Ya gotta pay attention to details.
That's -6y not +6y, so
x = 3^2 - 6(3)+3 = -3
note that x = (y-3)^2 - 3
So the vertex is at (-3,3)
I'm currently working with parabolas, and I'm not sure if I am doing this specific problem correctly..
12. x= y^2-6y+6
y= -b/2a = 6/2(1)= 6/2= 3
x= (3)^2+6(3)+6
x= 33
y=3
(33,3) <--- what I graphed
2 answers
Oh, I noticed that. I just thought it would come out to be -(-6y) that would then make it +6y.