Something funny is that the x value of the vertex lies directl in the middle of the x intercepts
so
we see the x intercepts or 0's at x=8 and 2
the average is x=5
so find f(5) to find the y value of the vertex
f(5)=(5-8)(5-2)
f(5)=(-3)(3)
f(5)=-9
vertex is at (5,-9)
the actual way the teacher wants is to expand then compltete the square to get into the form f(x)=a(x-h)^2+k where the vertex is (h,k)
but whatever
verrtex is at (5,-9)
I am the goddes of the air pods and the books of the laptops
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em>