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samalami
Group Title
Help on two Pre calc question pleaseee???
xoxo
attachment below
 one year ago
 one year ago
samalami Group Title
Help on two Pre calc question pleaseee??? xoxo attachment below
 one year ago
 one year ago

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

samalami Group TitleBest ResponseYou've already chosen the best response.0
I have no idea it came that way!! ;) haha any idea on how to solve them?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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 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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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:
 one year ago

samalami Group TitleBest 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 :(
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw: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
 one year ago

samalami Group TitleBest ResponseYou've already chosen the best response.0
Is the second graph a tangent function as well?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Sorry if I'm super duper slow :c also helping someone else.
 one year ago

samalami Group TitleBest 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 :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

samalami Group TitleBest ResponseYou've already chosen the best response.0
correct, I can see that :)
 one year ago

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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,\[\large \frac{\pi}{\color{#F35633}{b}}=2\pi \qquad \rightarrow \qquad \color{#F35633}{b}=\frac{1}{2}\]
 one year ago

samalami Group TitleBest ResponseYou've already chosen the best response.0
so the equation for the graph is \[y=\tan \frac{ 1 }{ 2 }x\] ??
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

samalami Group TitleBest ResponseYou've already chosen the best response.0
I thank you so much for the amazing help, it is appreciated!!! :D
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Np salami girl :3
 one year ago
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