Results 1 to 3 of 3

Math Help - Graphing relations

  1. #1
    Junior Member
    Joined
    Dec 2008
    Posts
    31

    Graphing relations

    I am really confused about this question, some clarity would be amazing.

    Graph the relation defined by the parametric equations when -2 ≤ t ≤ 3.
    x(t)=t^2-5
    y(t)=t-1

    Well a point/tips in the right direction would be great, I am completely clueless on where to start with this, thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    If you want to graph, there are two ways:

    (1) Assume arbitrary some values of t. Plot x and y corresponding to them.
    e.g. if t=1, x=-4 and y=0

    (2) The second method is to express y in terms of x..or x in terms of y.

    Spoiler:

    x=t^2-5
    y=t-1........or t=y+1

    x=y^2+2y+1-5=y^2+2y-4...either use this or

    y=\sqrt{x+5}-1

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644
    Hello, Murphie!

    You can simply plot-the-points . . .


    Graph the relation defined by the parametric equations when: . -2 \leq t \leq 3

    . . \begin{array}{ccc}x(t)&=& t^2-5 \\ y(t)&=&t-1 \end{array}

    Let t \:=\:\text{-}2,\text{-}1,0,1,2,3
    . . and we have some of the points of the graph.

    . . \begin{array}{|c||c|c||c|}\hline t & x & y & \text{Point}\\ \hline<br />
 \text{-}2 & \text{-}1 & \text{-}3 & (\text{-}1,\text{-}3)\\<br />
 \text{-}1 & \text{-}4 & \text{-}2 & (\text{-}4,\text{-}2)\\<br />
 0 & \text{-}5 & \text{-}1 & (\text{-}5,\text{-}1)\\<br />
 1 & \text{-}4 & 0 & (\text{-}4,0) \\<br />
 2 & \text{-}1 & 1 & (\text{-}1,1)\\<br />
 3 & 4 & 2 & (4,2) \\<br />
 \hline \end{array}



    Code:
                      |
                      |       *
                    * |
          - - * - - - + - - - - - - -
            *         |
              *       |
                    * |
                      |

    We have a portion of the parabola: . (y + 1)^2 \:=\:x+5

    Its vertex is (-5, -1) and it opens to the right.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relations and Functions - Inverse Relations Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 13th 2011, 12:20 PM
  2. Replies: 1
    Last Post: September 19th 2011, 01:09 PM
  3. Relations Help
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: May 2nd 2009, 06:56 AM
  4. relations
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: April 22nd 2009, 04:55 AM
  5. Relations
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: December 1st 2007, 06:25 PM

Search Tags


/mathhelpforum @mathhelpforum