$\displaystyle 9-4 \cdot 2 \cdot m \ge 0$

solve for m.

$\displaystyle 9-8 \cdot m \ge 0$

$\displaystyle -8 \cdot m \ge -9$

$\displaystyle m \le \dfrac{-9}{-8}\le \frac98$

**<--- sign error**
I'm always stuck with an wrong answer.

The graph of the function is a parabola opening downward. Choose c such that the vertex lies below the x-axis.

$\displaystyle y = -x^2+2x+c$

$\displaystyle y = -(x^2-2x+1)+1+c$

<- i understand the part where you factorize the minus sign out. but how did you manage to add this $\displaystyle .. +1)+1 ..$ in? **<-- actually I added zero, so no damage has been done**
$\displaystyle y = -(x-1)^2+1+c$ [COLOR="red"]

**The whole procedure is called "Completing the square". Google for it**
That means the vertex is at V(1, 1+c)

According to the question the y-coordinate of the vertex must be smaller than zero:

$\displaystyle 1+c<0$

solve for c.

Calculate the coordinates of the point of intersection between the straight line and the curve:

$\displaystyle (mx)^2 = x-c$

Solve for x.

**<--- That means apply the quadratic formula and go back to the first question. It's just the same procedure.**
You only get one single point of intersection (that's the tangent point!) if the discriminant equals zero. Solve for c.

okay i'm totally lost on this question. i think i'll need to wait for my tutorial class for an detailed explanation.