Finding coefficients in polynomial given three tangents I am stuck with a problem I simply cannot solve.
I have to find the coefficients of a quadratic polynomial given three tangents. The problem is stated as follows:
The three lines described by the equations
$y_1(x)=-4x-16.5$
$y_2(x)=2x-4.5$
$y_3(x)=6x-16.5$
are all tangents to a quadratic polynomial $p(x)=ax^2+bx+c$
Determine the values of the coefficients a, b & c.
I simply cannot solve this problem, I've been at it for a long time. Any help is greatly appreciated :)
Edit: I'm including the way I tried to solve it. I didn't get super far.
Given the polynomial $p(x)$ I know that $p'(x)=2ax+b$
Therefore, the following is true for the three points with x-values of $x_1, x_2 $ and $x_3$, where the lines $y_1, y_2$ and $y_3$ are tangent to the parabola:
$-4=2ax_1+b$
$2=2ax_2+b$
$6=2ax_3+b$
That's all I've managed to do. I've also found the points where the three lines intersect (well, three points two of the lines intersect), but I can't think of how to use that for anything.

Note that if a line and a quadratic are tangent $mx+d=ax^2+bx+c$ then the following quadratic will have discriminant zero
\begin{eqnarray*}
ax^2+(b-m)x+c-d=0.
\end{eqnarray*}
This will lead to $3$ equations for $a,b,c$ that are easily solved giving

 $(a,b,c)=(1/2,1,-4)$.

