f(x)=(x-2)^5 / (x+2)^3(x-4)^3

can someone not just show me the answer (i have that), but an easy way to get to it?

thanks!(Clapping)

- March 26th 2009, 08:36 PMcalcconfusedHow to find the horizontal asymptote of the following function
f(x)=(x-2)^5 / (x+2)^3(x-4)^3

can someone not just show me the answer (i have that), but an easy way to get to it?

thanks!(Clapping) - March 26th 2009, 09:20 PMmollymcf2009
Here are the three cases of horizontal asymptotes:

1) If the degree of the denominator is greater than the degree of the numerator, the HA is y=0

2) If the degrees of the numerator & denominator are the same, the HA is the ratio of the leading coefficients

3) If the degree of the numerator is larger than the denominator there is not a HA

In terms of finding which it is, you can factor it all out and see what the degrees of each will be.

Hope that helps! - March 26th 2009, 09:27 PMcalcconfused
that does help but can someone break it down for me a bit?

i dont see why 1/x^6 is put in both the num and denom