Question

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

Answer 1

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`

Answer 2

The x-value is -b/(2a) = -8/(2·4) = -1.

The corresponding y-value is ...

... 4(-1)² +8(-1) -8 = 4 -8 -8 = -12

The y-value of the vertex is -12.