First, simplify the expression of the function [tex]f(x)=x^2-4x+6[/tex] highlighting the perfect square:
[tex]f(x)=x^2-4x+4-4+6=(x-2)^2+2.[/tex]
Then find the vertwx of parabola. When x=2, f(2)=2 and the coordinates of the vertex are (2,2). The axis of symmetry passes through the vertex and is parallel to the y-axis. Therefore, its equation is x=2 (see attached graph for details).
Answer: x=2.
