A parabola equation written in vertex form is given by
[tex]f(x)=a(x-h)^2+k[/tex]
where (h, k) represents the coordinates of the vertex. Our equation is
[tex]f(x)=x^2-8x+80[/tex]
To rewrite this function in vertex form, we need to use the complete the square method. When we expand the binomial of a difference, we have
[tex](x-a)^2=x^2-2ax+a^2[/tex]
If we compare the expanded form with our function, we have the following correspondence
[tex]-8=-2a\implies a=4[/tex]
Then, we can rewrite our expression as
[tex]\begin{gathered} f(x)=x^2-8x+80 \\ =x^2-8x+16+64 \\ =(x)^2-2(4)(x)+(4)^2+64 \\ =(x-4)^2+64 \end{gathered}[/tex]
Comparing this equation to the vertex form, the vertex of our parabola is
[tex](4,64)[/tex]
And the equation is
[tex]f(x)=(x-4)^2+64[/tex]
The graph of this parabola is