TITLE: What are a , b and c?
QUESTION [5 upvotes]: $$y = ax^2 + bx + c$$
which is tangent at the origin with the line $y=x$, It is also tangential with the line $y=2x + 3$. Determine the function! Draw a figure!
My main question is this solvable? I am doubtful?

REPLY [3 votes]: Note: The method below is very similar to my answer to the question "Find equation of quadratic when given tangents?".
Since the derivative of $y=f(x)=ax^{2}+bx+c$ is $f^{\prime }(x)=2ax+b$, the
equations of the tangents to the graph of $f(x)$ at points $(x_{i},f(x_{i}))$, with $i=1,2$ are
$$\begin{eqnarray*}
y &=&f^{\prime }(x_{i})x-f^{\prime }(x_{i})x_{i}+f(x_{i}) \\
&=&\left( 2ax_{i}+b\right) x-\left( 2ax_{i}+b\right)
x_{i}+ax_{i}^{2}+bx_{i}+c.
\end{eqnarray*}$$
One of the points is $(x_{1},f(x_{1}))=(0,0)$. As the equation of the
tangent at $(0,0)$ is $y=x$ we must have
$$bx+c\equiv x.$$
Comparing coefficients we get $b=1,c=0$. Hence $f(x)=ax^{2}+x$. Similarly
for the tangent at $(x_{2},f(x_{2}))$ we must also have
$$\left( 2ax_{2}+1\right) x-\left( 2ax_{2}+1\right) x_{2}+ax_{2}^{2}+x_{2}\equiv 2x+3.$$
Comparing again coefficients, we get the following system in $a$ and $x_2$, which enables us to find $a$: 
$$\left\{ 
\begin{array}{c}
2ax_{2}+1=2\qquad\qquad\qquad \\ 
-\left( 2ax_{2}+1\right) x_{2}+ax_{2}^{2}+x_{2}=3.%
\end{array}%
\right. $$
From the first equation we get $x_{2}=1/(2a)$, which by substitution in the
second equation gives $a=-1/12$.
Therefore the quadratic equation $y=f(x)$ is 
$$y=-\frac{1}{12}x^{2}+x.$$
Below is the graph of $y=f(x)$ together with its two tangents at points $(0,0)$ and $(-6,-9)$.