Question

Vertex of y= 2x^2+8 +1​

Answer 1

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)

Answer 2

(-2,-7) is the vertex but it seems you missed an X