How do we use the picture to get the polynomial? Calculate the solution $p\in \mathbb{P}^4$ off the following interpolatin exercise: 
\begin{align*}&p(0)=2 , \ p'(0)=3, \ p''(0)=1 \\ &p(1)=2, \ p'(1)=0\end{align*} 
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I have done the following: 
The polynomial that we are looking for is of the form \begin{equation*}p(x)=ax^4+bx^3+cx^2+dx+e\end{equation*} 
The first two derivatives are \begin{align*}&p'(x)=4ax^3+3bx^2+2cx+d \\ &p''(x)=12ax^2+6bx+2c\end{align*} 
Using the given conditions we get the system: 
\begin{align*}\begin{cases}p(0)=2 \\ p'(0)=3 \\ p''(0)=1 \\ p(1)=2 \\ p'(1)=0\end{cases}  \Rightarrow \begin{cases}e=2 \\ d=3 \\ 2c=1 \\ a+b+c+d+e=2 \\ 4a+3b+2c+d=0\end{cases}  \Rightarrow \begin{cases}e=2 \\ d=3 \\ c=\frac{1}{2} \\ b=-10 \\ a=\frac{13}{2}\end{cases}\end{align*} 
Therefore we get the polynomial \begin{equation*}p(x)=\frac{13}{2}x^4-10x^3+\frac{1}{2}x^2+3x+2\end{equation*} 
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Now I saw that there is an easier way to solve this using the following picture, but I haven't really understood how we use that. Do you have an idea? 
 


You may solve this directly with Taylor:
$$p(x) =\sum_{k=0}^4\frac{p^{(k)}(0)}{k!}x^k$$
Pluggin in the given values you get
$$p(x) = p(0) + p'(0)x + \frac{p''(0)}{2}x^2 + \frac{p^{(3)}(0)}{6}x^3 + \frac{p^{(4)}(0)}{24}x^4 $$ $$= 2 + 3x + \frac{1}{2}x^2 + ax^3+bx^4$$
Now, use the other two conditions to find $a$ and $b$:
$$0 = p'(1) =3+1+3a+4b$$
$$2 = p(1) = 2+3+\frac{1}{2} + a + b$$
$$\Rightarrow a=-10, b=\frac{13}{2}$$
$$\Rightarrow p(x) = 2 + 3x + \frac{1}{2}x^2 -10x^3+\frac{13}{2}x^4$$
