How do you find the vertical, horizontal or slant asymptotes for #f(x) = (4x) / (x^2+1)#?

Question

Madelyn Anderson · Accepted Answer

horizontal asymptote at y = 0

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes. solve : #x^2+1=0rArrx^2=-1# This has no real solutions hence there are no vertical asymptotes. Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide all terms on numerator/denominator by the highest power of x that is #x^2# #f(x)=((4x)/x^2)/(x^2/x^2+1/x^2)=(4/x)/(1+1/x^2)# as #xto+-oo,f(x)to0/(1+0)# #rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes. graph{(4x)/(x^2+1) [-10, 10, -5, 5]}

Chloe Alexander · Answer

undefined To find the vertical asymptotes of $ f(x) = \frac{4x}{x^2 + 1} $, set the denominator equal to zero and solve for $ x $. There are no real solutions, so there are no vertical asymptotes.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $.

To find the slant asymptote, perform long division of the numerator by the denominator. The result will be the equation of the slant asymptote, which in this case is $ y = 0 $.