What is the axis of symmetry for f(x)=3(x+4)^2+1

Give that f(x)=3(x+4)^2+1
This is a vertex form y=a(x-h)^2+k
where:
a=3
h=4
k=1

since a is positive, the parabola opens up
the vertex is (4,1)
axis of symmetry is at x=-b/2a
f(x)=3x^2+24x+49
thus
x=-(24/2*4)=-4
thus the axis of symmetry is at x=-4

Is there a picture because i cannot answer it sorry