1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Rational Expressions loooong BUT NEED HELP

  1. Dec 8, 2004 #1
    state the vertical asymptote(s), x-intercept(s), y-intercept(s), domain and range of f(x)=(1)/(4x^2)-1

    ok I factored the denominator and got (2x+1) (2x-1) I solved for x, so the x-intercepts are x=-1/2, and x=1/2 for only (4x^2)-1 the reciprocal function has no x-intercepts.

    Sub in x=0 to get y intercept, I got y=-1 for (4x^2)-1
    that means for f(x)=(1)/(4x^2)-1 it must be the same since -1 is an invariant point.

    Ok I think the vertex is (0,-1) for (4x^2)-1 this is also the vertex of f(x)=(1)/(4x^2)-1 but this function has a maximum instead of minimum.

    I gather the vertical asymptotes are x=-0.5, and x=0.5 for f(x)=(1)/(4x^2)-1

    The Range for f(x)=(1)/(4x^2)-1 {y:y does not equal 0 but is YER} if this is correct what is the proper notation.

    The Domain for f(x)=(1)/(4x^2)-1 {x:does not equal -1/2 or 1/2, but is XER}
    if this is correct what is the proper notation?

    Ok um I dont know if all that I have said is correct, can someone please Help me out?

    I have a sketch of the graph but the forum wont let me attach it it is saying it is too big....

    it sort of looks like JUL
    a parabola with a min is the original and then the reciprocal is a parabola with a max and in the top two quadrants a backwards L and L .
    Last edited: Dec 8, 2004
  2. jcsd
  3. Dec 9, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi Aisha,

    Is the function
    [tex]f(x)=\frac{1}{4x^2}-1[/tex] or [tex]f(x)=\frac{1}{4x^2-1}[/tex]?

    Your notation implied the former one, but then you cannot factor the denominator like [itex](2x-1)(2x+1)[/itex]. You can do so only in the latter one. I take it that's the one you meant.

    The notation you want for the range is :[tex]\mathbb{R}\setminus \{0\}[/tex], which means [itex]\mathbb{R}[/itex] minus the set [itex]\{0\}[/itex]. So minus the point 0.
    It's not correct though.
    The function looks more like [itex]J_\cap L[/tex] then JUL :rofl:
    Anyway, look for the extrema, you`ll find there's a local maximum at x=0.
    (0,-1), I guess you already found it. Never heard it being called a vertex though. That means the function is smaller than -1 in the interval (-1/2,1/2) and larger than 0 elsewhere in its domain.

    The domain is correct, a correct notation would be [itex]\mathbb{R}\setminus \{-\frac{1}{2},\frac{1}{2}\}[/itex]

    The asymptotes are ok too.
  4. Dec 11, 2004 #3
    Yes It Is The Second One!! I Still Need More Help Please!!! Anyone!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook