Answer:
[tex]f(x)=-(x-4)^{2}+5[/tex]
Step-by-step explanation:
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 a>0 -----> the parabola open upward (vertex is a minimum)
if a<0 -----> the parabola open downward (vertex is a maximum)
Verify each case
case A) [tex]f(x)=(x-4)^{2}+5[/tex]
The vertex is the point [tex](4,5)[/tex]
[tex]a=1[/tex]
a>0 -----> the parabola open upward (vertex is a minimum)
The range is the interval--------> [5,∞)
[tex]y\geq5[/tex]
case B) [tex]f(x)=-(x-4)^{2}+5[/tex]
The vertex is the point [tex](4,5)[/tex]
[tex]a=-1[/tex]
a<0 -----> the parabola open downward (vertex is a maximum)
The range is the interval--------> (-∞,5]
[tex]y\leq 5[/tex]
case C) [tex]f(x)=(x-5)^{2}+4[/tex]
The vertex is the point [tex](5,4)[/tex]
[tex]a=1[/tex]
a>0 -----> the parabola open upward (vertex is a minimum)
The range is the interval--------> [4,∞)
[tex]y\geq4[/tex]
case D) [tex]f(x)=-(x-5)^{2}+4[/tex]
The vertex is the point [tex](5,4)[/tex]
[tex]a=-1[/tex]
a<0 -----> the parabola open downward (vertex is a maximum)
The range is the interval--------> (-∞,4]
[tex]y\leq 4[/tex]
