Question

Could someone explain to me how I can put this equation into my graphing calculator? I have tried everything, and I just don't understand it.
y=-5secX I cant get it to graph. Any step by step explanation would help. Thank you!

Answer 1

What graphing calculator do you have?

Answer 2

I am using a TI-83 Plus

Answer 3

I know how to add equation but since it SEC I am somehow not entering the correct thing

Answer 4

Remember that sec(x)=1/cos(x)

Answer 5

Try entering y=-5/cos(x) instead; it should produce the same graph.

Answer 6

you are awesome! I actually see what I did wrong now. Thank you so much! Could you explain one more to me if you got time?

Answer 7

I could try. :)

Answer 8

y=cot(1/6x)

Answer 9

cot(x)=1/tan(x)

Answer 10

I got that far but what does it turn into? cot1/6x = 1/tan(6x) ?

Answer 11

Is it (1/6)*x or 1/(6x)?

Answer 12

y=cot(1/6x)

Answer 13

cot(1/6x)=1/tan(1/6x)

Answer 14

it gives me a wrong graph

Answer 15

You have to be very careful using writing out what you mean, use a lot of parenthesis. :D
What I wrote above can be interpreted two ways;
1/tan(1/(6x)) and 1/tan((1/6)*x)

Answer 16

\[\frac{1}{\tan(\frac{1}{6x})}\] or \[\frac{1}{\tan(\frac{1}{6}x)}\]

Answer 17

I tried both of them and neither looks like the correct graph on my homework. Its an online class and this is one of the questions I had wrong so I can see what the right graph is supposed to look like.

Answer 18

tried to ask my prof. but she was no help at all!

Answer 19

Maybe you got a strange scaling on?

Answer 20

Usually with calculators you have to change the intervals.

Answer 21

the pics on my calculator are nothing close to the image on the homework. let me try again tho..see if it changes anything

Answer 22

Check if the boundaries are set proper; if this does not yield any enlightenment, I'm not sure I can help you further. :(

Answer 23

..as I don't know what the problem could be then. :D

Answer 24

No you did great! It helped. I changed the window a couple times and got it to look how it should. It wouldn't do it on its own I had to input the max and min myself! Thank you so much!! Hopefully I can do better on my test on Monday now! :)

Answer 25

Glad I could help. Good luck with your test! :D

Answer 26

Thank you! :)