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

##### 2 Answers

#### Answer:

Vertical asymptotes at

Horizontal asymptote at

#### Explanation:

Vertical asymptotes will occur when the denominator equals

Since the degree of the numerator is *lesser* than that of the denominator, there will be a horizontal asymptote at **no slant asymptote. **

Hopefully this helps!

#### Answer:

vertical asymptotes at x =0 and x = - 1

horizontal asymptote at y = 0

#### Explanation:

Factorising 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))# Hence there is an excluded value at x = 1

Indicating a hole at x = 1 in f(x). This hole is not in the simplified f(x).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(1+x)=0rArrx=0" or " x=-1#

#rArrx=0" and " x=-1" are the asymptotes"# Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by x

#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"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2) Hence there are no oblique asymptotes.

graph{(-1)/(x(1+x) [-20, 20, -10, 10]}