Assume you have given $n+1$ points $(x_1,y_1),\cdots, (x_{n+1},y_{n+1}).$ (Of course, $x_i\ne x_j$ if $i\ne j.$) A polynomial of degree $n$ is of the form $p_n(x)=a_nx^n+\cdots+a_1x+a_0.$ To study the existence and uniqueness of such a polynomial consider the system of linear equations:
$$
\left\{\begin{array}{ccc}
a_nx_1^n+a_{n-1}x_1^{n-1}\cdots+a_1x_1+a_0 & =& y_1\\ \vdots & &\\
a_nx_{n+1}^n+a_{n-1}x_{n+1}^{n-1}\cdots+a_1x_{n+1}+a_0 & =& y_{n+1}
\end{array}\right.
$$
We write the system as
$$
\begin{pmatrix}x_1^n & x_1^{n-1} &\cdots & x_1 & 1 \\ \vdots & \vdots & \ddots & \vdots \\ x_{n+1}^n & x_{n+1}^{n-1}& \cdots & x_{n+1} & 1\end{pmatrix} \begin{pmatrix} a_n \\ \vdots \\ a_0 \end{pmatrix}=\begin{pmatrix} y_1 \\ \vdots \\ y_{n+1} \end{pmatrix}
$$
Since the matrix of coefficients of the system is non singular (it is a Vandermonde matrix (see Vandermonde)) the system has a unique solution, that is, there exists one polynomial of degree $n$ through the $n+1$ given points, and it is unique.
The big trick in polynomial interpolation problems uses the function:
$$g(x)=(x-x_1)(x-x_2)\dots(x-x_n)$$
Note that this function is zero at each $x_n$ (and nowhere else). Also, $g'(x)=\sum_{i=1}^n \frac{g(x)}{x-x_i}$ (where the division is intended to cancel out the extra factor, so that it is defined even at each $x_i$).
If I divide $g(x)$ by $g(x^*)$ for some $x^*$ which is not equal to any $x_n$, I get a function that is $1$ at $x^*$ and $0$ at $x_i$. Adding functions of this form multiplied by scalars allows us to solve the problem of finding a polynomial which goes through $x_i,y_i$ (just take $\sum_{i=1}^ny_ig_i(x)$ where $g_i$ is zero for $x_j,j\ne i$ and one at $x_i$).
To get the derivative right at some specified $x^*$, just add to this expression $a\frac{g^*(x)}{{g^*}'(x^*)}$ where $g^*(x)$ is zero at each $x_i$ and at $x^*$; since $g^*(x)$ has a single root at $x^*$ its derivative there is nonzero, so the derivative of this part of the function is $a$, and we can set $a$ to be the desired derivative minus the derivative of the point-interpolation part of the function.
To give an example, let's find a polynomial such that $f(-2)=1$, $f(-1)=0$, $f'(0)=4$, $f(2)=2$. (I just made these numbers up.) First let's ignore the derivative part and construct functions that are zero at two of the points and one at the remaining point:
$$g_1(x)=\frac{(x+1)(x-2)}{(-2+1)(-2-2)}=\frac{x^2-x-2}{4}$$
$$g_2(x)=\frac{(x+2)(x-2)}{(-1+2)(-1-2)}=\frac{-x^2+4}{3}$$
$$g_3(x)=\frac{(x+2)(x+1)}{(2+2)(2+1)}=\frac{x^2+3x+1}{12}$$
Then $h(x)=g_1(x)+0g_2(x)+2g_3(x)=\frac{5x^2+3x-2}{12}$ satisfies $h(-2)=1$, $h(-1)=0$, $h(2)=2$. But it probably has the wrong derivative: $h'(x)=\frac{10x+3}{12}$, so $h'(0)=\frac14$ is not what we want. So let's construct a function which won't affect our other values, but has a nonzero derivative at $0$:
$$g^*(x)=(x+2)(x+1)(x-2)=x^3+x^2-4x-4\implies {g^*}'(x)=3x^2+2x-4$$
So ${g^*}'(0)=-4$. (It is possible for this to come out zero, in which case you can multiply the formula by $x-x^*$ to guarantee nonzero derivative at the expense of a higher degree.) Now the $h$ function is off from what we want by $4-\frac14=\frac{15}4$, while the adjustment function has derivative $-4$, so $-\frac{15}{16}{g^*}(x)$ has derivative $4-\frac14$ at $0$ and $f(x)=h(x)-\frac{15}{16}{g^*}(x)=-\frac{15}{16}x^3-\frac{25}{48}x^2+4x+\frac{43}{12}$ should do the trick.
Best Answer
HINT
Let $P(x) = ax^{2} + bx + c$. According to the given data, we can conclude that
\begin{align*} \begin{cases} P(1) = a + b + c = 3\\\\ P'(1) = 2a + b = 1\\\\ P'(2) = 4a + b = 5 \end{cases} \end{align*}
Can you take it from here?