We must complete the square to put the equation into vertex form: #y=a(x-h)^2+k#, where #(h,k)# is the vertex.
#y=-7(x^2+2/7x+color(red)(?))+3#
We must complete the square. In order to do this, we must recall that #(x+a)^2=x^2+2ax+a^2#, so the middle term, #2/7x#, is #2x# times some other number, which we can determine to be #1/7#. Thus, the final term must be #(1/7)^2#.
#y=-7(x^2+2/7x+color(red)(1/49))+3+color(red)(1/7)#
Note that we had to balance the equation—we can add numbers randomly. When the #1/49# was added, we must realize that it is actually being multiplied by #-7# on the outside of the parentheses, so it is actually like adding #-1/7# to the right side of the equation. In order to balance the equation we add a positive #1/7# to the same side.
Now, we can simplify:
#y=-7(x+1/7)^2+22/7#
Since the vertex is #(h,k)#, we can determine its location is #(-1/7,22/7)#. (Don't forget the #h# value switches signs.)
