Question 192356
y=8x^2+4x-1   to  form  y=a(x-h)^2+k

y=8(x^2+x/2)-1

now  must  complete  square  of  bracket  terms

y=8(x^2 +x/2 +1/16) -1  -8(1/16)
simplify
y=8(x+1/4)^2-1.5

this  fits  form  with  

a=(8)
h=(-1/4)
k=(-1.5)

checking
let  x=3  solve  for  y
initial  eqn  y=8x^2+4x-1= 8(3^2) +4(3)-1=72+12-1=83
final  form  y=8(x+1/4)^2-1.5=8(3.25)^2-1.5= 8(10.5625)-1.5=84.5-1.5=83    ok

to  complete  the  square  we  take  coefficient  of  middle  term,  divide  by  two,  and  square  it
ie  (1/2)  ,  (1/2)/2=1/4,  (1/4)^2=(1/16)
this  becomes  constant  term  of  "square"
since  we  added  this  to  complete  square  terms  we  must  subst  from  constant  term  to  balance  eqn.  In  this  case  we  added 8(1/16)  in  total  therefore  we  substract  (8/16) from  constant