Do polynomial functions have asymptotes? If yes, how do you find them?
1 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.
Explanation:
To show this, let's consider a generic polynomial function
where
Now we search for the possible asymptotes. The only boundary points of the domain are
For
If what stated here is true, then the asymptote is the line given by the equation
Let's show that
-
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
-
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.