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.
 A: If we have a parabola $$ y = a x^2 + b x + c  $$ and a line $$ y = mx + d,$$
they are tangent if
$$  b^2 - 4ac = -4da + 2mb  - m^2  $$ as then the parabola
$$ y = a x^2 + (b-m) x + (c -d)  $$ has a double root, i.e. is a constant times a square, $$ a (x+p)^2 $$
Might as well write this: I fixed
$$  \Delta = b^2 - 4 a c $$
and then had three equations
$$   -4da + 2mb = m^2 + \Delta $$
by plugging in the values from the three lines $ y = mx + d.$ I expected bigger problems, but just taking the differences of two of the equations cancels the extra unknown $\Delta,$ alowing us to find $a,b$ quickly. From that, we finally get a value for $\Delta,$ after which we have one equation for $c$ 
$$ 66a - 8 b = 16 + \Delta \; ,    $$
$$ 66a + 12 b = 36 + \Delta \; ,    $$
$$ 18a + 4 b = 4 + \Delta \; .    $$
Second minus first gives $b.$ Plug in the $b$ value, then subtract second minus third, which gives $a.$ Plug both into any of the three to find $\Delta.$ Finally, $c = \frac{b^2 - \Delta}{4a}$
A: 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

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

A: Given any two tangents to a parabola of the form $y = f(x) = ax^2 + bx + c,$ the $x$-coordinate of the intersection of the tangent lines will be midway between the $x$-coordinates of the tangent points.
Working out the intersection points of the three lines, we can see that
the intersection of $y = y_1(x)$ and $y = y_2(x)$ occurs at $x = -2$
and the intersection of $y = y_2(x)$ and $y = y_3(x)$ occurs at $x = 3.$
Let 


*

*$x = x_1$ at the tangent point with $y = y_1(x)$;

*$x = x_2$ the tangent point with $y = y_2(x)$; and

*$x = x_3$ the tangent point with $y = y_3(x).$


Due to the $y$-coordinates and slopes at the intersection points, it is clear that $x_1 < -2 < x_2 < 3 < x_3$;
moreover, $-2$ is midway between $x_1$ and $x_2$ and $3$ is midway between $x_2$ and $x_3.$
It follows that 
$$x_3 - x_1 = 2\left(\frac{x_3 + x_2}2 - \frac{x_2 + x_1}2\right)
= 2(3 - (-2)) = 10.$$
The tangency condition implies that $f'(x_1) = y_1'(x) = -4$ and
$f'(x_3) = y_3'(x) = 6.$
But $f'(x) = 2ax + b,$ 
so $20a = 2a(x_3 - x_1) = f'(x_3) - f'(x_1) = 10,$ 
and therefore $a = \frac12.$
It is then a relatively straightforward exercise to find the other two coefficients.
