Results 1 to 4 of 4

Math Help - Graphing

  1. #1
    Newbie
    Joined
    Jul 2006
    Posts
    23

    Graphing

    Hello I would appreciate any help with the following problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The graph of \sqrt{x+5}-2

    Can you see what the domain is?.
    Last edited by galactus; November 24th 2008 at 05:39 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    quick overview: hopefully you know what this y=\sqrt{x} looks like,

    now when you see an equation like this y=\sqrt{x+5} you move the graph of y=\sqrt{x} left 5 units, if you see an equation like this y=\sqrt{x}-2 you move the graph of y=\sqrt{x}-2 down 2 units.

    f(x)=(x+2)(x-2)^2

    the x-intercepts happen when f(x)=0 therefore we have: 0=(x+2)(x-2)^2

    now anything multiplies by zero is zero, so only one of the two groups has to be zero for the entire equation to go, so we figure out when either of them are zero:

    \begin{array}{cc}x+2=0\Rightarrow x=-2\\x-2=0\Rightarrow x=2\end{array}

    so the y-intercepts are when x=\pm2
    Attached Thumbnails Attached Thumbnails Graphing-2-graphs.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,802
    Thanks
    691
    Hello, pashah!

    Sketch the graph and state the domain and range.

    . . y \;= \;\sqrt{x+5} - 2

    Since x + 5 must be nonnegative: . x \geq -5 (domain)

    The minimum value of y occurs when x = -5
    Hence, the range is: .  y \geq -2


    We have: . y + 2 \:=\:\sqrt{x+5}
    Square the equation: . (y+2)^2 \:=\:x + 5
    We have a parabola with vertex (-5,-2) which opens to the right: \subset
    And we want only the upper half.
    Code:
                      |
                      |      *
                      *
          - - - - * - + - - - - -
               *      |
             *        |
            *         |
         (-5,-2)      |
    Graph and identity all intercepts.

    . . y \;= \;(x+2)(x-2)^2

    If x = 0,\;y\:=\:2(-2)^2 \:=\:8 . . . y-intercept: (0,8)

    If y = 0,\;x\:=\:\pm2 . . . x-intercepts: (\pm2,0)


    This is a cubic graph which rises to the right.
    Since (x-2)^2 has an even exponent,
    . . the graph is tangent to the x-axis at x = 2
    Code:
                      |
                      * *         *
               -2  *  |   *     *
          - - - * - - + - - -*- - -
              *       |      2
             *        |
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. graphing
    Posted in the Algebra Forum
    Replies: 6
    Last Post: May 19th 2011, 05:43 AM
  2. Graphing
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: May 23rd 2009, 05:31 PM
  3. graphing of sin and cos
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 28th 2008, 09:50 AM
  4. Graphing
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: January 16th 2008, 11:18 AM
  5. help with graphing
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 9th 2007, 09:23 AM

Search Tags


/mathhelpforum @mathhelpforum