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

Answer:

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

Step-by-step 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 $-12$