Find a polynomial $f$ of lowest possible degree such that $f(x_{1})=a_{1}$, $f(x_{2})=a_{2}$, $f'(x_{1})=b_{1}$, $f'(x_{2})=b_{2}$ where $x_{1} \neq x_{2}$ and $a_{1}, a_{2}, b_{1}, b_{2}$ are given real numbers.
I understand that the polynomial can be at most of degree 3, but I believe it to be a quadratic. Not sure how to incorporate $a_{1}, a_{2}, b_{1}, b_{2}$ into it.
On
\begin{matrix} x_1 & a_1&\\\\ & & b_1& \\ x_1 & a_1&&&\frac{\frac{a_2-a_1}{x_2-x_1}-b_1}{x_2-x_1}\\\\ & & \frac{a_2-a_1}{x_2-x_1}&&&\frac{\frac{b_2-\frac{a_2-a_1}{x_2-x_1}}{x_2-x_1}-\frac{\frac{a_2-a_1}{x_2-x_1}-b_1}{x_2-x_1}}{x_2-x_1}\\ x_2 & a_2&&&\frac{b_2-\frac{a_2-a_1}{x_2-x_1}}{x_2-x_1}\\\\ &&b_2\\ x_2 & a_2&\\ \end{matrix}
$$P(x)=a_1+b_1(x-x_1)+\frac{\frac{a_2-a_1}{x_2-x_1}-b_1}{x_2-x_1}(x-x_1)^2+ \frac{\frac{b_2-\frac{a_2-a_1}{x_2-x_1}}{x_2-x_1}-\frac{\frac{a_2-a_1}{x_2-x_1}-b_1}{x_2-x_1}}{x_2-x_1}(x-x_1)^2(x-x_2) $$
Hint: You have $4$ parameters to fit in your polynomial ($a_1,a_2,b_1,b_2$), so in general you will need $4$ free parameters to move around, and your polynomial will have to be of degree $3$.
Try to write down the general formula for a polynomial of degree $3$ and to show that you can find coefficients solving your problem. It will be easy to understand when you can take the polynomial to be of degree $2,1$ or $0$ (constant). Find an example of $a_1,a_2,b_1,b_2$ so that no degree $2$ polynomial can solve the problem.