Hello, uniquereason81!
Sketch the grah of a continuous function $\displaystyle f(x)$
that satisfies all of the stated conditions.
$\displaystyle f(0) \,=\,2;\;\;f(\text{}2) \,=\, f(2)\,=\,0\;\;{\color{blue}[1]}$
$\displaystyle f'(\text{}2) \,= \,f'(0)\,=\,f'(2)\:=\:0\;\;{\color{blue}[2]}$
$\displaystyle f'(x) \;=\;\bigg\{\begin{array}{cccc}> 0 & \text{if }\text{}2 < x <0 & {\color{blue}[3]}\\< 0 & \text{if }x<\text{}2\:\cup\: x>0 & {\color{blue}[4]}\end{array}$
$\displaystyle f''(x) \;=\; \bigg\{\begin{array}{cccc}> 0 & \text{if }x < \text{}1 \:\cup\:1 < x < 2 & {\color{blue}[5]}\\ < 0 & \text{if }\text{}1 < x < 1 \:\cup \:x > 2 & {\color{blue}[6]}\end{array}$
[1] gives us three points: .$\displaystyle (2,0),\0,2),\2,0)$
[2] tells that the tangents are horizontal at those three points Code:
2
=o=



 =o=    +    =o=  
2  2

[3] tells us that the curve is uphill on [2.0]
[4] says it is downhill everywhere else. Code:

*
/  \
\ /  \
\ /  \
 *    +    * 
2  2 \
 \

[5] tells us the curve is concave up to the left of 1, and on [1,2] Code:

* *
/  \
* /  *
* *  *
  * +  +  + * 
2 1  1 2\
 \

[6] says the curve is concave down everywhere else. Code:

* *
*  *
* *  *
* *  *
  * +  +  + *   
2 1  1 2 *
 *
 *

There!