Question

Graph the function in the interval from 0 to 2p:

Answer 1

\[y = -4 \cos \frac{ 1 }{ 2 }q\]

Answer 2

I think your questions are a little screwy. Firstly, there's no \(\theta\) in the function, but there's a \(\theta\) on the graph, the horizontal axis.

I will assume, however, that \(\theta = q\ \pi\), because I think that's what works out...

Answer 3

\(\Large \theta = \frac{q}{\pi}\), rather..

Answer 4

Well, @ExplainItLikeImFive , what's the amplitude?

Answer 5

Working on it........

Answer 6

aaaaaaaaaaaaaand Im lost in my own work

Answer 7

i dont think it did this right at all

Answer 8

What work did you do?

Answer 9

I tried plugging the equation in, coulndt do it on my own so I gabbed my calculator, and then presumed to error my calculator.

Answer 10

Haha, sounds like a nightmare!

So, what's the amplitude? And can that rule out any options?

Answer 11

Yeah I cant figure out the amplitude. But I can rule out B, because thats what I answered and it got marked wrong.

Answer 12

\[y = -4 \cos \frac{ 1 }{ 2 }q
\\\ \ \ \ \ \ \ \ \ \uparrow\]

Answer 13

The horizontal stretching or shrinking comes from what you multiply the function by. Otherwise it would be \(1\). Amplitude is the absolute value of that multiplier. Your amplitude is....

Answer 14

4

Answer 15

Right! And that doesn't help, because all of the choices have an amplitude of 4.

But there is something. There is no horizontal shift. So, this function would be what when \(x=0\)?

Answer 16

There should be any horizontal movement?

Answer 17

I was actually going to say that - no! Because there is no extra term in the \(cos()\) function! If there was, then yes, but there's not!

Answer 18

The \(\Large\frac{1}{2}\) would cause a stretching, though!

Answer 19

And I think \(q=\theta\), simply. I'm sort of tired! :P

Answer 20

Me to. This is my last question, don'y worry.

Answer 21

Haha, alright, nice!

Answer 22

i'm going with A

Answer 23

Why?

Answer 24

Because its the right answer

Answer 25

Is it really?

Answer 26

Yes

Answer 27

Nuh-uh.

Answer 28

Because its 1Am where I am, therfore it is A.

Answer 29

im totally kidding, its B.

Answer 30

Haha, put \(x=0\) into the function and use a calculator.

Answer 31

B

Answer 32

Really?!

Answer 33

Yes

Answer 34

It's 2am here....

It's either B or D after plugging in \(x=0\), and the \(\Large\frac{1}{2}\) tells you it does cycle normally.

Answer 35

B

Answer 36

The period would be \(4\pi\). Is that so with B?

Answer 37

No

Answer 38

So its D

Answer 39

\[\frac{2\pi}{\frac{1}{2}}=4\pi\]

Ahem, final answer?

Answer 40

D

Answer 41

By the way, you've guess all but C. I just found it funny. And I think it's D. Goodnight!

Answer 42

I cannot thank you enough for all the time you spent. I'll let you know what I make on this test, tomorrow, after a good night sleep.

Answer 43

Hopefully you didn't cheat, and you know this stuff! But it seemed like an online "open book" "open internet" test, so what are you going to do? Take care! :)