we have
[tex]f(x)=-(x+3)(x-1)[/tex]
[tex]f(x)=-(x^{2} -x+3x-3) \\ \\f(x)=-(x^{2} +2x-3)\\ \\f(x)=-x^{2}-2x+3[/tex]
Convert into vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-3=-(x^{2}+2x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)-3-1=-(x^{2}+2x+1)[/tex]
[tex]f(x)-4=-(x^{2}+2x+1)[/tex]
Rewrite as perfect squares
[tex]f(x)-4=-(x+1)^{2}[/tex]
[tex]f(x)=-(x+1)^{2}+4[/tex] -------> equation in vertex form
This is the equation of a vertical parabola open down
The vertex is a maximum---------> [tex](-1,4)[/tex]
therefore
the domain of the function is the interval---------> (-∞,∞)
Domain is all real numbers
The range of the function is the interval --------> (-∞,4]
Range is all real number less than or equal to [tex]4[/tex]
Using a graphing tool
see the attached figure to better understand the problem
