vertex of y=3x^2+6x-2

Question

vertex of a parabola y=3x^2+6x-2.

Answer 1

Given parabola y = 3x^2+6x-2

Compare it standard parabola form y = ax^2+bx+c

a = 3, b = 6, c = -2.

Axis of symmetry x = -b/2a  = -6/6

x = -1

To find vertex substitute x value in given equation.

y = 3(-1)^2+6(-1)-2

y = 3-6-2

y = -5

Vertex of given parabola = (-1,-5).