Question 42660
{{{f(x) = a(x-h)^2 + k}}} 
{{{f(x) = a(x)^2 + k}}}
vertex: (0,k)
If {{{k}}} is negative, you have no intersection with the x-axis.
If {{{k}}} is zero, you have one intersection with the x-axis.
If {{{k}}} is positive, you have two intersections with the x-axis.
Since the value of {{{a}}} is negative, the parabola opens downward.
The answer is (B).