Given:
The equation describing the rocket's height after t seconds is
[tex]h(t)=-16t^2+128t+6[/tex]
To find:
The maximum height and time when it is maximum.
Solution:
We have,
[tex]h(t)=-16t^2+128t+6[/tex]
Leading coefficient is -16, which is negative. So, it is a downward parabola.
Vertex of a downward parabola is the point of maxima.
If a quadratic function is [tex]f(x)=ax^2+bx+c[/tex], then
[tex]Vertex =\left(\dfrac{-b}{2a},f\left(\dfrac{-b}{2a}\right)\right)[/tex]
In the given function, a=-16, b=128 and c=6.
[tex]\dfrac{-b}{2a}=\dfrac{-128}{2(-16)}[/tex]
[tex]\dfrac{-b}{2a}=4[/tex]
Putting x=4 in the given function, we get
[tex]h(4)=-16(4)^2+128(4)+6[/tex]
[tex]h(4)=-16(16)+512+6[/tex]
[tex]h(4)=-256+518[/tex]
[tex]h(4)=262[/tex]
So, vertex of the function is (4,262).
Therefore, maximum reached by the rocket is 262 feet it 4 seconds.
