Question

Given a rational function where the degree of the numerator is greater than the denominator. Find a hypothetical equation and list all the asymptotes. under the following conditions:
(DO NOT WORK BACKWARDS)

1. The denominator is a quartic function and is not factorable
2. The numerator is a quadratic function and is not factorable

Answer 1

woops reverse numerator and denominator

Answer 2

numerator is quartic and denominator is quadratic

Answer 3

how?

Answer 4

Do you know the rules for asymptotes of rational functions?

Answer 5

yea

Answer 6

Oh, I apologize, youre right.

Answer 7

Im thinking backwards :D But yeah, no asymptotes on that one.

Answer 8

\[\frac{ x^{4}+1 }{ x^{2}+1 }\] no asymptotes, my apologies.

Answer 9

hmm how would you be able to graph this without technology

Answer 10

What class is this for?

Answer 11

Pre calc 12

Answer 12

i can find the oblique asymptote, but how would you know x asymptotes

Answer 13

There are no asymptotes in this equation. An oblique asymptote requires the numerator to be exactly one degree higher than the denominator, but since it is 2 higher, we do not have an oblique. We can only have a horizontal asymptote if the degree of the numerator and denominator are the same or if the denominator is of a higher degee, so we do not have a horizontal asymptote. Vertical asymptotes can occur at x-values in which the denominator is equal to 0. But since we have x^2 + 1, there would only be complex solutions to x^2 + 1 = 0, so that means no vertical asymptotes either.

Answer 14

so there's a parabolic asymptote?

Answer 15

There is no asymptote at all.

Answer 16

And then the next question is harder, it bumps the degree of the numerator and denominator both up by 1.

Answer 17

Were you still concerned about how we might graph it, or do you want to go to the next question?

Answer 18

I'm just needing help on both

Answer 19

It just clicked that i could use the quadratic equation

Answer 20

on the first one.. but how bout the second

Answer 21

Let me clarify, are the two conditions above meant to be in the same equation or in separate equations?

Answer 22

\[\frac{ x^5+1 }{x^3+1 }\]

Answer 23

Is the condition still needed that they cannot be factored?

Answer 24

yup

Answer 25

Both of those can be factored, though.

Answer 26

im just giving an example of what i mean

Answer 27

We have to do a little bit more work to make them not factorable. \[\frac{ x^{5} +2 }{ x^{3}+x+3 }\]Something like that should work.

Answer 28

hmm that works

Answer 29

but how would you find the asymptotes

Answer 30

Well, there are no horizontal or oblique asymptotes. There is a vertical asymptote, but thats the tricky part, having a cubic that doesnt factor yet still has easy to find asymptotes. Anything that has nice numbers for vertical asymptotes will factor. So basically you're forced to have an equation with nasty asymptotes.

Answer 31

so you "force factor" x^3 +x +3 ??

Answer 32

There is no force factoring. If there was, itd be factorable and would break the conditions of your question :) With the techniques you have available, youd have to approximate a solution. It'd just be tedious.

Answer 33

so there's no way you could find the exact asymptote

Answer 34

without tech

Answer 35

There wouldnt be an exact asymptote, itd be some sort of irrational number. But you could get an answer very close with use of a basic calculator.

Answer 36

hmm

Answer 37

Anything with an exact asymptote would be somethin that could be factorable. It might be a nasty factoring, but itd be factorable. If I have a vertical asymptote of, let's say, 2.35. Well, then all that means is that (x-2.35) is a factor. But we DONT want factors. So that forces you to have solutions that aren't nice and pretty. For this particular cubic I made up, the vertical asymptote would be approximately x = -1.2134. With precalculus techniques, you're using a primitive approximation method, but it could be done. With calculus, there's something called Newton's Method that could get you that answer in not too long with a basic calculator. But no matter what, since its a cubic equation, you're going to have at least one real asymptote, its just it has to be an irrational one in order to make sure its not factorable.

Answer 38

thank you

Answer 39

You're welcome :)