Results 1 to 4 of 4

Math Help - solving a function

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    142

    solving a function

    Here is a rational function:

     f(x)= \frac {3x^2}{4-x^2}

    Find:
    domain: (- infinity, infinity)
    Range: (- infinity, -2)U(-2,2)U(2, infinity)
    x-intercept:
    y-intercept: none
    roots:
    Asymptotes:

    For y-intercept:
    Let x=0
     f(0)= \frac {3(0)^2}{4-0^2}
     f(x)= 0 so, Y=0 (No Y-intercept)

    For x-intercept:

     f(x)= \frac {3x^2}{4-x^2}=0
    I factor the denominator:
     f(x)= \frac {3x^2}{(x-2)(x+2)}


    What is next ? >.<
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2010
    Posts
    20
    there is actually an intercept at the point (0,0)
    as u said:




    therefore f(0)= 0/4 =0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    142
    Quote Originally Posted by chrizzle View Post
    there is actually an intercept at the point (0,0)
    as u said:




    therefore f(0)= 0/4 =0

    I found the Vertical Asymptotes X= 2 and X= -2

    Horizontal Asymptotes is y=3
    thanks!
    Last edited by Anemori; March 17th 2010 at 12:00 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,780
    Thanks
    1521
    Quote Originally Posted by Anemori View Post
    Here is a rational function:

     f(x)= \frac {3x^2}{4-x^2}

    Find:
    domain: (- infinity, infinity)
    Range: (- infinity, -2)U(-2,2)U(2, infinity)
    x-intercept:
    y-intercept: none
    roots:
    Asymptotes:

    For y-intercept:
    Let x=0
     f(0)= \frac {3(0)^2}{4-0^2}
     f(x)= 0 so, Y=0 (No Y-intercept)

    For x-intercept:

     f(x)= \frac {3x^2}{4-x^2}=0
    I factor the denominator:
     f(x)= \frac {3x^2}{(x-2)(x+2)}
    There is no point in factoring the denominator. A fraction, a/b, is equal to 0 if and only if the [b]numerator, a, is equal to 0. For this function, both x and y intercepts are at (0, 0).


    [/quote]What is next ? >.<[/QUOTE]
    Yes, there are vertical asymptotes at x= -2 and x= 2 but the horizontal asymptote is at y= -3, not y= 3.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving inverse function
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 19th 2010, 10:27 PM
  2. Trigonemetric Function - Solving
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: November 7th 2009, 06:46 PM
  3. Solving for x with a trig function
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 29th 2009, 11:41 PM
  4. Function Solving Help
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 12th 2009, 03:05 PM
  5. solving this for x (zero of the function)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 20th 2009, 11:12 AM

Search Tags


/mathhelpforum @mathhelpforum