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