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

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Discussion Overview

The discussion revolves around finding the asymptotes of the function f(x) = x/sqrt(4x-1). Participants explore the definitions and calculations related to vertical and horizontal asymptotes, including the implications of the square root in the function.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant expresses uncertainty about finding the vertical asymptote, suggesting that setting the denominator to zero (4x - 1 = 0) leads to x = 1/4, but questions whether the square root affects this.
  • Another participant confirms that the vertical asymptote is at x = 1/4 based on definitions.
  • For the horizontal asymptote, one participant describes the process of taking the limit as x approaches infinity and dividing by the highest power of x, leading to a proposed horizontal asymptote of y = 1/4.
  • A later reply suggests that the horizontal asymptote is actually y = 0, indicating a different interpretation of the limit process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the horizontal asymptote, as one suggests it is y = 1/4 while another claims it is y = 0. The vertical asymptote at x = 1/4 appears to be more consistently accepted.

Contextual Notes

The discussion includes uncertainty regarding the influence of the square root on the vertical asymptote and the calculations for the horizontal asymptote, with differing interpretations of the limit process.

m0286
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Hello,
Well I am 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! :smile:
 
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What is an asymptote?
(I know; but show that you have thought a bit upon the problem)
 
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 that's 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 I am 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!
 
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
 

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