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sasogeek
Group Title
how to find the coordinates of holes in a rational function....
 one year ago
 one year ago
sasogeek Group Title
how to find the coordinates of holes in a rational function....
 one year ago
 one year ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.4
what happens when you remove a part of a line or curve?
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
see where the function is discontinuous
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
how do i find that if i haven't graphed it?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
lets start with big concepts and then refine them for this
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
what happens when you remove a part of a line or curve?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
dw:1348750115514:dw
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
okay, that part is either missing or doesn't satisfy the function....
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
now relate this to removing something from a fraction; what enables us to cancel things top and bottom?
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
if they're like terms of can be factored?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
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

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
ok so for example, \(\huge \frac{3x^3}{x^21}\) my hole will be 1,+1 ? same as the asymptotes?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
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

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
\[\frac{3xxx}{(x+1)(x1)}\] there are no common factors in this; so no factors can be removed; therefore no holes are created
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
I'll need more practice on this but thanks :) I'll just state that there's no holes xD
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
more practice is good ;)
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
to be clear though, can you give me an example function that has holes?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
\[\frac{(x+2)(x3)}{x(x3)(x+7)}\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
can you tell me all the bad xs?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
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

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
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

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
\[y=\frac{(3+2)\cancel{(x3)}}{3\cancel{(x3)}(3+7)}\] \[y=\frac5{30}=\frac16\] so the coordinate of the hole is (3, 1/6)
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
you said it is one of the holes... are there more?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
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

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
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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