Question

What is the vertex form of y=2x^2+8x+1

Answer 1

So, the vertex form of your function is
`y(x)=2*(x+2)^2-7`.

The vertex is at
`(-2|-7)`

Explanation:

Given function:

`y(x)=2*x^2+8x+1`

Steps:

`y(x)=2*x^2+8x+1\\`

`y(x)=2(x^2+4x+(1)/(2) )` (Factor out)

`y(x)=2(x^2+4x+2^2-2^2+(1)/(2) )` (Complete the square)

`y(x)=2((x+2)^2-2^2+(1)/(2) )` (Use the binomial formula)

`y(x)=2((x+2)^2-(7)/(2) )` (Simplify)

`y(x)=2*(x+2)^2-7` (Expand)

Answer 2

y = a ( x − h ) 2 + k is vertex form.

-b/2a to get x of vertex (x , y)

-8/4 = -2

So far we have (-2, y)

plug -2 into equation:

y = 2(-2)^2 - 16 + 1

y = -7

(-2, -7)

so 2(x+2)^2 - 7