You need to notice that the graph of function `f(x)=x^2-6x+4` is a parabola, hence the axis of symmetry of parabola passes through vertex of parabola.
You need to remember that vertex of parabola `f(x)=ax^2+bx+c` may be found using the next formulas such that:
Equating the coefficients of like powers of `f(x)=ax^2+bx+c` and `f(x)=x^2-6x+4` yields:
You need to substitute `1` for `a` , `-6` for`b ` and `4` for`c` in `x=-b/(2a), y=(4ac-b^2)/(4a)` such that:
`x=3 , y=-5`
Hence, the axis of symmetry is `x=3` , it is parallel to y axis and it passes through vertex of parabola `(3,-5).`