Answer
See below.
Work Step by Step
The vertex of the quadratic function $f(x)=ax^2+bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$, so the vertex of $f(x)=x^2-6x+8$ is at $\left(-\frac{-6}{2}, f(-\frac{-6}{2})\right)=(3,f(3))=(3,3^2-6\cdot3+8=(3,9-18+8)=(3,-1)$
