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sasogeek Group Title

how to find the coordinates of holes in a rational function....

  • one year ago
  • one year ago

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  1. amistre64 Group Title
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    what happens when you remove a part of a line or curve?

    • one year ago
  2. bahrom7893 Group Title
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    see where the function is discontinuous

    • one year ago
  3. sasogeek Group Title
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    how do i find that if i haven't graphed it?

    • one year ago
  4. amistre64 Group Title
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    lets start with big concepts and then refine them for this

    • one year ago
  5. amistre64 Group Title
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    what happens when you remove a part of a line or curve?

    • one year ago
  6. amistre64 Group Title
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    |dw:1348750115514:dw|

    • one year ago
  7. sasogeek Group Title
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    okay, that part is either missing or doesn't satisfy the function....

    • one year ago
  8. amistre64 Group Title
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    now relate this to removing something from a fraction; what enables us to cancel things top and bottom?

    • one year ago
  9. sasogeek Group Title
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    if they're like terms of can be factored?

    • one year ago
  10. amistre64 Group Title
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    correct, so any factor that creates a zero in the denominator of a rational function; if it can be removed from the equation, it creates a hole

    • one year ago
  11. sasogeek Group Title
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    ok so for example, \(\huge \frac{3x^3}{x^2-1}\) my hole will be -1,+1 ? same as the asymptotes?

    • one year ago
  12. amistre64 Group Title
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    a hole is not "the same" as an asymptote. they have similar effects on the equation (they zero out a denominator); but a hole can be removed. an asymptote cannot.

    • one year ago
  13. amistre64 Group Title
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    \[\frac{3xxx}{(x+1)(x-1)}\] there are no common factors in this; so no factors can be removed; therefore no holes are created

    • one year ago
  14. sasogeek Group Title
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    I'll need more practice on this but thanks :) I'll just state that there's no holes xD

    • one year ago
  15. amistre64 Group Title
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    more practice is good ;)

    • one year ago
  16. sasogeek Group Title
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    to be clear though, can you give me an example function that has holes?

    • one year ago
  17. amistre64 Group Title
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    \[\frac{(x+2)(x-3)}{x(x-3)(x+7)}\]

    • one year ago
  18. amistre64 Group Title
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    can you tell me all the bad xs?

    • one year ago
  19. sasogeek Group Title
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    x=3 ?

    • one year ago
  20. amistre64 Group Title
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    x=3 is one of the bad xs, yes and since it can be removed from the setup; it creates a hole at x=3

    • one year ago
  21. sasogeek Group Title
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    so basically since you can remove that from the function, that makes a hole.... now it's even clearer :) i think i'm getting it :) not quite there yet though :P

    • one year ago
  22. amistre64 Group Title
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    \[y=\frac{(3+2)\cancel{(x-3)}}{3\cancel{(x-3)}(3+7)}\] \[y=\frac5{30}=\frac16\] so the coordinate of the hole is (3, 1/6)

    • one year ago
  23. sasogeek Group Title
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    you said it is one of the holes... are there more?

    • one year ago
  24. amistre64 Group Title
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    can we remove any of the other factors from the equation? if not, then there are no more holes to be found. the rest of the zeros in the denominator would create vertical asymptotes

    • one year ago
  25. sasogeek Group Title
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    okay :) thanks. can i assume then that if the zero in the denominator cancels out something in the numerator, it's a hole, if not, it is a vertical asymptote?

    • one year ago
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