Question

Which equation is y = –6x2 + 3x + 2 rewritten in vertex form?

Answer 1

The answer is
`y = -6( x- (1)/(4)) ^(2)+3`

Regular form: y = ax² + bx + c
Vertex form: y = a(x - h)² + k
(h, k) - vertex


`y = -6 x^(2) +3x+2 \\ y -2=-6 x^(2) +3x \\ y-2+6*=-6* x^(2) +6* (1)/(2) x \\ \\ y-2-6* (1)/(16) =-6* x^(2) +6* (1)/(2) x -6* (1)/(16) \\ \\ y -2 -(6)/(16) =-6( x^(2) -(1)/(2) x+(1)/(16)) \\ \\ y- (2*16)/(16) -(6)/(16) =-6( x^(2) -(1)/(2) x+((1)/(4))^(2) ) \\ \\ y - (32)/(16) -(6)/(16) = -6( x- (1)/(4)) ^(2) \\ \\ y - (32+6)/(16) = -6( x- (1)/(4)) ^(2) \\ \\ y - (48)/(16) = -6( x- (1)/(4)) ^(2) \\ \\`


`y - 3= -6( x- (1)/(4)) ^(2) \\ \\ y = -6( x- (1)/(4)) ^(2)+3`

Answer 2

we have

`y = -6x^(2) + 3x + 2`

we know that

the vertex form of the vertical parabola equation is equal to

`y=a(x-h)^(2) +k`

where

(h,k) is the vertex of the parabola

To find the equation rewritten in vertex form let's factor the equation

Factor the leading coefficient

`y = -6(x^(2) - 0.5x) + 2`

Complete the square. Remember to balance the equation by adding the same constants to each side

`y = -6(x^(2) - 0.5x+0.0625-0.0625) + 2`

`y = -6(x^(2) - 0.5x+0.0625) + 2 +0.375`

`y = -6(x^(2) - 0.5x+0.0625) + 2.375`

Rewrite as perfect squares

`y = -6(x-0.25)^(2) + 2.375`

therefore

the answer is

`y = -6(x-0.25)^(2) + 2.375`