A rock is thrown vertically upward from the top of a cliff the equation representing this H=-5(t+3) (t-9)

i need to know how to calculate starting height, when the rock hits the ground, and the maximum height t=time h=height

Starting height = H - int = -5*3*(-9) = 135

When the rock hits the ground: On the 9th second

Because -5*(t + 3)*(t - 9) = 0, t = 9

Maximum heigtt = Vertex = (3, 180)

diff(-(5*(t+3))*(t-9), t) = - 10*t + 30

solve(- 10*t + 30 = 0)

t = 3, at this time, the rock reaches its maximum height, who is

H(3) = - 5*6*(-6) = 180

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On "Axes", change the range of Y edge from - 10 to 200, then "OK".

On "Function I Insert function", type -5*(t + 3)*(t - 9), then "OK".

Starting height is the height at t = 0, H(0) = -5(0+3)(0-9) = -5(-27) = 135

The rock hits the ground at H(t) = 0: 0 = -5(t+3)(t-9) or t = -3, 9. Since t ≥ 0, discard t = -3. The rock hits the ground at t = 9

Maximum height is when H'(t) = 0 = -5[(t-9)+(t+3) = 2t -6 or t = 3
Maximum height is H(3) = -5(6)(-6) = 180