Question

given the guadratic function, y=3x^2+6x-2, find the vertex
using the vertex formula x= -b over 2a

A. (-1,-5)
B. (2,-3)
C. (-2,-3)
D. (-1,5)

Answer 1

The vertex is where the derivative is zero.
\[\frac{dy}{dx}=6x+6=0\]
Find this x. Substitute into original equation to find y. Done!

Answer 2

okay you want to complete the square

if you label your quadratic in terms of A B and C
3=A
6=B
-2=C


first of all you want to take out a factor of 3

3(x^2+2x-2/3)
now we complete the square into (x+b/2)^2-(b/2)^2
3((x+1)^2 -1-2/3)

multiply it all by 3
3(x+1)^2 -3-2

3(x+1)^2 -5

your vertex should be 1, -5

Answer 3

that should be -1,-5