Tuesday
July 29, 2014

Homework Help: math

Posted by Lily on Saturday, April 28, 2007 at 11:56am.

Can someone please help me with these few problems here?

Find the holes and points of discontinuity:

12x^4+10x-3
-----------
x-1

and

12x+24
------
x^2+2x

In the first, examine x=1

and in the second, examine x=-2. Look at the numerator also at that same point.

Oh, I am so sorry, I typed the wrong thing. For the first, I meant:

12x^4+10x-3
-----------
3x^4

Wouldn't the pt of discontinuity be 0 because you set 3x^4 equal to 0.


And for the second, I got x=-2 as the hole and -2 and 0 as the pts of discontinuity.

This is my work:
12x+24=6(2x+4)
=6(2x-2)(x+2)
So you have:
6(2x-2)(x+2)
----------
x^2+2x

which equals

6(2x-2)(x+2)
-----------
(x+2)(x+0)




so I am assuming you typed the second one correctly, but your work is not correct.

(12x+24)/(x^2+2x)
=12(x+2)/(x(x+2))
=12/x

the (x+2)/(x+2) cancelled, which would be 0/0 when x=-2,

so there would be a "hole" when x=-2

your final expression contains x as a denominator, we cannot divide by 0, so the expression is discontinuous when x=0




I did say the hole was at x=-2. So there are no points of discontinuity, then?

to be speaking of points you have to have a function.
is the expression in the form y = ... or f(x)=...?

you merely gave an expression.
If a function is discontinuous there is usually an asymptote, which is a line that the graph will approach but can never ever reach.

in the case of your second expression, that asymptote is x=0, the y-axis.

so then since you are saying my second expression has an asymptote that means it is a function? so then doesn't that mean there are points of discontinuity?

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