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

##### 2 Answers

#### Answer:

The line

The line

So the line

#### Explanation:

Now let's find first the vertical asymtodes of

Because

So the line

Also the line

Now for the oblique and horizontal ones :

If

Then the line

If

So the line

#### Answer:

#### Explanation:

#"simplifying f(x)"#

#f(x)=(x-1)/(x(1-x^2))=-(cancel((1-x)))/(xcancel((1-x))(1+x))#

#rArrf(x)=-1/(x(1+x))#

#"the removal of the factor "(1-x)" from the "#

#"numerator/denominator indicates a hole at "x=1# 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" and "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 the highest power of x. that is

#x^2#

#f(x)=-(1/x^2)/(x/x^2+x^2/x^2)=-(1/x^2)/(1/x+1)# as

#xto+-oo,f(x)to0/(0+1)#

#rArry=0" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.

graph{-(1)/(x(x+1) [-10, 10, -5, 5]}