Answer:
The minimum is the point (-3,2)
Step-by-step explanation:
we have
[tex]x^{2} +6x+11[/tex]
This is a vertical parabola open upward (because the leading coefficient is positive)
The vertex is a minimum
Convert the equation in vertex form
Complete the square
[tex]f(x)=(x^{2} +6x+3^2)+11-3^2[/tex]
[tex]f(x)=(x^{2} +6x+9)+11-9[/tex]
[tex]f(x)=(x^{2} +6x+9)+2[/tex]
Rewrite as perfect squares
[tex]f(x)=(x+3)^{2}+2[/tex] ----> equation in vertex form
The vertex is the point (-3,2)
therefore
The minimum is the point (-3,2)
