help? when i graphed this the graph turned out really wild, and i dont know how else to solve it Where are the asymptotes of f(x) = tan(4x − π) from x = 0 to x = pi over 2 ? (1 point) x = pi over 4 , x = 3 pi over 4 x = 0, x = pi over 4 x = pi over 2 , x = 3 pi over 2 x = 3 pi over 8 , x = 5 pi over 8
asympotes are where the input is equal to \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) solve this by solving \[4x-\pi=-\frac{\pi}{2}\] and \[4x-\pi=\frac{\pi}{2}\]
steps to solve both are the same a) add \(\pi\) to both sides then b) divide both sides by \(4\)
here is a nice picture if you need one http://www.wolframalpha.com/input/?i=tan(4x-pi)+domain+0..pi%2F2
@satellite73 ok so for the first one i got 1.78539.... and .39269..... are these right??
i would leave the answers with pi's in them, since those are your choices
oh ok so how would it be written as a fraction?
lets take it slow
\[4x-\pi=-\frac{\pi}{2}\]solve for \(x\) step one add \(\pi\) ti both sides
\[4x=\pi-\frac{\pi}{2}\] so i guess the question is, what is \(\pi-\frac{\pi}{2}\) do you know it ? (no is a fine answer, just asking)
i know as a decimal it would be 1.57079.....
i meant as a fraction
here is an easier question, what is \[1-\frac{1}{2}\]
1/2, right?
yes, right
so by the same arithmetic \[\pi-\frac{\pi}{2}=\frac{\pi}{2}\]
hmm now that i look at your answer choices, i am not sure i like any of them, so i wonder what is going on allow me to solve the first one for you, \[4x-\pi=-\frac{\pi}{2}\] first add \(\pi\) to get \[4x=-\frac{\pi}{2}+\pi=\frac{\pi}{2}\] then divide by \(4\) to get \[x=\frac{\pi}{8}\]
but that is not one of your answer choices, so lets try \[4x-\pi=\frac{\pi}{2}\] add \(\pi\) get \[4x=\frac{\pi}{2}+\pi=\frac{3\pi}{2}\]
divide by \(4\) get \(x=\frac{3\pi}{8}\)
which at least has the benefit of being one of your choices i wonder what happened to \(\frac{\pi}{8}\)?
would that be the 5pi over 8???
i guess it you solve \[4x-\pi=\frac{3\pi}{2}\] you get it , yes
question has an error in it , don't fret about it, not worth your effort pick the last one and move on
oh, ok!! thank you for all your help!!! do you think you can also help me with another one im having trouble with?
sure no problem always happy to help a squid
ok!! ill post it as a separate question so i can give you another medal for it
not necessary, i have plenty
but you can if you like
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