Trying to enter into a TI-84 -16t^2+96t+5?

Question

The height (s) of a ball (in feet) thrown with an initial velocity of 96 feet per second from an initial height of 5 feet is given as a function of the time t(in seconds) by s(t) = -16t^2+96t+5.

When I try and enter that equation as a Y1= I just get a straight line vice a curve as the graph in the example depicts. What am I doing wrong?
Thank you in advance for your help.

What should my Tableset be then? I have it set as Table Start 1, Delta TBL .1

Answer 1

Try clicking all the keys shown below;

[on] [y=] [(-)] [1] [6] [x,t,o,n] [^] [2] [+] [9] [6] [x,t,o,n] [+] [5] [enter] [window] [0] [enter] [1] [0] [enter] [1] [enter] [0] [enter] [2] [5] [0] [enter] [graph]

Now on screen you should have a negative parabola with the ball reaching a maximum height when t = 3 seconds.

Answer 2

It worked out good for me. It is an inverted parabola.
Use this in matlab

clc;
for t=0:.001:10
s = -16*t^2+96*t+5;
plot (t,s);
hold on;
if s<=0
t
break;
end;
end

Answer 3

Try pressing the 'trace' button. It should display the equation in the upper right hand corner. Also, use the 'table' key to see what's there.

Answer 4

you gotta zoom out the graph, thats a parabole...whatever (i am spanish speaking so its hard to translate math concept) but either way the lowest part is so low that you only see a specific point.

Answer 5

Use [Math] > 0:Solver (or B:Solver). enter the equation solved for 0, hit [enter], and ask your self in a guess. as quickly as you do, hit [Alpha][enter](resolve), and the calculator will provide its ultimate guess as to what the variable is comparable to.

Answer 6

Your window is messed up, it is a parabola but you probably wont see it unless u fiddle with the window settings

Answer 7

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