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amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4what happens when you remove a part of a line or curve?

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.0see where the function is discontinuous

sasogeek
 2 years ago
Best ResponseYou've already chosen the best response.0how do i find that if i haven't graphed it?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4lets start with big concepts and then refine them for this

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4what happens when you remove a part of a line or curve?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4dw:1348750115514:dw

sasogeek
 2 years ago
Best ResponseYou've already chosen the best response.0okay, that part is either missing or doesn't satisfy the function....

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

sasogeek
 2 years ago
Best ResponseYou've already chosen the best response.0if they're like terms of can be factored?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4correct, 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
 2 years ago
Best ResponseYou've already chosen the best response.0ok so for example, \(\huge \frac{3x^3}{x^21}\) my hole will be 1,+1 ? same as the asymptotes?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4a 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
 2 years ago
Best 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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4more practice is good ;)

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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4can you tell me all the bad xs?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4x=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
 2 years ago
Best ResponseYou've already chosen the best response.0so 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
 2 years ago
Best 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)

sasogeek
 2 years ago
Best ResponseYou've already chosen the best response.0you said it is one of the holes... are there more?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4can 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
 2 years ago
Best ResponseYou've already chosen the best response.0okay :) 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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