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sasogeek

  • 2 years ago

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

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

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

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

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

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

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

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

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

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

  10. amistre64
    • 2 years ago
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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

  11. sasogeek
    • 2 years ago
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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?

  12. amistre64
    • 2 years ago
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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.

  13. amistre64
    • 2 years ago
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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

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

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

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

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

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

  19. sasogeek
    • 2 years ago
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    x=3 ?

  20. amistre64
    • 2 years ago
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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

  21. sasogeek
    • 2 years ago
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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

  22. amistre64
    • 2 years ago
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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)

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

  24. amistre64
    • 2 years ago
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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

  25. sasogeek
    • 2 years ago
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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?

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