we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex
if [tex]a>0[/tex] ------> the parabola open upward ( the vertex is a minimum)
if [tex]a<0[/tex] ------> the parabola open downward ( the vertex is a maximum)
in this problem we have
[tex]g(x)=x^{2}+2x-1[/tex]
convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]g(x)+1=x^{2}+2x[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]g(x)+1+1=x^{2}+2x+1[/tex]
[tex]g(x)+2=x^{2}+2x+1[/tex]
Rewrite as perfect squares
[tex]g(x)+2=(x+1)^{2}[/tex]
[tex]g(x)=(x+1)^{2}-2[/tex] --------> equation in vertex form
the vertex is the point [tex](-1,-2)[/tex]----> is a minimum (parabola open upward)
using a graphing tool
see the attached figure
