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 $A$.
Your condition means that $y=0.1x$ is the linear approximation of your quadratic polynomial at $x=0$. Thus the "error" made by replacing the one by the other is quadratic.

Otherwise said, you are looking for an equation of the form:

$$y=0.1x + \underbrace{A x^2}\_{\text{error term}}$$

with a general parameter $A$, with $B=0.1$ and $C=0$.

It is difficult for me to understand the second part of the problem. Could you give more precision ?