we know that
The equation in vertex form of a vertical parabola is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex of the parabola
and the axis of symmetry is equal to the x-coordinate of the vertex
so
[tex]x=h[/tex] -----> equation of the axis of symmetry
In this problem we have
[tex]y-4x=7-x^{2}[/tex]
Convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]y-7=-x^{2}+4x[/tex]
[tex]y-7=-(x^{2}-4x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]y-7-4=-(x^{2}-4x+4)[/tex]
[tex]y-11=-(x^{2}-4x+4)[/tex]
Rewrite as perfect squares
[tex]y-11=-(x-2)^{2}[/tex]
[tex]y=-(x-2)^{2}+11[/tex]
the vertex is the point [tex](2,11)[/tex]
therefore
the answer is
The axis of symmetry is
[tex]x=2[/tex]
see the attached figure to better understand the problem
