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/sci/ - Science & Math

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4798980 No.4798980 [Reply] [Original]

so, /sci/, I was taking a test yesterday and the following problem came up:

Find the vertical asymptotes of the following equation:

f(x) = (x^2 - 6x + 5)/(x^2 - 2x - 15)

I said that the answer was x = 5 and x = -3.

Apparently this was wrong, but after poring over the problem for quite a while, I'm still unable to figure out why. I'm sure I'm retarded and I'm just forgetting something basic, but I have no clue what. Any help?

>> No.4798987

1. Factor the numerator.
2. Simplify terms.
3. Find the asymptotes.

Hint: The factor (x-5) appears on top and bottom.

>> No.4798995

limit of equation as x approaches 5 is 0.5, not +- infinity

>> No.4798996

-3 is correct, 5 isn't because it's also a root of the quadratic in the numerator.

So if x/0 for nonzero x indicates an asymptote (YES I know this isn't rigorous) then 0/0 is indeterminate, in this case you get a removable discontinuity (a single undefined point on the graph).

>> No.4798993 [DELETED] 

>>4798980
Graph it?

It's 1

>> No.4798994

You have to check the numerator isn't 0. The function may not tend to infinity if it is.

>> No.4798998
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4798998

Don't they let you use graphing calculators?

>> No.4799000

>>4798996
that isn't rigorous.

>> No.4798997

>>4798993
that isn't an explanation you pleb

>> No.4799001

there's no asymptote at 5

the numerator factors to (x-5)(x-1) and the denominator to (x-5)(x+3). so the x-5's cancel: you merely have a hole at x=5.

>> No.4799006

>>4798998
>>4798995

> it isn't an asymptote because it doesn't go to infinity

HURR DURR TAUTOLOGY DURR

>> No.4799008

-3 is fine. 5 is not.

See, since both the numerator and denominator are 0 at x=5 you have to check that shit out.

>> No.4799032

OP here.

>>4798998
No, they don't. I probably wouldn't have gotten that anyway, though, because I would have just jumped into table mode and left it alone once I saw "ERROR."

>>4799001
Okay, thanks. I was completely unaware that there's a difference between a hole in a graph and an asymptote. I guess my math teacher in High School just skipped over it, or something.

>> No.4799045

>>4799032
> Okay, thanks. I was completely unaware that there's a difference between a hole in a graph and an asymptote. I guess my math teacher in High School just skipped over it, or something.

wait... wtf do you think an asymptote is then?

>> No.4799058

>>4799045
I had just thought that an asymptote was when the denominator = 0. I probably just forgot because it was an issue that almost never came up when I was learning it.

>> No.4799066

>>4799058
...okay. well no. an asymptote is where the graph goes to infinity. it's just that, with quotients of polynomials, the denominator has be 0 for this to happen.

>> No.4799076

>>4799066
Thank you for clarifying that, anon. My understanding of asymptotes is now greatly improved.

>> No.4799366

>>4799066
Then define "removable asymptote."

>> No.4799382

>>4799366
There's no such thing. You're thinking of a removable discontinuity.