If the evaluate of a is negative and the vertex is at (13, -2) then how many x-interceps will the graph?

a) 2
b) it cannot be determined from the information
c)0
d) 1

Y=a(x-h)²+k
vertex is (h,k)
and when a is positive then the parabola opens up and the vertex is lowest point

so since vertex is (13,-2) then it is below the x axis (-2<0)
and since a is negative, it opens down
so the vertex is the highest point
since the highest point is already below the x axis, there cannot be any x intercepts

0 intercepts
answer is C