Sketching a graph help

[nickp657]

Hi guys I have this set of parameters

Domain is (-00,00)
The only intercept x-intercept is x=0
Lim x->00 f(x)=-00 lim x->-00 f(x)=0
f(x) is negative for x>0 and f(x) is positive for x<0 Moreover f(1)=-9 and f(-2)=2 and f(-1)=3
f'(x) is 0 only at x = -1 Moreover, f'(-4) positive while f'(4) is negative
f''(x) is 0 only at x=-2 Moreover, f''(-6) is positive while f''(4) is negative

Now I know how the graph goes but once I hit (1,-9) I am confused, also the other problem is I can't tell if this graph has any asymptoes or local mins. Any help would be great thanks.

 

[mmm4444bot]

Domain [of function f] is (-00,00)

Lim x->00 f(x)=-00 lim x->-00 f(x)=0

f`(x) is 0 only at x = -1

:idea: These three lines provide key information about asymptotes and extrema :idea:



I can't tell if this graph has any asymptotes or local mins Here are some questions for you to consider.

Regarding local minimums or maximums [extrema], do you know what it means graphically when a function's first derivative equals zero ?

Regarding horizontal asymptotes, if there were any, what would they do to the function's global behavior ?

Regarding vertical asymptotes, if there were any, what would they do to the domain of the function ?



I know how the graph goes ? I do not know what this statement means.



once I hit (1,-9) I am confused Can you be more specific ?

 

[soroban]

Hello, nickp657!

There is a lot of information in thse statements. . .

\(\displaystyle \text{Given a function }f(x):\)

\(\displaystyle \text{Domain is: }\-\infty,\:\infty)\)
The graph exists for all values of x.

\(\displaystyle \text{The only }x\text{-intercept is: }\:x = 0\)
The graph crosses the x-axis only at the Origin.

\(\displaystyle \lim_{x\to\infty} f(x) \:=\:-\infty\)
To the far right, the graph falls lower and lower . . .
\(\displaystyle \lim_{x\to\text{-}\infty} \:=\:0\)
To the far left, the graph approaches the x-axis.

\(\displaystyle f(x) < 0 \text{ for }x>0\)
On the right, the graph is below the x-axis.
\(\displaystyle f(x) > 0 \text{ for }x<0\)
On the left, the graph is above the x-axis.
\(\displaystyle f(1 )=\text{-}9,\;f(\text{-}2)=2,\; f(\text{-}1)=3\)
We are given three points on the graph.

\(\displaystyle f'(x) = 0\,\text{ only at }\,x = \text{-}1\)
The only horizontal tangent is at (-1,3).
\(\displaystyle f'(\text{-}4) > 0\)
At x = -4, the graph is increasing.
\(\displaystyle f'(4) < 0\)
At x = 4, the graph is decreasing.

\(\displaystyle f''(x) = 0\,\text{ only at }\,x=\text{-}2\)
The only inflection point is (-2,2).
\(\displaystyle f''(\text{-}6) > 0\)
At x = -6, the graph is concave up.
\(\displaystyle f''(4) < 0\)
At x = 4, the graph is concave down.

I believe the graph looks like this . . .

                   (-1,3)             |
                      o               |
         (-2,2) *           *         |
             o                  *     |
          *                        *  |
     *                               *|
  - - - - - - - - - - - - - - - - - - o - - - - -
                                      |*
                                      |
                                      | * 
                                      |
                                      |
                                      | -*

*
[/size]