General form of a quadratic equation is
[tex]ax^2+bx+c=0[/tex]another form is
[tex](x+h)(x+k_{})=0[/tex]where h and k are the number opposite by the sign of the solutions, then on this case the values of h and k are 2 and -6
[tex](x+2)(x-6)=0[/tex]Our equatio is a parabola then if we find the vertex we are finding the axis of simmetry
to find the vertex we trasnforme ou equation to the general form of a quadratic equation multipliying parenthesis
[tex]\begin{gathered} (x\times x)+(x\times-6)+(2\times x)+(2\times-6)=0 \\ x^2-6x+2x-12=0 \\ x^2-4x-12=0 \end{gathered}[/tex]now take the equation and derivate
[tex]\begin{gathered} 2x-4-0=0 \\ 2x-4=0 \end{gathered}[/tex]if we solve x we find the coordinate x of the vertex and the axis of simmetry
then
[tex]\begin{gathered} 2x-4=0 \\ 2x=4 \\ x=\frac{4}{2} \\ \\ x=2 \end{gathered}[/tex]axis of Symmetry is x=2