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

1. 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!

2. 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       |
*    |
*  |
* |
|
*|
|

3. Originally Posted by Ahe
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!