2 ways
1. get into vertex form y=a(x-h)^2+k
2. use math hacks
1.
vertex is (h,k) in y=a(x-h)^2+k
complete square
y=(2x^2+4x)+1
y=2(x^2+2x)+1
y=2(x^2+2x+1-1)+1
y=2((x+1)^2-1)+1
y=2(x+1)^2-2+1
y=2(x+1)^2-1
y=2(x-(-1))^2+(-1)
vertex is (-1,-1)
2.hack way
in form y=ax^2+bx+c
x value of vvertex is -b/2a
given
y=2x^2+4x+1
b=4, a=2
xvaluevertex=-4/(2*2)
xvaluevertex=-4/4=-1
sub back to find y value
y=2(-1)^2+4(-1)+1
y=2(1)-4+1
y=2-4+1
y=-1
y value is -1
vertex is (-1,-1)
those are 2 ways to find vertex which is (-1,-1)
