Crossing The Line

$\begin{aligned} h(x) & = f(x) + g(x) \\ h(a) & = f(a) + g(a) = 0 + 0 = 0 & \implies (a,0) \text{ is on }h(x) \\ h(b) & = f(b) + g(b) = f(b) + 0 = f(b) & \implies \color{#3D99F6}{(b,f(b))} \text{ is on }h(x) \\ h(c) & = f(c) + g(c) = f(c) + 0 = f(c) & \implies (c,f(c)) \text{ is on }h(x) \end{aligned}$

The only point appears in the choice options is $\boxed{\color{#3D99F6}{(b,f(b))}}$ .

First let's find the zeros of both f(x) and g(x) for x = a, b, and c. Since f(x) contains $(x-a)^2$ , when x=a, f(x) will be 0. Since g(x) contains $(x-a)$ , $(x-b)$ and $(x-c)$ ; when x = a, b, or c, g(x) will equal 0.

so for h(x), we can check the values of h(x) by plugging in the appropriate value.

(b, f(b) ) can't be right since g(x) is not 0 when x = b.

(a , f(b) / g(b) ) can't be the correct answer because g(b) is 0, and division by 0 is not defined.

(c, g(c) - f(c) ) can't be the correct answer because h(x) = f(x) + g(x), not g(x) - f(x).

This leaves only (b , f(b) ), which is the correct answer because when x=b, g(x) = 0, and f(x) = f(b). Therefore, h(b) = f(b).