graphing a function f (please help if you can) ASAP

[thomcart8]

With the following info
f(x) = 0 at x = -5, x=0, and x = 5
lim f(x) = infinity, with x approaching negative infinity
lim f(x) = -3, with x approaching infinity
f'(x) = 0 at x = -3, x = 2.5, and x = 7

can anyone at least point me in the right direction on how to graph this problem?
I have gotten this far, I just need some help.

ok so i have the curve starting above the x axis going down and passing through -5 on the x axis then to -3 curving back to go through 0 then going up to positive 3 then curving back down through 5 then curving back up at -3 again.

 

[soroban]

Hello, thomcart8!

$\displaystyle \text{Sketch a graph of }f(x)\text{ where:}$

. . $\displaystyle [1]\;f(\text{-}5) \:=\: f(0) \:=\: f(5) \:=\: 0$

. . $\displaystyle [2]\;f'(\text{-}3) \:=\: f'(2.5) \:=\: f'(7) \:=\: 0$

. . $\displaystyle [3]\;\lim_{x\to\:\!\text{-}\,\!\infty} f(x) \:=\:\infty$

. . $\displaystyle [4]\;\lim_{x\to\infty} f(x) \:=\: -3$

$\displaystyle \text{[1] says the }x\text{-intercepts are: }\:x \:=\:-5,\:0,\:5$

$\displaystyle \text{[2] says there are horizontal tangents at: }\:x \:=\:-3,\:2.5,\:7$

$\displaystyle \text{[3] says that the graph rises infinitely to the far left.}$

$\displaystyle \text{[4] says that the graph approaches the horizontal line }y \,=\,-3\text{ to the far right.}$


$\displaystyle \text{This is sufficient information to make a rough sketch.}$

                      |
        *             |     o
                      |  *  :  *
         *     -3     |*    :    *5       7
      ----o-----+-----o-----+-----o-------+--------------
         -5*    :    *|    2.5            :
             *  :  *  |            *      :
                o   -3+ - - - - - - - - - : - - - - - - -
                      |             *     :         *
                      |              *    :     *
                      |               *   :   *
                      |                   o
                      |

 

[thomcart8]

thank you. we basically got the same thing. I am glad to see that.