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

1 Answer
Dec 22, 2017

Answer:

#x=0# and #x+y=3#

Explanation:

First of all one need to see, whether there areany holes or not, i.e. those values of #x# for which value of function exists but function is not defined. This happens when there are common factors between numerator and denominator. For example if #x-alpha# is a common factor both in numerator and denominator, though they cancel out, function may not be defined. Here, we have none.

Further, if denominator can be factorized, we have vertical asymptotes. If factors are #(x-alpha)(x-beta)#, then vertical asymptotes are #x=alpha#and #x=beta#. For example, here we have just #x# as factor in denominator (with repetition) and hence only vertical asymptote is #x=0#,.

Horizontal asymptotes appear when degree of numerator and denominator are same. It is not so. However as degree of numerator is just one more than that of denominator, there is a slant or oblique asymptote.

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

and as #x->oo#, #f(x)->3-x#

Hence slanting asymptote is #y=3-x# or #x+y=3#

graph{(1+3x^2-x^3)/x^2 [-19.17, 20.83, -1.44, 18.56]}