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anonymous

  • 4 years ago

horizontal shift of: -240sin(4θ-360) and 4cos(1/2θ-15)-2.5 Please explain how you got those answers! Thanks :)

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  1. PaxPolaris
    • 4 years ago
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    If original is y= f(x) and final is, \[y-k = a \cdot f(x-h)\]or y = a⋅f(x−h) + k h is horizontal shift k is vertical shift a is stretch/ squish

  2. anonymous
    • 4 years ago
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    So in this case, -360 would be considered the horizontal shift, and it doesn't consist of a vertical shift?

  3. PaxPolaris
    • 4 years ago
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    +360 is horizontal shift from -240sin(4θ)

  4. anonymous
    • 4 years ago
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    Why is it positive?

  5. PaxPolaris
    • 4 years ago
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    if f(x) becomes f(x-h) because now f(1) is moved to f(1+h) ... anyways if θ is in degrees it does not matter because -240sin(4θ-360) will look exactly like -240sin(4θ)

  6. anonymous
    • 4 years ago
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    So basically what ever h's sign is in the brackets, should be reversed? ex: -3cos(θ-15) would be positive 15?

  7. PaxPolaris
    • 4 years ago
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    yep

  8. anonymous
    • 4 years ago
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    Ahhh! Okay Thanks a lot!

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