Results 1 to 3 of 3

Math Help - Sketching a graph using given characteristics...I'm stumped

  1. #1
    Ahe
    Ahe is offline
    Newbie
    Joined
    Oct 2008
    Posts
    4

    Sketching a graph using given characteristics...I'm stumped

    Ok so maybe this is easy, maybe it's not. At least for me it seems impossible; I'm absolutely horrible with graphs. But here's the problem. Help please?

    Sketch a graph with the following characteristics: f(-2),f'(-2), and f''(-2) are all negative; f'(0) does not exist; f(x) has a local maximum at the point (2,3); there is exactly one point of inflection on the interval (0,2); and the limit of f(x) as x goes to infinity is equal to -1

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,550
    Thanks
    542
    Hello, Ahe!

    Just "talk" your way through it . . .


    Sketch a graph with the following characteristics:

    (1) f(\text{-}2),\;f'(\text{-}2),\text{ and }f''(\text{-}2) are all negative.

    (2) f'(0) does not exist.

    (3) f(x) has a local maximum at (2,3).

    (4) There is exactly one point of inflection on the interval (0,2).

    (5) \lim_{x\to\infty} f(x) \:=\:\text{-}1

    (1) f(\text{-}2) is negative. .At x = \text{-}2 the graph is below the x-axis.

    . . . f'(-2) is negative. .The graph is decreasing.

    . . . f''(-2) is negative. .The graph is concave down.

    It is shaped like this:
    Code:
                -2
       - - - - - + - - - - -
          *      :
              *  :
                 o
                   *
                    *

    (2) f'(0) does not exist.
    . . .There is a vertical asymptote at x=0.


    (3) There is a local maximum at (2,3)


    (4) There is one inflection point on (0,2).
    Since the curve is concave down at the maximum (2,3),
    . . it was concave up to the left of the inflection point.
    It looks something like this:
    Code:
            |            (2,3)
            |*            o
            |          *  :  *
            | *      o    :      *
            |  *    *     :
            |    *        :
          --+-------------+--------
            0             2

    (5) There is a horizontal asymptote: . y \:=\:\text{-}1


    I would guess that the graph looks like this:
    Code:
                       |
                       |*             o
                       |           *  :  *
                       | *      o     :    *
                       |  *    *:     :      *
                       |    *   :     :        *
              -2       |        :     :           *
      ---------+-------+--------+-----+--------------*---------
               :       |        1     2                    *
     - - - - - : - - - + - - - - - - - - - - - - - - - - - - -
      *        :       |                    y = -1
           *   :       |
               o       |
                  *    |
                    *  |
                     * |
                       |
                      *|
                       |
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,992
    Thanks
    1129
    Quote Originally Posted by Ahe View Post
    Ok so maybe this is easy, maybe it's not. At least for me it seems impossible; I'm absolutely horrible with graphs. But here's the problem. Help please?

    Sketch a graph with the following characteristics: f(-2),f'(-2), and f''(-2) are all negative; f'(0) does not exist; f(x) has a local maximum at the point (2,3); there is exactly one point of inflection on the interval (0,2); and the limit of f(x) as x goes to infinity is equal to -1

    Thanks!
    Do you understand that the problem says "a" graph because there are an infinite number of such graphs.

    f(-2) is negative: mark some point (-2, y) with y any negative number you like. f'(-2) is negative: draw a short line segment with negative slope (upper left to lower right) at that point. f"(-2) is negative: draw a little arc at that point, tangent to the line segment, curving slightly downward.

    f'(0) does not exist: make a "corner", like |x|, at x=0.

    f(x) has a local maximum at the point (2, 3). Mark the point (2,3) and draw a small arc with vertex there, opening downward.

    There is exactly one inflection point on the interval [0, 2]. Since you already have that the graph is 'concave downward' (curving downward) at (2,3) draw an arc at some point (0, y) with y between 3 and the "y" at f(-2) curving upward and connect those two arcs.

    The limit of f(x), as x goes to infinity, is -1. Draw a horizontal line at y= -1 to the right of your coordinate systems. Draw a graph getting closer and closer to that as you go to the right.

    Finally, connect all of those pieces together in any way you like!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Graph Theory Question - Totally stumped
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: May 12th 2010, 10:38 PM
  2. Sketching the graph
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 2nd 2009, 09:36 AM
  3. Characteristics of the graph
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 15th 2009, 02:16 AM
  4. Graph sketching
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 18th 2009, 06:08 AM
  5. Sketching sin graph
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 9th 2008, 05:46 AM

Search Tags


/mathhelpforum @mathhelpforum