Hello, littlejodo!
I'll do part of the first one . . .
Sketch the graph of a function $\displaystyle y = f(x)$
satisfying: .$\displaystyle \begin{Bmatrix}f(0)&=&1 \\
f '(0) &=& \text{}3 \\ f''(0) &=& 6 \\ f'''(0) &=& \text{}6 \end{Bmatrix}$ .in the vicinity of $\displaystyle x = 0.$
Examine what each statement tells us . . .
$\displaystyle f(0) = 1$
The graph has the point (0, 1)
$\displaystyle f'(0) = \text{}3$
The slope is negative; the function is decreasing: $\displaystyle \searrow$
$\displaystyle f''(0) = +6$
The graph is concave up: $\displaystyle \cup$
$\displaystyle f'''(0) = \text{}6$
The concavity is decreasing (it is "flattening out").
So, near $\displaystyle x = 0$, the graph might look like this: Code:

* 

* 
* 
1*
 *
   +     *      
 *
