Hi,

I need help with a problem. I don't understand it.

Describe the condition that will produce a rational function with a graph that has no vertical asymtote.

Thanks for any help!

Michele

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- Apr 21st 2009, 04:32 PMmgannon93Radical function with no vertical asymptote
Hi,

I need help with a problem. I don't understand it.

Describe the condition that will produce a rational function with a graph that has no vertical asymtote.

Thanks for any help!

Michele - Apr 21st 2009, 04:39 PMReckoner
- Apr 21st 2009, 04:45 PMmgannon93
- Apr 21st 2009, 04:50 PMReckoner
- Apr 21st 2009, 04:55 PMmgannon93
- Apr 21st 2009, 05:03 PMReckoner
I meant in the denominator. If the vertical asymptotes of a rational function only occur when the denominator is zero, then what happens if the denominator has no zeroes?

Although technically, if the numerator and denominator are simultaneously zero at a point, it is possible that there is no vertical asymptote there; it depends on the multiplicity of the factors. - Apr 21st 2009, 05:14 PMmgannon93
- Apr 21st 2009, 05:17 PMReckoner
- Apr 21st 2009, 05:22 PMmgannon93
OH! So equations that give you imaginary numbers!

Wow I feel like an idiot. Thanks. Is this explained alright?

--

Describe the condition that will produce a rational function with a graph that has no vertical asymptote.

Rational functions with no rational zeroes will have no vertical asymptote, so an equation like x^2+1 would be a rational function with no vertical asymptotes.

Is that right? - Apr 21st 2009, 05:25 PMReckoner
No, you aren't quite getting it. Yes, the function $\displaystyle f(x)=x^2+1$ has no vertical asymptotes, and we may consider it a rational function. But what I was trying to get you to see was that you can put it in the denominator, and the denominator will always be nonzero (that is, the denominator doesn't have to be a constant).

$\displaystyle g(x)=\frac1{x^2+1}$

has no vertical asymptotes. - Apr 21st 2009, 06:04 PMmgannon93