The height in feet of a water rocket launched from the ground is given by the function h(t) = -16t^2 + 96t where t represents the number of seconds after launch. what is the maximum heigh attained by the rocket?

The biggest height is 144 ft at t = 3 seconds.

Let's imagine this function in real life. We the rocket is launched, it will follow a path similar to a negative parabola (a 'frown', as the coefficient of x^2 is negative).

So, we can work out the answer by finding the vertex of this of this 'graph', as this represents the highest point that the rocket will reach during its journey. To find t, we use the expression -b/2a (these letters are from the format of the quadratic equation ax^2+bx+c, so in this case a= -16 and b= 96):
-b/2a
-(96)/2(-16)
-96/-32
t=3

Now, we can plug 3 back into the function:
h(t)= -16t^2+96t
h(3)= -16(3^2)+96(3)
h(3)= -16(9)+96(3)
h(3)= -144+288
h(3)= 144

Therefore, the maximum height attained by the rocket is 144ft, at t = 3 seconds.

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.