How do find the vertex and axis of symmetry, and intercepts for a quadratic equation f(x)=x^2+12x+36?

$f \left(x\right) = {x}^{2} + 12 x + 36 = {\left(x + 6\right)}^{2}$
x-intercepts ->$x = - 6$ -> Double root. The parabola is tangent to the x axis at x = -6. (Minimum)