Results 1 to 5 of 5

Thread: HELP!!! With graphs (It might be hyberbolic--unsure!)

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    7

    HELP!!! With graphs (It might be hyberbolic--unsure!)

    I need to know how to sketch graphs of the form y= 1/ x+c:

    (i) y= 1/ x+2
    (ii) y= 1/ x-4
    (iii) y= 1/ x+1

    Please help!! I'm clueless. I think they are hyberbolas in the xy=k thing but i cant draw or work it out. Pls show the steps

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    I suppose you know how $\displaystyle f(x)=\frac{1}{x}$ looks like

    The cases you posted are equal to this one in shape, mind you they have another position (they are horizontal translations).

    The function $\displaystyle f(x)=\frac{1}{x+c}$ has a vertical asymptote (the only one) in $\displaystyle x=-c$, so what you have to do is to move (horizontally) $\displaystyle \frac{1}{x}$ so that it has its aymptote there

    Remember that they have horizontal asymptotes on both sides

    Attached Thumbnails Attached Thumbnails HELP!!! With graphs (It might be hyberbolic--unsure!)-graph-1x.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    7
    Quote Originally Posted by PaulRS View Post
    I suppose you know how $\displaystyle f(x)=\frac{1}{x}$ looks like

    The cases you posted are equal to this one in shape, mind you they have another position (they are horizontal translations).

    The function $\displaystyle f(x)=\frac{1}{x+c}$ has a vertical asymptote (the only one) in $\displaystyle x=-c$, so what you have to do is to move (horizontally) $\displaystyle \frac{1}{x}$ so that it has its aymptote there

    Remember that they have horizontal asymptotes on both sides

    But how come the x+3 is situated on top of the x-intercept 2?? and what if it is y= 1/ x-4

    the minus sign.....
    im so sleepy
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2007
    Posts
    7
    PLS HELP!!!!!!!
    i dunno./.............

    im gonna b in big truoouble...omg its so late..........
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Let's go over the cases.

    First $\displaystyle f(x)=\frac{1}{x+2}$

    It has a vertical asymptote in $\displaystyle x=-2$ (what would $\displaystyle f(-2)$ be? )

    Apart from that, it has the same shape the ones in the other post have.

    Now $\displaystyle g(x)=\frac{1}{x-4}$

    The vertical asymptote is $\displaystyle x=4$(what would $\displaystyle f(4)$ be? )


    Finally $\displaystyle h(x)=\frac{1}{x+1}$

    Asymptote $\displaystyle x=-1$

    The important thing is that you realise that: $\displaystyle f(x)=g(x+6)$ for example
    Just for you to see$\displaystyle f(1)=\frac{1}{1+2}$ and $\displaystyle g(7)=\frac{1}{7-4}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. unsure with the terms
    Posted in the Statistics Forum
    Replies: 4
    Last Post: Feb 14th 2010, 02:58 AM
  2. hyberbolic manipulations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 25th 2009, 10:31 PM
  3. unsure? + hyperbola!
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: Nov 23rd 2006, 03:19 AM
  4. unsure
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Feb 2nd 2006, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum