Asked by Riley
How would you show your work when graphing one cycle of the function y = 3sin(-2x+(pi/3))-1? Thank you!
Answers
Answered by
Reiny
recall that sin(-A) = -sinA
y = 3sin(-2x+(pi/3))-1
= -3sin(2x - π/3) - 1
= -3sin(2(x - π/6) - 1
so my parent curve is y = -3sin (2x) , which you should be able to sketch, one full period from 0 to π
then move it to the right π/6, then drop it down -1
so the range would be -4 ≤ y ≤ 2
The "greenish" colour graph is your final version,
https://www.wolframalpha.com/input/?i=y+%3D+-3sin%282x%29%2C+y+%3D+-3sin%282x+-+%CF%80%2F3%29%2C+y+%3D+-3sin%282x+-+%CF%80%2F3%29+-+1
or compared to the original given equation, they coincide
https://www.wolframalpha.com/input/?i=+y+%3D+-3sin%282x+-+%CF%80%2F3%29+-+1%2C+y+%3D+3sin%28-2x%2B%28%CF%80%2F3%29%29-1
y = 3sin(-2x+(pi/3))-1
= -3sin(2x - π/3) - 1
= -3sin(2(x - π/6) - 1
so my parent curve is y = -3sin (2x) , which you should be able to sketch, one full period from 0 to π
then move it to the right π/6, then drop it down -1
so the range would be -4 ≤ y ≤ 2
The "greenish" colour graph is your final version,
https://www.wolframalpha.com/input/?i=y+%3D+-3sin%282x%29%2C+y+%3D+-3sin%282x+-+%CF%80%2F3%29%2C+y+%3D+-3sin%282x+-+%CF%80%2F3%29+-+1
or compared to the original given equation, they coincide
https://www.wolframalpha.com/input/?i=+y+%3D+-3sin%282x+-+%CF%80%2F3%29+-+1%2C+y+%3D+3sin%28-2x%2B%28%CF%80%2F3%29%29-1
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