Long- 4,5 part questions (Who's up for the challenge) ^~^
determining the period of sec/csc function is|dw:1521817537805:dw| little tricky but
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it looks like one repeating unit of the question starts at -pi/2 and ends at about pi, what would the period be?
just count in between right
actually just a minute, we would start at -pi/2 and go all the way until the graph extends to -infinity, completing the cycle
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|dw:1521817748454:dw|
so it's really just -pi/2 to pi/2, then continuing to pi/2 to -pi/2 wrapping around, so adding up all the pieces we get 2pi as the period (this is kind of hard to see but you would sum up the x-values of all the parts since only one period is being represented so yeah period = 2pi
anyway, for part II, let's try drawing straight lines where the period starts and ends like so:
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so if it the period starts at -pi/6 and ends at 2pi/3 what's the period?
4
Ugh
the entire period is just ending point - starting point so (2pi/3) - (-pi/6) = ?
5pi/6
actually my mistsake, the period ends at (5pi/6) to be a bit more precise making the period (5pi/6) - (-pi/6) = pi so the period is pi
okay so for part 3 it would be sec ?
yup, I'm not really sure what they'd want for the explanation but you could mention it has the same period as cos, has asymptotes where the zeroes of cos would be, has the same symmetry, etc.
Got it for part 4 it would be 3?
yup, 3 since the new vertex is (0,3)
for part 5 i'm a little stumped
well, if the period changed from 2pi to pi, what coefficient of x would we need to produce a period of pi instead of 2pi?
remember: period = 2pi/b where b is the coefficient of x
2
awesome, so the dotted function = sec(2x) which is all they need for part V
for part 6 i would just use the line right
what line
oh nvm
I think we mentioned it before but the solid line is sec(x) so the recprical would be cos(x) = your answer
Yay thank you
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