Question

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Answer 1

\[f(x)=\dfrac{g(x)}{x^2+1}\]

Let g(x) be some function of x that's not x²+1. As you should be able to tell, the denominator is positive for all x, so no real value of x will make f(x) undefined. Thus, no vertical asymptotes.

Why not g(x) = x²+1? Because, then you have g(x)/g(x) = 1, which is not a rational function.

Answer 2

And also, I'm wondering is that the only equation with no vertical asymptotes and holes, or are there more?

Answer 3

Your answer should be an example of a function, so yes, your final answer should be a function.

As for your most recent question: Mine is only an example. As long as you have a function in the denominator that cannot be 0 for any input value, you'll have a rational function without (vertical) asymptotes.

Answer 4

Great, thanks! Just to check, there are also no holes in the equation you gave, right?

Answer 5

Nope. Holes only occur when a factor in the numerator can be divided by a factor in the denominator, and the factor in the denominator happens to be one that would otherwise cause the presence of an asymptote.

As an example,
\[f(x) = \frac{g(x)}{x+1}\]
has a hole at x = -1 if g(x) = x + 1, or if g(x) contains a factor whose root is -1. Does that make sense?

Answer 6

Yes, Thanks!