Question

can someone tell me how to graph this equation on my graphing calculator: 300(2+sin(4sqrt x+0.15)) for 0 15 years ago

Answer 1

You just do it...
What type of calculator do you have?

Answer 2

ti 83

Answer 3

Y= --> Enter: 300(2 + sin(4 sqrt(x + 0.15))
Then, to see the region from 0<x<20:
Window (at the top) --> Xmin = 0; Xmax = 20; Ymin = -20 Ymax = 20

Answer 4

Might have to change the Ymin / Ymax values - I'm just guessing :D

Answer 5

Become a fan DX

Answer 6

give me a sec and let me try it

Answer 7

what should mt table set be

Answer 8

Do you need the table? If so --> 2nd Window (for table settings)
Tbl start = 0; triangleTable (change in table) = 1 or 0.5

Answer 9

it didnt work

Answer 10

Why's that - what did you enter on the calculator?

Answer 11

Zoom out then.

Answer 12

Oh yeah, our Ymax / Ymin values are not right at all.
Make Ymax = 900 and Ymin 200

Answer 13

nothing is graping

Answer 14

Do what I said above.

Answer 15

Btw is that sqrt(x) + 0.15 or sqrt(x + 0.15)
I'm guessing it's the former.

Answer 16

hey it worked good job but why did u choose 200 and 900

Answer 17

Cause on the table it starts out over 200 and then it goes up to a max of 900 ish.

Answer 18

I have attached the graph. This is the accurate graph, and is drawn to scale

Answer 19

What do you open up that file with?

Answer 20

May be you couldn't open the file. Ok here it is

Answer 21

hey ty u for sending it over but i do not understand how to choose my max and min

Answer 22

for the y

Answer 23

Could anyone open the file?

Answer 24

yah i did i saw it ty

Answer 25

does anyone know how to choose 200 and 900

Answer 26

why it has to be 200 and 900

Answer 27

are you asking me?

Answer 28

yah

Answer 29

For that you will have to find the derivative, and equate it to zero

Answer 30

are you getting me?

Answer 31

yah . do u know how to find the integral on the calculator

Answer 32

No I don't, but I can provide you that here

Answer 33

If you wish.........

Answer 34

if u know plez show

Answer 35

Here it is attached

Answer 36

And I am attaching the graph in jpg format, as last time it was not openeable

Answer 37

yah i saw it

Answer 38

Did your problem got solved or it increased? :)

Answer 39

kind of ty