Answer:
All the intervals that contain the number [tex]-2[/tex] are solution of the problem
Step-by-step explanation:
we know that
The equation of a vertical parabola into vertex form is equal to
[tex]f(x)=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex of the parabola
In this problem we have
[tex]f(x)=(x+3)(x+1)[/tex]
Convert to vertex form
[tex]f(x)=x^{2}+x+3x+3[/tex]
[tex]f(x)=x^{2}+4x+3[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-3=x^{2}+4x[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]f(x)-3+4=x^{2}+4x+4[/tex]
[tex]f(x)+1=x^{2}+4x+4[/tex]
Rewrite as perfect squares
[tex]f(x)+1=(x+2)^2[/tex]
[tex]f(x)=(x+2)^2-1[/tex] -------> equation in vertex form
The vertex is the point [tex](-2,-1)[/tex]
The x-coordinate of the vertex is [tex]-2[/tex]
