Answer:
\(y=2(x-3)^2-12\)
Step-by-step explanation:
The vertex form of a parabola is
\(y=a(x-h)^2+k\) ...(i)
where, a is a constant and (h,k) is vertex.
It is given that vertex of the parabola is (3,-12).
Substitute h=3 and k=-12 in equation (i).
\(y=a(x-(3))^2+(-12)\)
\(y=a(x-3)^2-12\) ...(ii)
It is given that the parabola passes through the point (0,6). It means the equation of parabola must be true for (0,6).
Substitute x=0 and y=6 in equation (ii).
\(6=a(0-3)^2-12\)
\(6+12=9a\)
\(18=9a\)
\(2=a\)
Substitute a=2 in equation (ii).
\(y=2(x-3)^2-12\)
Therefore, the equation of parabola is \(y=2(x-3)^2-12\).
