We have the parabola:
[tex]y=x^2-10x+28[/tex]There is one expression that gives you the x-coordinate of the vertex:
[tex]-\frac{b}{2a}[/tex]We know that the standard form of a quadratic function is: ax^2+bx+c, then:
[tex]-\frac{(-10)}{2(1)}=\frac{10}{2}=5[/tex]Now, having x=5 we can substitute it on the original equation:
[tex]\begin{gathered} y=(5)^2-10(5)+28 \\ y=25-50+28 \\ y=3 \end{gathered}[/tex]The vertex of the parabola is (5, 3)
