Question

H=-16t^2 + 104t + 56​

Answer 1

The graph in the attached figure

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