertical asymptotes if its written in the form y=f(x) you can make the denominator =0 and figure it out from there.. and I understand doing that however not really witha square root???? Would it just be 4x-1=0, x=1/4??? and thats the vertical asymptote, or does the square root play more in that????

For horizontal asymptotes, i know you make it so
lim x/sqare root(4x-1)
x->infinity
Then you divide all parts by the highest power of x. so i know x/x is 1 so it would be 1/square root(4x-1) This im not sure about now.. how do i divide square root (4x-1) by x??? does the square root matter or would it be 4x/x-1/x which is 1/4-0, which is 1/4 as the horizonatl asymptote???

THANKS!!

4. May 19, 2005

heman

you are thinking right i guess....
For horizontal asymptote it will be y=0and for vertical it will be x=1/4 as per definitions