we know that
The equation of a vertical parabola in vertex form is equal to
\(y=a(x-h)^{2}+k\)
where
(h,k) is the vertex
if \(a > 0\) -----> then the parabola open upward (the vertex is a minimum)
if \(a < 0\) -----> then the parabola open downward (the vertex is a maximum)
The axis of symmetry is equal to
\(x=h\)
In this problem let's analyze two cases
First case
\(f(x)=3(x-2)^{2}+4\)
the vertex is the point \((2,4)\)
\(a=3\)
so
\(3 > 0\) -----> then the parabola open upward (the vertex is a minimum)
The axis of symmetry is equal to
\(x=2\)
Second case
\(f(x)=3(x-2)^{2}-4\)
the vertex is the point \((2,-4)\)
\(a=3\)
so
\(3 > 0\) -----> then the parabola open upward (the vertex is a minimum)
The axis of symmetry is equal to
\(x=2\)
