How do you find vertical, horizontal and oblique asymptotes for #F(x) = (x - 1)/(x - x^3)#?

Question

Serenity Jones · Accepted Answer

Vertical asymptotes at #x = 0, x = -1 and x= 1#
Horizontal asymptote at #y = 0# Vertical asymptotes will occur when the denominator equals #0#. #x - x^3 = 0# #x(1 - x^2) = 0# #x(1 + x)(1 - x) = 0# #x = 0, -1 and 1# Since the degree of the numerator is lesser than that of the denominator, there will be a horizontal asymptote at #y = 0# and no slant asymptote. I hope this is useful!

Serenity Jones · Answer

vertical asymptotes at x =0 and x = - 1
horizontal asymptote at y = 0

Factorization of f(x) #f(x)=(x-1)/(x(1-x^2))=(x-1)/(x(1-x)(1+x))# #=(-(1-x))/(x(1-x)(1+x))=(-(cancel(1-x))^1)/(x(cancel(1-x))(1+x))=(-1)/(x(1+x))# As a result, at x = 1, there is an excluded value that indicates a hole in f(x) that is not present in the simplified f(x). The values that x cannot be are obtained by equating the denominator to zero and solving the problem; if the numerator is non-zero for these values, they are vertical asymptotes. The denominator of f(x) cannot be zero since this would render f(x) undefined. solve: #x(1+x)=0rArrx=0" or " x=-1# #rArrx=0" and " x=-1" are the asymptotes"# As horizontal asymptotes arise, #lim_(xto+-oo),f(x)toc" ( a constant)"# split terms by x on the numerator and/or denominator #f(x)=(-1/x)/(x/x+x^2/x)=(-1/x)/(1+1/x)# as #xto+-oo,f(x)to0/(1+0)# #rArry=0" is the asymptote"# graph{(-1)/(x(1+x) [-20, 20, -10, 10]} There are no oblique asymptotes because the degree of the denominator here (numerator-degree 0, denominator-degree 2) is not greater than the degree of the numerator.

Ariana Alvarez · Answer

undefined To find the vertical asymptotes of a function, we identify the values of x that make the denominator equal to zero. In this case, the denominator of F(x) is (x - x^3). We set the denominator equal to zero and solve for x:

x - x^3 = 0

Factoring out x, we get:

x(1 - x^2) = 0

Setting each factor equal to zero, we find:

x = 0 and x^2 - 1 = 0

The solutions to x^2 - 1 = 0 are x = 1 and x = -1.

Therefore, the vertical asymptotes of F(x) occur at x = 0, x = 1, and x = -1.

To find the horizontal asymptote, we analyze the behavior of the function as x approaches positive or negative infinity. In this case, the highest power of x in the numerator and the denominator is 1. Therefore, we divide the leading coefficient of the numerator by the leading coefficient of the denominator:

Leading coefficient of numerator: 1
Leading coefficient of denominator: 1

Therefore, the horizontal asymptote is y = 1.

To find oblique asymptotes, we divide the numerator by the denominator using polynomial long division. The result of this division will give us the oblique asymptote, if it exists.

Performing polynomial long division, we get:

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

Since the degree of the numerator is less than the degree of the denominator, there is no oblique asymptote for this function.