sasogeek
  • sasogeek
how to find the coordinates of holes in a rational function....
Mathematics
schrodinger
  • schrodinger
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amistre64
  • amistre64
what happens when you remove a part of a line or curve?
bahrom7893
  • bahrom7893
see where the function is discontinuous
sasogeek
  • sasogeek
how do i find that if i haven't graphed it?

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amistre64
  • amistre64
lets start with big concepts and then refine them for this
amistre64
  • amistre64
what happens when you remove a part of a line or curve?
amistre64
  • amistre64
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sasogeek
  • sasogeek
okay, that part is either missing or doesn't satisfy the function....
amistre64
  • amistre64
now relate this to removing something from a fraction; what enables us to cancel things top and bottom?
sasogeek
  • sasogeek
if they're like terms of can be factored?
amistre64
  • amistre64
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
sasogeek
  • sasogeek
ok so for example, \(\huge \frac{3x^3}{x^2-1}\) my hole will be -1,+1 ? same as the asymptotes?
amistre64
  • amistre64
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.
amistre64
  • amistre64
\[\frac{3xxx}{(x+1)(x-1)}\] there are no common factors in this; so no factors can be removed; therefore no holes are created
sasogeek
  • sasogeek
I'll need more practice on this but thanks :) I'll just state that there's no holes xD
amistre64
  • amistre64
more practice is good ;)
sasogeek
  • sasogeek
to be clear though, can you give me an example function that has holes?
amistre64
  • amistre64
\[\frac{(x+2)(x-3)}{x(x-3)(x+7)}\]
amistre64
  • amistre64
can you tell me all the bad xs?
sasogeek
  • sasogeek
x=3 ?
amistre64
  • amistre64
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
sasogeek
  • sasogeek
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
amistre64
  • amistre64
\[y=\frac{(3+2)\cancel{(x-3)}}{3\cancel{(x-3)}(3+7)}\] \[y=\frac5{30}=\frac16\] so the coordinate of the hole is (3, 1/6)
sasogeek
  • sasogeek
you said it is one of the holes... are there more?
amistre64
  • amistre64
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
sasogeek
  • sasogeek
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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