Hello there.
To answer this question, we need to remember some properties about the vertex of quadratic functions.
Let [tex]f(x)=ax^2+bx+c,~a\neq0[/tex]. Its vertex can be found on the coordinates [tex](x_v,~y_v)[/tex] such that [tex]x_v=-\dfrac{b}{2a}[/tex] and [tex]y_v=-\dfrac{b^2}{4a}+c[/tex], in which [tex]y_v=f(x_v)[/tex].
Using the coefficients given by the question, we get that:
[tex]x_v=-\dfrac{8}{2\cdot(-1)}\\\\\\ x_v=-\dfrac{8}{-2}\\\\\\ x_v=4[/tex]
Thus, we have:
[tex]y=f(x_v)=f(4)[/tex]
So the statement is true, because the [tex]x[/tex] coordinate of the vertex of the function is equal to [tex]4.~~\checkmark[/tex]
