If the points $(1,y_1)$ and $(-1,y_2)$ lie on the graph of $y=ax^2+bx+c$, and $y_1-y_2=-6$, then $b$ equals:
$\text{(A) } -3\quad \text{(B) } 0\quad \text{(C) } 3\quad \text{(D) } \sqrt{ac}\quad \text{(E) } \frac{a+c}{2}$
Because the two points are on the quadratic, $y_1 = a + b + c$ and $y_2 = a - b + c$.  Because $y_1 - y_2 = -6$,
\[(a+b+c)-(a-b+c)=-6\]
\[2b=-6\]
\[b=-3\]
The answer is $\boxed{\textbf{(A)}}$.