Somebody can you please help?

Question

Ok, I've been sitting here for a while now and I'm getting cramps...I'm about to rip my paper because I don't know how to find the horizontal asymptote for this function:

f(x)= (3x-x^2)/(4-x)

SOMEONE please find the horizontal asymptote for this function and show me how to do it using the highest degree.

Answer 1

Ok, here's how i think you do it.

to get the horizontal asymptote, you have to get the limit of the functon when x tends to be infinite, so you would get something like this

lim (3x-x^2)/(4-x)
x→∞

then you substitute ∞ on the x's and you get

∞/∞, that's not defined, so what you use is l'hopital law where you derive both the nominator and the denominator of a fraction, so you will get

lim 3-2x/-1
x→∞

now you can substitute ∞ and you get

(3-2*∞)/-1 =∞

and you do the same with -∞ and you get as an answer -∞

which means there is no horizontal asymptote.

I hope this helps.

Answer 2

How to find a Horizontal Asymptote

Given a rational function, how do you know if there’s a horizontal asymptote?

The Procedure

•Check the highest power of x in the numerator – call it “n”

•Check the highest power of x in the denominator – call it “d”

•Now look at the ratio of n to d…in other words make a fraction with n in the numerator and d in the denominator. You will need to note the size of the fraction:

n/d > 1

n/d = 1

n/d < 1


A)if n/d is bigger than 1, then there is no horizontal asymptote… you will find out later that there might be a kind of asymptote called an oblique asymptote, but no horizontal one


B)if n/d is one, then the asymptote is the ratio of the coefficients in the natural order

C)if n/d is less than one, then the horizontal asymptote is the x axis.

In your case:

3x-x^2
----------
4-x

n=2
d=1

n/d=2/1=2 >1 ==> there is no horizontal asymptote

Answer 3

enter your formula in any good search engine.

Answer 4

http://www.analyzemath.com/quadraticg/quadraticg.htm