Help on two Pre calc question pleaseee???
xoxo
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- anonymous

Help on two Pre calc question pleaseee???
xoxo
attachment below

- schrodinger

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- anonymous

##### 1 Attachment

- zepdrix

Hmmm they both look like tangent functions. What is that silly little electricity looking break on the first graph? XD

- anonymous

I have no idea it came that way!! ;) haha any idea on how to solve them?

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## More answers

- zepdrix

So in the first one, we can figure out the period of the function based on how long it takes to repeat.
I guess the easiest point to look at would just be where it crosses the x-axis.
So it crosses at \(x=6\pi\) then again at \(x=7\pi\).
So that function would have a period of \(\pi\).
Hmm you'll want to remember some of your basic trig graphs.
I think this one lines up with the tangent function.
But ummmm let's think about what's happening by the origin so we can make sure it lines up correctly.
Tangent goes through the origin. Would this function as well? :o

- zepdrix

Yah it looks like it. It'll go through \(5\pi\), \(4\pi\), \(3\pi\), \(2\pi\), \(\pi\) and \(0\pi\).
I'm pretty sure the first graph is just \(y=\tan x\).
It doesn't appear to have any transformations applied to it. Hmmmm D:

- anonymous

@zepdrix that would make sense actually, we've been doing a lot of work on finding equations from Graphs, it's just hard for me to understand :(

- zepdrix

Hmm :c

- zepdrix

|dw:1358117529784:dw|This first one is kinda tricky. I think they wanted you to try and figure out which function it is without knowing exactly what it looks like about the origin.
If you know what it looks like about the origin you can probably just match it up with one of the trig graphs that you remember.
So in this first one, they want you to notice the SHAPE that it has (it still is the tangent function based on it's shape) and label any transformations that have been applied to it (shifts to the right, left, up, down, or scaling).
Hmmmmmmmm let's check out this second one a sec :O

- anonymous

okay! :D

- anonymous

Is the second graph a tangent function as well?

- zepdrix

Yah it has that same shape doesn't it? Hmm
But we have a small problem, instead of passing through the origin, the function has an asymptote at the origin, which means it has been altered a little bit.
It'll still end up being \(y=\tan x\), but with some alteration applied to it :D Let's see if we can figure it out.

- zepdrix

Sorry if I'm super duper slow :c also helping someone else.

- anonymous

@zepdrix It's no problem I understnad, and you're helping me so much I'm willing to wait :)

- zepdrix

Let's look at the point where it crosses the origin again.
It crosses at \(-\pi\) and then repeats at \(\pi\).
So THIS function has a period of \(2\pi\).
Twice the size of the last one.

- anonymous

correct, I can see that :)

- zepdrix

Hmm this one is a bit tricky :d Let's seeeeeee....

- zepdrix

Remember how to integrate the period into the function?\[\large y=\tan \color{#F35633}{b}x\]The period of this function is, \(\large \frac{\pi}{\color{#F35633}{b}}\)
Andddd we determined that the period is \(2\pi\),
Solving for b gives us,\[\large \frac{\pi}{\color{#F35633}{b}}=2\pi \qquad \rightarrow \qquad \color{#F35633}{b}=\frac{1}{2}\]

- anonymous

so the equation for the graph is \[y=\tan \frac{ 1 }{ 2 }x\] ??

- zepdrix

Wellllll it looks like it also has a shift applied to it, shifting it to the right (or left I suppose..). Instead of passing through the origin it's passing through \(\pi\), so it has been shifted to the right \(\pi\).
\[\large y=\tan\left[\frac{1}{2}\left(x-\pi\right)\right]\]I think this is the answer we're trying to get. Ugh I'm not 100% confident in that. I wish we had an answer key to check it against :D

- anonymous

I thank you so much for the amazing help, it is appreciated!!! :D

- zepdrix

Np salami girl :3

- anonymous

;)

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