# Asymtotes and polynomials

Ok, I have a final in Pre-Calc comming up, and im still a bit confused on finding asymtotes (vertical, horizontal, and slant) could someone help me with equations i can use to find the asymtotes or how i do? im just really confused.

Heres the problem. Find the vertical asymtote(s): F(x) = X+3 / (X-2)(X+5)

I dont even have a start because im so confused Also, I am having trouble with finding a fourth degree polynomial that has a set of given zeros. How might i go about solving one of those? Im very confused, please help me

The problem Find a fourth Degree polynomial that has zeros: 1, -3, 2i

Once again i have no clue where to start...a little help please?

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The vertical asymptotes occur when F(x) tends to infinity. You should be able to see these clearly as they're when the denominator is 0. There are two of them; one at x = 2 and one at x = -5.

For the second bit, you need to use the factor theorem. You know that if f(a) = 0, then (x - a) is a factor. You also know that if one complex number is a root, then its complex conjugate is also a root. Can you go from there?

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Vertical asymptotes - look for the number that makes the denominator = 0

Horizontal Asymptotes - If the degree of the powers are equal, take the coefficients of them and you have y = a/b. If the power in the denominator is larger than the one in the numerator, then you have a H.A. at y=0

Slant Asymptotes - If the power in the numerator is 1 degree larger than the denominator, then you divide the bottom into the top. (ex. x^2 / (x-1))

If the power in the numerator is more than 1 degree higher than the denominator, then there is no H.A.!

Tjl
HawKMX2004 said:
Ok, I have a final in Pre-Calc comming up, and im still a bit confused on finding asymtotes (vertical, horizontal, and slant) could someone help me with equations i can use to find the asymtotes or how i do? im just really confused.

Heres the problem. Find the vertical asymtote(s): F(x) = X+3 / (X-2)(X+5)

I dont even have a start because im so confused Also, I am having trouble with finding a fourth degree polynomial that has a set of given zeros. How might i go about solving one of those? Im very confused, please help me

The problem Find a fourth Degree polynomial that has zeros: 1, -3, 2i

Once again i have no clue where to start...a little help please?
There is an easier method to finding these. It covers basically what was said above but uses more conventional methods.

Vertical Asymptote(s): X values that make the denominator zero.

Just reverse factor the zero's and you will end with the original equation. Although be careful, with imaginary numbers. => 2i they have special conditions.

Tjl
Nylex said:
The vertical asymptotes occur when F(x) tends to infinity. You should be able to see these clearly as they're when the denominator is 0. There are two of them; one at x = 2 and one at x = -5.

For the second bit, you need to use the factor theorem. You know that if f(a) = 0, then (x - a) is a factor. You also know that if one complex number is a root, then its complex conjugate is also a root. Can you go from there?
They dont occur when F(x) tends to infinity, that is the limit as x approaches positive infinity and as a way to find the Horizontal asymptote. You need to specify the use of only the highest degree terms in the numerator and denominator.

Tjl said:
They dont occur when F(x) tends to infinity, that is the limit as x approaches positive infinity and as a way to find the Horizontal asymptote. You need to specify the use of only the highest degree terms in the numerator and denominator.
Not sure I understand. You find vertical asymptotes where the denominator is zero and division by zero should be infinity?