Question

What is the vertex form of `y= 6x^2 + 11x + 4`?

Answer 1

the vertex form of the equation is
`y = 6(x + 0.916666667)^2 -1.041666667`

Explanation:

The general form of a quadratic equation is
`y = ax^2 +bx+c`

the vertex form of a quadratic equation is
`y = a(x-h)^2 +k`

where `(h, k)` is the vertex of the line

for a standard quadratic the vertex of the line can be found where the slope of the line is equal to 0
The slope of a quadratic is given by the its first derivative
in this case
`(dy)/(dx) = 12x +11`

the slope is `0` when `x = -11/12 or -0.916666667`

The original equation
`y = 6x^2 +11x + 4`

Substitute in what we know
`y = 6*(-11/12)^2 + 11*(-11/12) +4 = -1.041666667`

The vertex is at `(-0.916666667, -1.041666667)`

Thefore
the vertex form of the equation is
`y = 6(x + 0.916666667)^2 -1.041666667`