Question: I've been trying to find a solution to this problem but I'm not too sure how to go about solving it.

> 
> I need to find the unknown values of $A, B$ and $C$ in the parabola
>  equation $y(x) = Ax^2 + Bx + C$ given that the parabola passes through
>  point (0,0) and is tangent to the line $y1(x) = 0.1x$ which also
>  passes through the point (0,0).
> 
> 
> I also need to find an unknown point (x-coordinate, y-coordinate) on
>  the parabola which is tangent to $y2(x) = -0.08x + 10$ given that
>  $y2(x) = -0.08x + 10$ passes through (200,-6).
> 
> 
> 

I would greatly appreciate any help with solving this problem. So far I've tried to use the $y1(x)$ line to determine the vale of $B$ and $C$ in the parabola:

As the parabola passes through (0,0) I tried,

$$y(0) = 0$$ so 
$$A(0)^2 + B(0) + C = 0\\
C = 0$$

As the line $y1(x) = 0.1x$ is tangent to the parabola at (0,0) I tried,

$$ y'(0) = 0.1$$ so  

$$ 2A(0) + B = 0.1\\ 
B = 0.1$$

However when trying to find a point on the parabola where the parabola is tangent to $y2(x)$ it seems to me that the value for B would be different, so this has me really confused on how to solve this problem.

Once again, any help with this would be very very appreciated :)

Answer: 
Hint:
Let the tangent meet the parabola at $(x\_0,y\_0)=(x\_0,Ax\_0^2+0.1x\_0)$. The slope of the tangent is $$-0.08 = 2Ax\_0+0.1, \tag 1$$
and the equation of the line implies $$Ax\_0^2+0.1x\_0 = -0.08x\_0+10. \tag 2$$
Solve them simultaneously to compute $x\_0$ and $A$.
Since it passes through $(0,0)$ then $C=0$. 

Since it has a tangent line on x=0, we know that its derivate must match the slope of the line it is tangent to. 

So we have $y(x) = ax^2 + bx$ 

$y'(x) = 2ax + b$ 

$y'(0) = b = 0.1$

So know we can write:

$y(x) = ax^2+0.1x$ 

We are still missing the value of $a$ we can use the next chunk of data to see if it resolves or not.

According the next piece of information the parabola needs to intercept somewhere $(x\_0,y\_0)$ the tangent $y\_2(x)$ this means that:

$y\_0=ax\_0^2+0.1x\_0$
Because the line needs to touch we know:
$y\_2(x\_0)=y\_0 = ax\_0^2+0.1x\_0 = -0.08x\_0 +10$

This allows us to calculate the missing a based on the value of $x\_0$. I however suspect the problem is not well written and probably they are giving you the tangent point as well when they mention $(200,6)$. 

In this case $x\_0= 200$ and $y\_0=6$ 

$a200^2+0.1\cdot200 = 6$ 

Which makes: 

$a=\frac{-7}{100^2}$