Answer:
[tex]y\geq -1[/tex]
All real numbers greater than or equal to [tex]-1[/tex]
Step-by-step explanation:
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]f(x)=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 we have
[tex]f(x)=(x-4)(x-2)[/tex]
Convert to vertex form
[tex]f(x)=(x-4)(x-2)=x^{2} -2x-4x+8\\f(x)=x^{2}-6x+8[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-8=x^{2}-6x[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]f(x)-8+9=x^{2}-6x+9[/tex]
[tex]f(x)+1=x^{2}-6x+9[/tex]
Rewrite as perfect squares
[tex]f(x)+1=(x-3)^{2}[/tex]
[tex]f(x)=(x-3)^{2}-1[/tex] -------> equation in vertex form
The vertex is the point [tex](3,-1)[/tex]
[tex]a=1[/tex]
therefore
[tex]a> 0[/tex] ----> the parabola open upward ( vertex is a minimum)
The range of the function is the interval-------> [-1,∞)
[tex]y\geq -1[/tex]
All real numbers greater than or equal to [tex]-1[/tex]
see the attached figure to better understand the problem
