, 2012

Since the graphs of $f(x)$ and $h(x)$ has exactly one common point, the equation $ax^2 + bx + c = cx + b$, i.e. the equation
$$ax^2 - (c-b)x + (c-b) = 0, \quad (1)$$
has exactly one root. So, its discriminant is equal to $0$, i.e. $(c-b)^2-4a(c-b) = 0 \Leftrightarrow (c-b)(c-b-4a) = 0$. By condition, $c \neq b$, i.e. $c-b \neq 0$, so we have $c-b-4a=0$ or
$$c - b = 4a. \quad (2)$$

From (1) and (2) it follows $ax^2 - 4ax + 4a = 0$, whence $x^2 - 4x + 4 = 0 \Leftrightarrow (x-2)^2 = 0 \Leftrightarrow x = 2$. Thus $x = 2$ is an abscissa of the common point of all three graphs.

Since the graphs of $g(x)$ and $h(x)$ has also exactly one common point, the equation $\frac{cx+b}{cx+a} = cx+b$, i.e. the equation $cx+b = (cx+b)(cx+a)$ has exactly one root. Therefore, $x = -\frac{b}{c} = -\frac{1-a}{c}$. But $x = 2$, so
$$-\frac{b}{c} = 2, \quad \text{and} \quad 2 = \frac{1-a}{c}. \qquad (3)-(4)$$
From (2), (3) and (4) we obtain $a = \frac{3}{11}$, $b = -\frac{8}{11}$ and $c = \frac{4}{11}$. It is easy to see that for these values of $a$, $b$ and $c$ the graphs of all three functions have exactly one common point.