Question

What is the y value of the vertex 4x^2+8x-8

Answer 1

To start with, you need to find the vertex, you need to find the axis of symmetry: x=(-b/2a).  b=8.  a=4.  x=(-8/2(4)).  x=-1.  Plugin -1 into the original equation to find y.  y=-12.

Answer 2

The y-value of the vertex is
`-12`

Explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

`f(x)=a(x-h)^(2)+k`

where

(h,k) is the vertex of the parabola

In this problem we have

`f(x)=4x^(2)+8x-8` -----> this a vertical parabola open upward

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`f(x)+8=4x^(2)+8x`

Factor the leading coefficient

`f(x)+8=4(x^(2)+2x)`

Complete the square. Remember to balance the equation by adding the same constants to each side

`f(x)+8+4=4(x^(2)+2x+1)`

`f(x)+12=4(x^(2)+2x+1)`

Rewrite as perfect squares

`f(x)+12=4(x+1)^(2)`

`f(x)=4(x+1)^(2)-12`

The vertex is the point
`(-1,-12)`

The y-value of the vertex is
`-12`