Question 323500
Complete the square to put the equation into vertex form, {{{y=a(x-h)^2+k}}}, where (h,k) is the vertex.
{{{y=x^2-6x+10}}}
{{{y=x^2-6x+9+10-9}}}
{{{y=(x-3)^2+1}}}
The vertex is (3,1).
The axis of symmetry is x=3 so all point equidistant from the axis are symmetric.
{{{x=4}}} and {{{x=2}}}
{{{y=(4-3)^2+1=2}}}
(4,2) and (2,2)
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{{{drawing(300,300, -5,11,-3,13, circle(3,1,.2),circle(4,2,.2),circle(2,2,.2),grid(1),blue(line(3,-20,3,20)),graph(300,300,-5,11,-3,13, (x-3)^2+1))}}}