Question

what is a function that meets these requirments, no vertical asymptote, horizontal of y=0, y intercept (0,1/4)?

Answer 1

OK, what part of this is giving you trouble?

Answer 2

the no vertical asymptote part

Answer 3

No vertical asymptote implies that the domain is all real numbers.

Answer 4

forgot to mention its a rational function

Answer 5

Rational.... aaah....

Answer 6

Well, then there should be no way for the bottom to be 0.

Answer 7

x will never be in the denominator.

Answer 8

@EulersEquation What about \[\frac{x^2+\frac{5}{4}x+\frac{1}{4}}{x+1}\]

Answer 9

Try \(y = \dfrac{1/4}{x^{2}+ 1}\)

Answer 10

https://www.desmos.com/calculator/oearcetoau

Answer 11

e.mccormick, x + 1 cannot go in the denominator, otherwise you have a vertical asymptote at x = -1. However, x^2 + 1 is okay. That would be in keeping with a horizontal asymptote at y = 0.

Answer 12

@e.mccormick, x + 1 cannot go in the denominator, otherwise you have a vertical asymptote at x = -1. However, x^2 + 1 is okay. That would be in keeping with a horizontal asymptote at y = 0.

Answer 13

@EulersEquation Nope. It is a removable. No asymptote.

Answer 14

I was just giving hints. But tkhunny gave the actual solution, so i added it to the graph.

Answer 15

@e.mccormick I like it. You could not have done that at x = 0, as this would violate the Intercept requirement.

Right, after the bad idea that 'x' could not appear in the denominator, it seemed prudent to get past that.

Answer 16

\[y = \frac{ 1 }{ x^2 }\] is a good candidate. @tkhunny beat me to it, however.

Answer 17

@tkhunny gets a medal.

Answer 18

No -- mine is wrong because that gives a vertical asymptote at x = 0!

Dang!

Answer 19

That's what the +1 is for. The denominator cannot be zero for Real x.

Answer 20

Seeing that I am a 17th century mathematician, I am quite old. Getting too old for this sort of thing.

Answer 21

ha! You are not the 17th century mathematician. You are just his brain child! =P

Answer 22

a wannabe