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korosh23

  • one year ago

Pre-calculus 12 question!

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  1. korosh23
    • one year ago
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  2. korosh23
    • one year ago
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    I know about reflections and transformations. But, I do not know what is the original quadrant for the y= f(x)

  3. jim_thompson5910
    • one year ago
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    |dw:1442628165997:dw|

  4. jim_thompson5910
    • one year ago
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    #15 a) y = f(x) is in Q4 completely |dw:1442628178659:dw| so draw any function that is just in the lower right quadrant

  5. jim_thompson5910
    • one year ago
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    since y = f(x), and because we go from f(x) to -f(x), this means that the y coordinates will flip in sign. If they are negative, they become positive, and vice versa

  6. korosh23
    • one year ago
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    How shoud I know if it is in Quadrant 4?

  7. jim_thompson5910
    • one year ago
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    it's given in the instructions

  8. korosh23
    • one year ago
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    Oh. Right

  9. jim_thompson5910
    • one year ago
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    for example, if we know (2,-10) lies on f(x), then (2,10) lies on -f(x). All I did was flip the sign of the y coordinate |dw:1442628273298:dw|

  10. korosh23
    • one year ago
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    ok

  11. korosh23
    • one year ago
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    it would become quadrant 1?

  12. jim_thompson5910
    • one year ago
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    imagine doing this for EVERY point on f(x) |dw:1442628349334:dw|

  13. jim_thompson5910
    • one year ago
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    yeah it will all be in Q1 |dw:1442628394556:dw|

  14. korosh23
    • one year ago
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    yes. Vertical reflection

  15. jim_thompson5910
    • one year ago
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    yep reflection over the x axis

  16. korosh23
    • one year ago
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    I can do the rest from now on. Thank you for your help @jim_thompson5910 :)

  17. jim_thompson5910
    • one year ago
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    no problem

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