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

Question

amistre64 · Answer

what happens when you remove a part of a line or curve?

bahrom7893 · Answer

see where the function is discontinuous

sasogeek · Answer

how do i find that if i haven't graphed it?

amistre64 · Answer

lets start with big concepts and then refine them for this

amistre64 · Answer

what happens when you remove a part of a line or curve?

amistre64 · Answer

|dw:1348750115514:dw|

sasogeek · Answer

okay, that part is either missing or doesn't satisfy the function....

amistre64 · Answer

now relate this to removing something from a fraction; what enables us to cancel things top and bottom?

sasogeek · Answer

if they're like terms of can be factored?

amistre64 · Answer

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 · Answer

ok so for example,

$\huge \frac{3x^3}{x^2-1}$

my hole will be -1,+1 ? same as the asymptotes?

amistre64 · Answer

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 · Answer

$$\frac{3xxx}{(x+1)(x-1)}$$
there are no common factors in this; so no factors can be removed; therefore no holes are created

sasogeek · Answer

I'll need more practice on this but thanks :) I'll just state that there's no holes xD

amistre64 · Answer

more practice is good ;)

sasogeek · Answer

to be clear though, can you give me an example function that has holes?

amistre64 · Answer

$$\frac{(x+2)(x-3)}{x(x-3)(x+7)}$$

amistre64 · Answer

can you tell me all the bad xs?

sasogeek · Answer

x=3 ?

amistre64 · Answer

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 · Answer

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 · Answer

$$y=\frac{(3+2)\cancel{(x-3)}}{3\cancel{(x-3)}(3+7)}$$
$$y=\frac5{30}=\frac16$$
so the coordinate of the hole is (3, 1/6)

sasogeek · Answer

you said it is one of the holes... are there more?

amistre64 · Answer

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 · Answer

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?