Complete the square to find the vertex
of this parabola.
2
y +6y+8x+1=0
([?], [ ]

vertex at (1, -3)

Step-by-step explanation:

When x = 0

y² + 6y + 1 = 0

y² + 6y +   = -1

y² + 6y + 9  = -1 + 9

(y + 3)² = 8   or    (-y - 3)² = 8

y + 3 = √8    or    -y - 3 = √8

y = - 3 +√8    or    y = -3 - √8

(0, - 3 +√8)   and (0, -3 - √8)

The mid point between these two is the average

y = ( - 3 + √8 +  -3 - √8) / 2 = - 3

y² + 6y + 8x + 1 = 0

(-3)² + 6(-3) + 8x + 1 = 0

9 - 18 + 1 = -8x

- 8 = -8x

x = 1

you would use (d.) the law of cosines or a^2 = b^2 + c^2 – 2bc(cos a)

, b^2 = a^2 + c^2 – 2ac(cos b), or c^2 = a^2 + b^2 - 2ab(cos c). you wouldn't use b. the law of sines as it is mostly effective when 1side and 2angles are given. a and c aren't the answer for they are the improper versions of the law of cosines. hope this is .