Question

Conceptual question about algebra graphs

Answer 1

when you have a function such as :\[\frac{ (x-2)(x-3) }{ x^2 (x-2)^2 }\]
and you are trying to decide if there are holes in the graph and where vertical asymptotes are, do you first simplify the fraction? cancelling common factors? so I guess what I am asking is do you first consider where holes are and then decide where asymptotes are? If so what happens in the case where you have a multiple of that factor left in the denominator? can you have a hole where an asymptote is?

Answer 2

its easy to find asymptotes from the formula as given
any value of x which makes the denominator = zero gives vertical asymptote at that point

Answer 3

For holes, I would simplify. A hole is where you have a 0/0 form. If you don't simplify, it LOOKS like x = 2 could be a hole. Since this would make the numerator and denominator 0. But the (x-2) term cancels out of the numerator, so x=2 turns out to not result in a hole, since you're left with only (x-3) in the numerator after the simplification.

Answer 4

right but a hole is when you have a common factor in the numerator and the denominator.

Answer 5

Right, my point is that you wouldnt KNOW this unless you DID simplify and cancel out the terms. As written, it APPEARS that x = 2 is a hole. After you get rid of common factors, it turns out that it is NOT a hole, it's asymptotic there.

Answer 6

First find any values of x that would cause division by zero; those are your discontinuities.
Then cancel factors if you can. Factors that cancel show where the holes are, the factors that don't cancel are the asymptotes.

Answer 7

There are discontinuities at x=0 and at x=2 because of the x^2 and (x-2)^2 in the denominator.

You can get one of the (x-2)'s on the bottom to cancel, but there will still be one left.

Since (x-2) didn't go away completely, and there's nothing you can do about that x^2, both x=0 and x=2 are vertical asymptotes.

Answer 8

Make sense, @chrisplusian ?

Answer 9

ok so after cancellation if there are factors in the denominator left that have already been canceled then there is a discontinuity rather than a hole

Answer 10

Both holes and asymptotes are types of discontinuities. If the factor on the bottom cancels out completely, then it is a hole, if that factor won't go away no matter how hard you try to cancel, then it is an asymptote.

Answer 11

e.g. If it started with (x-2) on the bottom instead of (x-2)^2, then that factor will cancel completely with the one on top and there will be a hole at x=2 instead of an asymptote.

Answer 12

Cool thanks. I am actually tutoring a young man and today at our meeting he asked this, and I was amazed that I could not answer it. I told him I would call him when I could answer it.

Answer 13

Another important point is that the overall shape of the graph depends on the simplified (after all common factors have been canceled) form.
i.e. y=[(x-2)(x-3)]/[x^2(x-2)^2]
looks identical to y=(x-3)/[x^2(x-2)] on a graph.

Answer 14

However, y=[(x-2)(x-3)]/[x^2(x-2)] will not; it will look identical to
y=(x-3)/(x^2) on a graph, except it'll have that hole at x=2 that you can't tell from just looking at the simplified equation, y=(x-3)/(x^2) .

Answer 15

Thanks that is real helpful. I am employed by Florida Gateway College as a tutor up to calculus level, but I typically tutor trigonometry. I am better at trig than most subjects and I forget the stuff from college algebra sometimes.