8. For the quadratic function $y=a x^{2}+b x+c(a \neq 0)$, the vertex of its graph is in the first quadrant, and it passes through the points $(0,1)$ and $(-1,0)$. Then the range of values for $S=a+b+c$ is ( ).
(A) $0<S<1$
(B) $0<S<2$
(C) $1<S<2$
(D) $-1<S<1$
8. B.

Solution one: Let $x=0, y=1$ and $x=-1, y=0$, we get
$c=1, a=b-1$.
$\therefore S=a+b+c=2b$.
From the problem, we know that $-\frac{b}{2a}>0$, and $a<0$.
Also, from $b=a+1$ and $a<0$, it leads to a contradiction. Therefore, (D) can be excluded.

Solution three: Exclude (A) and (C) as above. If $a+b+c=0$, then when $x=-1, x=1$, the $y$ values are both 0, which means the y-axis is the axis of symmetry of the parabola. However, this contradicts the fact that the vertex of the parabola is in the first quadrant, hence (D) can be excluded.