Answer:
\(y=6(x+1)^{2}-16\)
Step-by-step explanation:
we know that
The equation of a vertical parabola in vertex form is equal to
\(y=a(x-h)^{2}+k\)
where
(h,k) is the vertex of the parabola
In this problem we have
\(y=6x^{2}+12x-10\)
Convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(y+10=6x^{2}+12x\)
Factor the leading coefficient
\(y+10=6(x^{2}+2x)\)
Complete the square. Remember to balance the equation by adding the same constants to each side
\(y+10+6=6(x^{2}+2x+1)\)
\(y+16=6(x^{2}+2x+1)\)
Rewrite as perfect squares
\(y+16=6(x+1)^{2}\)
\(y=6(x+1)^{2}-16\) -------> equation in vertex form
The vertex is the point \((-1,-16)\)
