Answer:
[tex]y+6=(x+1)^{2}[/tex]
Step-by-step explanation:
we have
[tex]y=x^{2}+2x-5[/tex]
This is the equation of a vertical parabola open upward (because the leading coefficient is positive)
The vertex is a minimum
The equation of a vertical parabola into vertex form is
[tex]y-k=a(x-h)^2[/tex]
where
(h,k) is the vertex of the parabola
Convert the equation into vertex form
Move the constant term to the left side
[tex]y+5=x^{2}+2x[/tex]
Complete the square
[tex]y+5+1=x^{2}+2x+1[/tex]
[tex]y+6=x^{2}+2x+1[/tex]
Rewrite as perfect squares
[tex]y+6=(x+1)^{2}[/tex]
therefore
[tex]a=1\\h=-1\\k=-6[/tex]
The vertex is the point (-1,-6)
