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 \(-12\)
