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sasogeek Group Title

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

  • 2 years ago
  • 2 years 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?

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

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

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

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

    • 2 years ago
  6. amistre64 Group Title
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    |dw:1348750115514:dw|

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

    • 2 years 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?

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

    • 2 years 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

    • 2 years 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?

    • 2 years 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.

    • 2 years 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

    • 2 years 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

    • 2 years ago
  15. amistre64 Group Title
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    more practice is good ;)

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

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

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

    • 2 years ago
  19. sasogeek Group Title
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    x=3 ?

    • 2 years 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

    • 2 years 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

    • 2 years 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)

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

    • 2 years 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

    • 2 years 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?

    • 2 years ago
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