SOLUTION: if y=8x^2 + 4x-1 is expressed in the form y=a(x-h)^2+k, where a, h, and k are constants, what is the value of k ?

Question 192356: if y=8x^2 + 4x-1 is expressed in the form y=a(x-h)^2+k, where a, h, and k are constants, what is the value of k ?
Found 2 solutions by Mathtut, RAY100:
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
to complete square had to add 8(1/16) therefore had to subtract that value outside the parenthesis to keep equation eqivalent.
:

:
k=-3/2
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
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
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