How do you find the asymptotes for #f(x) = (x^2 - 3x)/(x^2 + 1)#?

2 Answers
Feb 4, 2016

Answer:

#y=1#

Explanation:

To find the vertical asymptotes set the denominator equal to 0 then solve for #x#. For this example in particular however, their are no such values for #x# so there will be no vertical asymptotes.

Any other asymptotes will either be horizontal or oblique. As the degree of the polynomial on the numerator is the same as or less than the degree of the polynomial on the denominator then we will have a horizontal asymptote.

To find this see what happens when #x# gets very large. We see that #x^2# term is the dominating term on both the numerator and the denominator. The #3x# term and the #1# term will become insignificant compared the x^2, so we can say that as #x# tends towards infinity:

# (x^2-3x)/(x^2+1) -> x^2/x^2 = 1 #

Thus there is a horizontal asymptote along #y=1#

Feb 4, 2016

Answer:

First observation: there can be no vertical asymptote, as the lower part of the fraction will never be zero.

Explanation:

So let's see what happens if we make #x# as large as we want -- this is the way to get a horizontal asymptote.
The #+1# below will have no meaning anymore, and when we cancel the #x#'s above and below, we're left with:
#(x+3)/x# and of course the #+3# will lose its significance as #x# becomes greater and greater, and we're left with:
#lim_(x->+-oo) f(x)=x/x=1#
The function does cross the #y=1#-value at #x=-1# but further on it's an asymptote.
graph{(x^2-3x)/(x^2+1) [-46.2, 46.27, -23.1, 23.16]}