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anonymous

  • one year ago

Precalculus help please?

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  1. anonymous
    • one year ago
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  2. jim_thompson5910
    • one year ago
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    Draw a vertical line through x = -2 where does this vertical line cross the red curve?

  3. anonymous
    • one year ago
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    I already know part a is 1, -1, 3, 4 but I don't understand part b, c, or d

  4. jim_thompson5910
    • one year ago
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    b) the domain is the set of allowed x values so basically you state all of the x coordinates of each point as one big set

  5. jim_thompson5910
    • one year ago
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    the range is the possible set of y values that can be outputs of the function

  6. jim_thompson5910
    • one year ago
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    does that make sense?

  7. anonymous
    • one year ago
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    So the domain is [-3,4]?

  8. anonymous
    • one year ago
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    and the range is [-1,4]?

  9. jim_thompson5910
    • one year ago
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    very nice

  10. jim_thompson5910
    • one year ago
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    both are correct

  11. anonymous
    • one year ago
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    okay, what about part c and d?

  12. jim_thompson5910
    • one year ago
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    h(x) = 3 is the same as saying y = 3 draw a horizontal line through y = 3. Where does it cross the red curve?

  13. anonymous
    • one year ago
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    2

  14. jim_thompson5910
    • one year ago
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    I see two points. What are those points?

  15. anonymous
    • one year ago
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    oh, -3 also

  16. anonymous
    • one year ago
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    so would I just write -3 and 2? Or is there a special notation?

  17. jim_thompson5910
    • one year ago
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    well they just want x values, so that works

  18. jim_thompson5910
    • one year ago
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    actually there's one point I missed

  19. anonymous
    • one year ago
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    4?

  20. jim_thompson5910
    • one year ago
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    yeah so if the input is x = -3, x = 2, or x = 4, then the output is 3

  21. anonymous
    • one year ago
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    okay, and for part d is it -3, -2, 0, 2, 4?

  22. jim_thompson5910
    • one year ago
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    you'll need to use inequalities here

  23. anonymous
    • one year ago
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    um okay, so -3<=x<=2 and x=4?

  24. jim_thompson5910
    • one year ago
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    much better

  25. anonymous
    • one year ago
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    Thank you so much @jim_thompson5910 ! You've been helping me with math for years and I'm in college now and my Professor speaks very broken English :(

  26. jim_thompson5910
    • one year ago
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    I'm glad to be of help, and I'm sorry you and your teacher have a communication gap

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