Answer:
(-2,-7) is the vertex if you meant to write y=2x^2+8x+1
Explanation:
I think you missed an x next to 8
y=2x^2+8x+1
First step factor 2x^2+8x so that we get 1 directly in front of x^2
y=2(1x^2+4x) +1
y=2( x^2+4x ) +1 Now I will leave a space for completing the square
y=2( x^2+4x+(4/2)^2)+1-2(4/2)^2 Whatever you add in you must subtract out
y=2( x^2+4x+2^2)+1-2(2^2)
y=2( x+2)^2 +1 -2(4)
y=2( x+2)^2 +1-8
y=2( x+2)^2 -7
Vertex is (-2,-7)
Or You could find -b/(2a) then plug in that into y=f(x) to find the y-coodinate
So -b/(2a)=-8/(2*2)=-8/4=-2
2(-2)^2+8(-2)+1=8-16+1=-8+1=-7
So the vertex is (-2,-7)
