How Do I Find the vertex of 3x^2 + 3x + 2y = 0?

Question

I have to find it by completing the square

Answer 1

to complete the square, first divide through by a. (ax^2+bx+c format)

x^2+3x/3+2/3=0

put c on the other side

x^2+x= -2/3

on this part, get half of b and square it. then add the solution to both sides.

x^2+x+1/2= -2/3+1/2

factor the left, simplify the right.

(x+1/2)^2= -1/6

square root both sides

x+1/2= +/- 1/root6i

simplify

x=1i/root6+1/2 and x= -1i/root6+1/2

Answer 2

dang i did that last year...i just dont remember how to do it!! :(

sry

Answer 3

Isolate y on one side of the problem:

2y = -3x^2 - 3x

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

Then graph it on a graphing calculator and find the MIN.

I took that course such a long time ago, I forgot how to do it by hand.