Question

sketch 2 periods of the following function: y=tan(2x)+3. showing all critical points. I mostly need help with labeling the x-axis. I always get confused when there's a phase shift.

Answer 1

critical points? you doing caculus now?

Answer 2

yea i have a final exam tomorrow and this is one of the things i don't understand

Answer 3

basically she wants us to show all the points for the whole 2 periods

Answer 4

there is no phase shift here because it is \(\tan(2x)\) and not \(\tan(2x+c)\)

Answer 5

tangent is periodic with period \(\pi\) and has asymptotes at \(y=-\frac{\pi}{2}\) and \(y=\frac{\pi}{2}\)

Answer 6

if you want your new asymptotes, rather than memorizing some formula, set
\(2x=-\frac{\pi}{2}\) and \(2x=\frac{\pi}{2}\) and solve for \(x\). that will give them to you

Answer 7

you will see that your function looks just like tangent, only instead of having period \(\pi\) it has period \(\frac{\pi}{2}\) so you might want to label the \(x\) axis at the points \(-\frac{\pi}{4},\frac{\pi}{4}, \frac{\pi}{2}\) and graph the two periods there

Answer 8

of course it is also shifted up 3 units because of the \(+3\) at the end
here is a nice picture, notice the units along the x axis
http://www.wolframalpha.com/input/?i=tan%282x%29%2B3

Answer 9

why did u set up \[\Pi/2=2x\]

Answer 10

@satellite73 you there?

Answer 11

i set it up because i know the basic graph of \(y=\tan(x)\) it has asymptotes at \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\)

Answer 12

here is what i have in my mind
http://www.wolframalpha.com/input/?i=tan%28x%29

Answer 13

so starting with that basic graph, we now translate with
\[y=\tan(2x)+3\] the \(+3\) at the end means we shift up 3 units
as for the asymptotes, we have \(2x\) not \(x\) as the input, so i set \(2x=-\frac{\pi}{2}\) solve for \(x\) get \(x=-\frac{\pi}{4}\) so that is the left hand asymptote

Answer 14

right hand endpoint i put \(2x=\frac{\pi}{2}\) get \(x=\frac{\pi}{4}\) so that is the right hand asymptote
this also tells us the period is \(\frac{\pi}{2}\) the distance between the two points.
we could halve also used that the period of
\[\tan(bx)\] is \(\frac{\pi}{b}\)

Answer 15

*have

Answer 16

oh ok, i think i get it

Answer 17

check out both links i sent and make sure it is clear how to get from the second one, the basic graph, to the first one, your answer

Answer 18

ok, i mostly don't understand why the graph changed from \[\Pi/2\] to \[\Pi/4\]

Answer 19

ok you do understand that one asymptote of tangent is at \(\frac{\pi}{2}\) right? because tangent is sine over cosine, and cosine is undefined there

Answer 20

but you do not have \(\tan(x)\) you have \(\tan(2x)\)

Answer 21

so suppose \(x=\frac{\pi}{4}\) then
\[\tan(2x)=\tan(2\times\frac{\pi}{4})=\tan(\frac{\pi}{2})\] is undefined. so that is your new asymptote

Answer 22

why is it undefined? i know tangent is sine over cosine but i'm still confused

Answer 23

another way to think about it is that as \(x\) goes from \(-\frac{\pi}{4}\) to \(\frac{\pi}{4}\) we have \(2x\) goes from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) one full period for tangent

Answer 24

ok
\[\tan(x)=\frac{\sin(x)}{\cos(x)}\]
\[\tan(\frac{\pi}{2})=\frac{\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})}=\frac{1}{0}\]is undefined

Answer 25

because you cannot divide by 0
so \(\frac{\pi}{2}\) is not in the domain of tangent, and that is why it has a vertical asymptote there

Answer 26

oh ok i see

Answer 27

tangent is undefined anywhere cosine is 0 so it is undefined at
\(-\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, ...\)