Answer:
The correct options are A, B and E.
Step-by-step explanation:
The given function is
\(p(x)=6x^2+48x\)
Vertex form of the function is
\(p(x)=6(x+4)^2-96\) ..... (1)
The vertex form of a parabola is
\(f(x)=a(x-h)^2+k\) ..... (2)
where, (h,k) is vertex and x=h is the axis of symmetry.
If a>0, then parabola opens up and if a<0, then parabola opens down.
If |a|>1, then graph is narrower than the graph of \(f(x)=x^2\) and if |a|<1, then graph is wider than the graph \(f(x)=x^2\) .
From (1) and (2), it is clear that
\(a=6, h=-4, k=-96\)
So we conclude that
1. The axis of symmetry is the line x = –4.
2. The graph is narrower than the graph of \(f(x)=x^2\).
3. The parabola has a minimum.
4. The parabola opens up.
5. The value of k is –96, so there is a vertical shift down 96 units.
Therefore options A, B and E are correct.
