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