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anonymous
 4 years ago
Help on two Pre calc question pleaseee???
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anonymous
 4 years ago
Help on two Pre calc question pleaseee??? xoxo attachment below

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zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Hmmm they both look like tangent functions. What is that silly little electricity looking break on the first graph? XD

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have no idea it came that way!! ;) haha any idea on how to solve them?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1So 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 xaxis. 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
 4 years ago
Best ResponseYou've already chosen the best response.1Yah 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
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1358117529784:dwThis 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
 4 years ago
Best ResponseYou've already chosen the best response.0Is the second graph a tangent function as well?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Yah 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
 4 years ago
Best ResponseYou've already chosen the best response.1Sorry if I'm super duper slow :c also helping someone else.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@zepdrix It's no problem I understnad, and you're helping me so much I'm willing to wait :)

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Let'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
 4 years ago
Best ResponseYou've already chosen the best response.0correct, I can see that :)

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Hmm this one is a bit tricky :d Let's seeeeeee....

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Remember 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
 4 years ago
Best ResponseYou've already chosen the best response.0so the equation for the graph is \[y=\tan \frac{ 1 }{ 2 }x\] ??

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Wellllll 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
 4 years ago
Best ResponseYou've already chosen the best response.0I thank you so much for the amazing help, it is appreciated!!! :D
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