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

1 Answer
Nov 29, 2017

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