What are a , b and c? $$y = ax^2 + bx + c$$
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?
 A: This is the Graph of $f(x)= -\frac{1}{5}x^{2}+x$ which I graphed using KmPlot. The figure should give you an intuitive idea of how to go about solving.


*

*The Green line is $y=2x+3$.

*The Blue line is $y=x$.
If the line $y=2x+3$ and the parabola $y=ax^{2}+bx+c$ are going to be tangent at a given point then their slopes are equal. Let's find that out. Slope of line $y=2x+3$ is $2$ and we have $$2 = \frac{dy}{dx} = 2ax+1$$ So you have $x=\frac{1}{2a}$. Also we have 
\begin{align*}
2x+3 & = ax^{2} + x
\end{align*}
which says that $$2 \times \frac{1}{2a} + 3 = a \times \frac{1}{4a^{2}} + \frac{1}{2a}=\frac{3}{4a}$$ From this we have  $$\frac{1}{a} -\frac{3}{4a} = -3 \Longrightarrow a=-\frac{1}{12}$$ 

This is for the value $a=-\frac{1}{3}$

This is for the value $a=-\frac{1}{7}$.

A: Your problem is now that you have $y = ax^2 + x$ tangential to $y = 2x + 3$.  This means you have some number $n$ where $an^2 + n = 2n + 3$ (they meet at a point) and $2an + 1 = 2$ (they meet tangentially).
You have two equations and two unknowns; I'm sure you can solve from here.
A: 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 $(-6,-9)$.

