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