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Mathematics 16 Online
OpenStudy (katecc379):

Where are the asymptotes of f(x) = tan(4x − π) from x = 0 to x = pi over 2?

OpenStudy (katecc379):

OpenStudy (katecc379):

@TheSmartOne

OpenStudy (katecc379):

@Abhisar

TheSmartOne (thesmartone):

tan(4x - pi) is the same thing as tan(4x)

OpenStudy (katecc379):

okay

OpenStudy (katecc379):

i have it graphed i just don't know how to find asymptotes

TheSmartOne (thesmartone):

tan(x) has asymptotes as x = pi/2 every pi/2 tan(\(\omega\)x) has asymptotes at x = pi/(2*\(\omega\)) the period, (which is not needed for this question but helpful to know), would be \(\large\frac{\pi}{|\omega |}\)

OpenStudy (katecc379):

whats w

OpenStudy (katecc379):

would it be a?

TheSmartOne (thesmartone):

w is just like a variable, it's the greek letter omega

TheSmartOne (thesmartone):

it's \(\omega\) not w xD

TheSmartOne (thesmartone):

https://www.desmos.com/calculator/0vlxsqnuei we're looking for asymptotes in between x = 0 and x = pi/2 can you find it? :)

OpenStudy (katecc379):

haha i just didn't know how to type it but would this answer be a?

TheSmartOne (thesmartone):

There are no correct answer choices that fall within the range of x = 0 and x = pi/2

OpenStudy (katecc379):

so none of them are right? lol

TheSmartOne (thesmartone):

yeah o.O and A isn't close to being correct xD tan(\(\omega\)x) has asymptotes at x = pi/(2*\(\omega\)) tan(4x) has asymptotes at x = pi/(2 * 4) = pi/8 it has asymptotes every pi/4 start at pi/8

OpenStudy (katecc379):

are you too tired to help with 2 more its al i have left

TheSmartOne (thesmartone):

the closest answer choice for this one is D but 5pi/8 falls outside of the range but it's the closest we got xD

OpenStudy (katecc379):

ok thanks

OpenStudy (katecc379):

The tip of a 12-inch wiper blade wipes a path that is 30 inches long. What is the angle of rotation of the blade in radians to the nearest tenth?

TheSmartOne (thesmartone):

I can try. Make a new post in case I don't know how so that way someone else might be able to assist :)

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