Do polynomial functions have asymptotes? If yes, how do you find them?

Question

Elijah Abram · Accepted Answer

The only polynomial functions that have asymptotes are the ones whose degree is 0 (horizontal asymptote) and 1 (oblique asymptote), i.e. functions whose graphs are straight lines.

To show this, let's consider a generic polynomial function #p:mathbb(R) to mathbb(R)# of degree #n in mathbb(N)#. We can express it as follows: #p(x)=c_0+c_1 x+c_2 x^2 + ... + c_(n-1) x^(n-1) + c_n x^n# where #c_i in mathbb(R)# for #i=0,...,n#. Note that if the degree is #n#, this means that #c_n ne 0#. Now we search for the possible asymptotes. The only boundary points of the domain are #+infty# and #-infty#, because polynomials can always be defined on the whole real line. So there's no hope to find vertical asymptotes (the only ones that can be found in correspondence of finite boundary points). For #f:mathbb(R) to mathbb(R)# a general function, the conditions to have an asymptote at #+infty# (the case for #-infty# can be done by a slight modification of the following arguments) is that the two following limits exist and are finite #m:=lim_{x to +infty} f(x)/x#, #q:=lim_{x to +infty} [f(x)-mx]# If what stated here is true, then the asymptote is the line given by the equation #y=mx+q#. Let's show that #m# exists when #f# is a general polynomial #p# of degree #n# and it's finite if and only if #n le 1#: For #n = 0# we have #p(x)=c_0# and #m=lim_{x to +infty} [p(x)]/x=lim_{x to +infty} c_0/x=0# for all #c_0 in mathbb(R)# For #n=1# we have #p(x)=c_0 + c_1 x# and #m=lim_{x to +infty} [p(x)]/x=lim_{x to +infty} [c_0+c_1 x]/x=lim_{x to +infty} [c_0/x+c_1]=c_1#, #forall c_0,c_1 in mathbb(R)# For #n>1#, we get that #m=lim_{x to +infty} [p(x)]/x= lim_{x to +infty} [c_0+c_1 x+c_2 x^2 + ... + c_(n-1) x^(n-1) + c_n x^n]/x= lim_{x to +infty} ([c_0]/x+[c_1 x]/x+[c_2 x^2]/x + ... + [c_(n-1) x^(n-1)]/x + [c_n x^n]/x)= lim_{x to +infty} [c_1 +c_2 x + ... + c_(n-1) x^(n-2) + c_n x^(n-1)]# Now we can factor the polynomial in the following way: #lim_{x to +infty} c_n x^(n-1) [c_1/[ c_n x^(n-1)] +c_2/[c_n x^(n-2)] + ... + c_(n-1)/[c_n x] + 1]# All the denominators of the sum are positive powers of #x#, so their limit is zero and we get that #m=lim_{x to +infty} c_n x^(n-1) = pm infty# (signum depending on the signum of #c_n#, which can't be zero for the fact stated at the beginning). So asymptotes can exist only for #n=0,1#, because in the other cases the limit that defines #m# diverges. We should check the condition that determines #q# for #n=0,1#. For #n=0# we get #q=lim_{x to +infty} [p(x)-mx]=lim_{x to +infty} [c_0-0 x]=c_0# For #n=1# we get #q=lim_{x to +infty} [p(x)-mx]=lim_{x to +infty} [c_0+c_1 x-c_1 x]=c_0# This ends the proof and shows that the asymptotes at infinities exist if and only if the degree of the polynomial function is less or equal to 1. Their equations coincide with the equation of the function.

Adam Adkins · Answer

undefined Polynomial functions do not have asymptotes. Asymptotes typically occur in rational functions, which are the ratio of two polynomials. In polynomial functions, the degree of the polynomial determines the behavior of the function as $ x $ approaches positive or negative infinity. For example:

1. If the degree of the polynomial function is even, both ends of the graph tend to infinity in the same direction.
2. If the degree of the polynomial function is odd, one end of the graph tends to positive infinity while the other end tends to negative infinity.

To summarize, polynomial functions do not have asymptotes, but their behavior at positive and negative infinity depends on the degree and leading coefficient of the polynomial.