Question 4271
For a graph of a parabola that opens up or down in the form 

{{{y = ax^2 + bx + c}}}

the vertex and therefore the axis of symmetry is always at

{{{x = (-b)/2a}}}.  This formula comes from the quadratic formula!! {{{x=(-b+-sqrt (b^2-4ac))/(2a)}}}


Therefore, the axis of symmetry for 
{{{y = x^2 - 6x + 7}}}  where a=1, b=-6, and c=7
is {{{x=(-(-6))/(2(1))}}} or x=3


R^2 at SCC