Question

Write the equation in vertex form

y=3x^2 +8x+2

y= -2x^2+6x-2

Answer 1

Part 1)
`y=3(x+(4)/(3))^(2)-(10)/(3)`

Part 2)
`y=-2(x-(3)/(2))^(2)+(5)/(2)`

Explanation:

Part 1) we have

`y=3x^(2)+8x+2`

Factor the leading coefficient 3

`y=3(x^(2)+(8)/(3)x)+2`

Complete the square

`y=3(x^(2)+(8)/(3)x+(64)/(36))+2-(64)/(12)`

`y=3(x^(2)+(8)/(3)x+(64)/(36))-(40)/(12)`

Rewrite as perfect squares

`y=3(x+(8)/(6))^(2)-(40)/(12)`

simplify

`y=3(x+(4)/(3))^(2)-(10)/(3)`

The vertex is the point
`(-(4)/(3),-(10)/(3))`

Part 2) we have

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

Factor the leading coefficient -2

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

Complete the square

`y=-2(x^(2)-3x+(9)/(4))-2+(9)/(2)`

`y=-2(x^(2)-3x+(9)/(4))+(5)/(2)`

Rewrite as perfect squares

`y=-2(x-(3)/(2))^(2)+(5)/(2)`

The vertex is the point
`((3)/(2),(5)/(2))`