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anonymous

  • one year ago

how do you find the range and y intercept of y=3sin(2x)+2

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  1. jim_thompson5910
    • one year ago
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    to find the y intercept, plug in x = 0 and evaluate

  2. anonymous
    • one year ago
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    y=2?

  3. jim_thompson5910
    • one year ago
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    yes

  4. jim_thompson5910
    • one year ago
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    the range of sin(x) spans from -1 to +1 in other words, \[\Large -1 \le \sin(x) \le 1\] the x can be replaced with anything you want, so let's replace x with 2x \[\Large -1 \le \sin(2x) \le 1\]

  5. jim_thompson5910
    • one year ago
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    we can then multiply every side by 3 \[\Large -1*3 \le 3*\sin(2x) \le 3*1\] \[\Large -3 \le 3\sin(2x) \le 3\]

  6. jim_thompson5910
    • one year ago
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    and then finally add to 2 all sides \[\Large -3+2 \le 3\sin(2x)+2 \le 3+2\] \[\Large -1 \le 3\sin(2x)+2 \le 5\] so the range of y = 3*sin(2x)+2 spans from -1 to +5 (including both endpoints)

  7. anonymous
    • one year ago
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    Just to clarify, we must assume always that -1<sin(x)<1?

  8. jim_thompson5910
    • one year ago
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    yes sin(x) is a function that bounces up and down. It doesn't go past y = 1 or y = -1. It's boxed in so to speak in terms of the y direction

  9. jim_thompson5910
    • one year ago
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    |dw:1444537353436:dw|

  10. anonymous
    • one year ago
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    and how would we find the range?

  11. jim_thompson5910
    • one year ago
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    do you see my steps above and how they led to \(\LARGE \Large -1 \le 3\sin(2x)+2 \le 5\) all that work shows how to find the range for y = 3sin(2x)+2

  12. anonymous
    • one year ago
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    alright so it would be (-1,5)

  13. jim_thompson5910
    • one year ago
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    [-1,5] actually

  14. jim_thompson5910
    • one year ago
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    we're including the two endpoints

  15. anonymous
    • one year ago
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    ok, and why are we adding 2 on both sides? I missed that

  16. jim_thompson5910
    • one year ago
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    because we initially had just `3sin(2x)` in the middle (without the +2) adding 2 to all sides will have `3sin(2x)+2` in the middle (now with the +2)

  17. anonymous
    • one year ago
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    ok

  18. jim_thompson5910
    • one year ago
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    make sense?

  19. anonymous
    • one year ago
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    yep I just need to review it! Thanks again Jim :)

  20. jim_thompson5910
    • one year ago
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    no problem

  21. anonymous
    • one year ago
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    :-)

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