how do i label the axis of semmetry and the vertex?

Expert Answers
sciencesolve eNotes educator| Certified Educator

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:

`x=-b/(2a), y=(4ac-b^2)/(4a)`

Equating the coefficients of like powers of `f(x)=ax^2+bx+c`  and `f(x)=x^2-6x+4`  yields:

`a=1 `



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=-(-6)/2, y=(16-36)/4`

`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).`