Question

H=-16t^2 + 104t + 56​

Answer 1

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$h=-16t^{2}+104t+56$

Using a graphing tool

see the attached figure

This is the equation of a vertical parabola open downward

The vertex is the point (3.25,225) (a maximum)

The zero's of the function or the t-intercepts are the points (-0.5,0) and (7,0)

The h-intercept of the function is the point (0,56)

The domain of the function are all real numbers

The range of the function $h\leq225$ ----> All real numbers less than or equal to 225