The vertex form:
[tex]f(x)=ax^2+bx+c=a(x-h)^2+k\\\\h=\dfrac{-b}{2a},\ k=f(h)[/tex]
We have
[tex]f(x)=2x^2+4x+1\\\\a=2,\ b=4,\ c=1\\\\h=\dfrac{-4}{2(2)}=\dfrac{-4}{4}=-1\\\\k=f(-1)=2(-1)^2+4(-1)+1=2(1)-4+1=2-4+1=-1[/tex]
[tex]f(x)=2(x-(-1))^2+(-1)[/tex]
Answer: [tex]\boxed{f(x)=2(x+1)^2-1}[/tex]
