Respuesta :

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)