Why Does Graphing and Completing the Square Give Different Maximum Heights?

[Markd]

I don't think this is corrent...

Question By completiing the square determine the maximum height of the ball and how long it takes to reach the maximum height

h=-5t^2+30t+1.5

Work:
=t^2-6t+1.5/5
(t^2-6t+9)1.5/-5 + 9
(t-3)^2+43.5/5
That would mean the time of the ball to get to the max height is 3 seconds but when I graph it I get 6 seconds?

 

[dextercioby]

You've completed the square for another function... Not for the one you initially posted...Where did you put that "-5" you factored.??

Daniel.

 

[HallsofIvy]

I don't think this is corrent...

Question By completiing the square determine the maximum height of the ball and how long it takes to reach the maximum height

h=-5t^2+30t+1.5

Work:
=t^2-6t+1.5/5
(t^2-6t+9)1.5/-5 + 9

that should be h/5= (t^2- 6t+9)- 1.5/5+ 9

(t-3)^2+43.5/5
That would mean the time of the ball to get to the max height is 3 seconds but when I graph it I get 6 seconds?

What is h when t= 6?
h(6)= -5(36)+ 30(6)+ 1.5= -31+ 30+ 1.5= -1+ 1.5= 0.5 m. I hardly think that is the maximum height!
What exactly did you graph?

When I graph h= -5t²+ 30t+ 1.5, I get the maximum (h=46.5) at t= 3.