Answer:
Option B [tex]f(x)=-x^{2}+12x-32[/tex]
Step-by-step explanation:
we know that
For the given values, the quadratic function is a vertical parabola open downward (vertex is a maximum)
The equation in vertex form is equal to
[tex]f(x)=a(x-h)^{2} +k[/tex]
where
(h,k) is the vertex
a is a coefficient
we have
(h,k)=(6,4)
so
[tex]f(x)=a(x-6)^{2} +4[/tex]
Find the value of a
For x=8, y=0 -----> the y-intercept
substitute
[tex]0=a(8-6)^{2} +4[/tex]
[tex]0=a(2)^{2} +4[/tex]
[tex]0=4a +4[/tex]
[tex]a=-1[/tex]
substitute
[tex]f(x)=-(x-6)^{2} +4[/tex]
Convert to standard form
[tex]f(x)=-(x^{2}-12x+36) +4[/tex]
[tex]f(x)=-x^{2}+12x-36+4[/tex]
[tex]f(x)=-x^{2}+12x-32[/tex]
The graph in the attached figure
