The given parabola has its vertex at point (4,0)
Let us look at the options given and find the equation which will give us vertex (4,0)
The given equations are of the form y= [tex]ax^{2},+bx+c.[/tex]
The x of vertex = [tex]\frac{-b}{2a}[/tex]
The first equation is y=[tex]x^{2}-8x+16[/tex]
a=1,b=-8,c=16.
Vertex = [tex]\frac{-b}{2a}=\frac{-(-8)}{2}=\frac{8}{2}=4.[/tex]
The x of vertex is 4.
The y of vertex can be found by substituting the x value in the equation.
y=[tex]x^{2}-8x+16=4^{2}-8(4)+16=16-32+16=0.[/tex]
The vertex =(4,0)
The graph too has the same vertex.
The graph is represented by the equation y=[tex]x^{2}-8x+16.[/tex]
