Question 1159743
Consider the parabola 

{{{y=x^2-6x+4}}}

What is the axis of symmetry? 

The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola.

so, write your equation in vertex form: {{{y=a(x-h)^2+k}}}

{{{y=(x^2-6x)+4}}}....complete square

{{{y=(x^2-6x+b^2)-b^2+4}}}......{{{b=6/2=3}}}

{{{y=(x^2-6x+3^2)-3^2+4}}}

{{{y=(x-3)^2-9+4}}}

{{{y=(x-3)^2-5}}}->{{{h=3}}} and {{{k=-5}}}, so x -coordinate of the vertex is


{{{x=3}}}


Compute the coordinates of the vertex: done above

 ({{{3}}},{{{-5}}})


compute the y-intercept as a point: 

y-intercept where {{{x=0}}}

{{{y=0^2-6*0+4}}}
{{{y=4}}}

({{{0}}},{{{4}}})

find the point symmetric to the y-intercept

axis of symmetry is {{{x=3}}}, so axis is three units to the right from the y-intercept, and the point symmetric to the y-intercept will be three more units to the right from axis at:

({{{6}}},{{{4}}})

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(6,4,.12), locate(6,4,p(6,4)),green(line(3,10,3,-10)),
 graph( 600, 600, -10, 10, -10, 10, x^2-6x+4)) }}}