Question

Help please :)

Answer 1

is this calculus question?

Answer 2

I would assume you differentiate... I assume you know calculus?

Answer 3

If you can differentiate this becomes a lot easier to solve.

Answer 4

No, sadly i don't ( ._.)

Answer 5

Okay. Go to your calculator. Graph the function.

Answer 6

you dont need to differentiate...just find the two roots of the equation (the places where t=0) and divide their sum by 2 to get the t coordinate of the highest point. then plug t into the equation to get the height

Answer 7

there is a formula which gives vertex of a quadratic function, do u know that formula?

Answer 8

Do you have a graphing calculator?

Answer 9

@hartnn . THat's a lot harder >.< .

Answer 10

But Hartnn is correct you can compete the square.

Answer 11

@etemplin . Very clever. It works but I would have never thought of that.

Answer 12

*confused* o:

Answer 13

you can use calc or use the vertex equation, -b/2a
using calc: s'= -32t+128 = 0
-32t = -128
t = 4 seconds

using vertex equation
-128/2(-16) = -128/-32
t= 4seconds

Answer 14

the x coordinate is (-b)/(2a) of the vertex of a parabola. plug in the answer for this to get the y value

Answer 15

She can't use calculus.

Answer 16

he provided 2 ways...@emily9102 use the second way of @jayz657's post

Answer 17

Ohh I know :P .

Answer 18

I like the third way more.

Answer 19

the formula (-b)/(2a) is the easiest

Answer 20

By definition, the maximum point occurs halfway between the two solutions. (ie the vertex)

Answer 21

ok

Answer 22

Wait. That's wrong.

Answer 23

Ignore that.

Answer 24

My mistake.

Answer 25

For a quadratic of the form y=ax^2 + bx + c, the vertex occurs at x = -b/2a.

Answer 26

Simple algebra error hehe...

Answer 27

So the maximum point occurs at x=4.