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anonymous

  • one year ago

Create a graph with: Domain: -∞<x<∞ Range: -5<y<2

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  1. anonymous
    • one year ago
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    seems like you should fit your graph between the 2 dashed lines. Maybe use them as asymptotes|dw:1441670506468:dw|

  2. anonymous
    • one year ago
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    I've done that but i don't understand what to do next. The domain is confusing me.

  3. Hero
    • one year ago
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    Label the graph properly afterwards. Label your x axis, y axis, and label negative infinity, positive infinity and -5, and 2 on it properly.

  4. SolomonZelman
    • one year ago
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    Cool \((7/2)\cos(x)-1.5\)

  5. SolomonZelman
    • one year ago
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    7/2cos(x) makes it a range of 7 units long. and 1.5 shift down to adjust the interval

  6. SolomonZelman
    • one year ago
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    you can do same with sin(x) :)

  7. SolomonZelman
    • one year ago
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    oh, my bad.... 5 and 02 are not included. my apologizes

  8. anonymous
    • one year ago
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    the domain is (-∞, ∞) so the graph would go to the extents in both directions along the x-axis. I was thinking something shaped like the cube root function, but with asymptotes. Not sure exactly what function that would be

  9. anonymous
    • one year ago
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    |dw:1441671157497:dw| Would the graph look like this?

  10. Hero
    • one year ago
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    That's not between the two asymptotes. That's above and below the asymptotes.

  11. anonymous
    • one year ago
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    Oh yeah forgot about that :P

  12. anonymous
    • one year ago
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    |dw:1441671432757:dw|

  13. anonymous
    • one year ago
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    |dw:1441671544537:dw|

  14. Hero
    • one year ago
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    Actually, that's exactly right @peachpi

  15. anonymous
    • one year ago
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    :) Thanks for the help guys.

  16. anonymous
    • one year ago
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    you're welcome. thx @Hero

  17. Hero
    • one year ago
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    @peachpi Bonus points if you can come up with an algebraic representation of the graph.

  18. anonymous
    • one year ago
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    @Hero believe me I've been trying. Logistic function?

  19. Hero
    • one year ago
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    I only suggested because I'm stumped myself atm.

  20. anonymous
    • one year ago
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    I started with this \[f(x)=\frac{ a }{ 1+bc^{-x} }\] shifted it down 5, then a has to be 7 for the limit approaching ∞ to be 2. \[f(x)=\frac{ 7 }{ 1+bc^{-x} }-5\] I think b > 0 and c > 0 and not 1. @Hero

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