we have
[tex]f(x)=-(x+5)(x+1)[/tex]
we know that
The equation of a vertical parabola into vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex of the parabola
if [tex]a>0[/tex] ----> the parabola open upward (vertex is a minimum)
if [tex]a<0[/tex] ----> the parabola open downward (vertex is a maximum)
In this problem convert the quadratic equation into vertex form
so
[tex]f(x)=-(x+5)(x+1)\\f(x)=-(x^{2} +x+5x+5)\\f(x)=-( x^{2} +6x+5)[/tex]
[tex]f(x)=-x^{2}-6x-5[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)+5=-(x^{2}+6x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)+5-9=-(x^{2}+6x+9)[/tex]
[tex]f(x)-4=-(x^{2}+6x+9)[/tex]
Rewrite as perfect squares
[tex]f(x)-4=-(x+3)^{2}[/tex]
[tex]f(x)=-(x+3)^{2}+4[/tex]
This is a vertical parabola open down (vertex is a maximum)
the vertex is the point [tex](-3,4)[/tex]
The range is the interval--------> (-∞,4]
[tex]y\leq 4[/tex]
All real numbers less than or equal to [tex]4[/tex]
therefore
the answer is the option
All real numbers less than or equal to [tex]4[/tex]
