Version 1: Where I assume an #x# was accidentally missed on the #+16# term:
#y=4x^2+16x-9#
Vertex form of a quadratic is
#y = m(x-a)^2+b#
where the vertex of the parabola is at #(a,b)#
#y=4x^2+16x-9#
#=4(x^2+4x)-9 " extracting the "m" factor"#
#= 4(x^2+4x+2^2) -16 -9" [completing the square](http://socratic.org/algebra/quadratic-equations-and-functions/completing-the-square-1)"#
#=4(x+2)-25 " simplifying"#
#=4(x-(-2)) +(-25)" into vertex form"#
(The vertex is at #(x,y) =(-2,-25)#)
Version 2: Where the question was entered correctly (except for the missing #(y=)# which is needed to make it an equation
#y=4x^2+16-9#
which is equivalent to #y=4x^2+7#
This can be rearranged as
#y=4(x-0)^2+7#
for a parabola with a vertex at #(x,y)=(0,7)#
